3385. Minimum Time to Break Locks II

HardDepth-First SearchGraphArray
Leetcode Link

Problem Description

Bob needs to escape from a dungeon by breaking n locks in sequence. Each lock requires a certain amount of energy to break, given in an array strength where strength[i] represents the energy needed to break the i-th lock.

Bob uses a sword with special properties to break these locks:

  1. Initial State: The sword starts with 0 energy and a factor X = 1

  2. Energy Growth: Every minute that passes, the sword's energy increases by the current factor X

  3. Breaking Locks: To break the i-th lock, the sword must accumulate at least strength[i] energy

  4. Reset Mechanism: After successfully breaking a lock:

    • The sword's energy resets back to 0
    • The factor X increases by 1 (so it becomes 2 for the second lock, 3 for the third, and so on)

The goal is to find the minimum total time (in minutes) needed for Bob to break all n locks and escape.

For example, if there are 3 locks with strengths [3, 4, 1]:

  • For the 1st lock (strength 3): With X = 1, the sword gains 1 energy per minute, taking 3 minutes to reach strength 3
  • For the 2nd lock (strength 4): With X = 2, the sword gains 2 energy per minute, taking 2 minutes to reach strength 4
  • For the 3rd lock (strength 1): With X = 3, the sword gains 3 energy per minute, taking 1 minute to reach strength 1
  • Total time: 3 + 2 + 1 = 6 minutes

The key insight is that Bob can choose the order in which to break the locks, and different orders will result in different total times. The algorithm needs to find the optimal order that minimizes the total time.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

  • Yes: While the problem doesn't explicitly mention a graph, we can model it as a bipartite graph matching problem. We have n locks and n time slots (positions in the breaking sequence), and we need to find the optimal assignment of locks to positions.

Is it a tree?

  • No: The problem structure is not a tree. It's more of a complete bipartite graph where any lock can be assigned to any position.

Is the problem related to directed acyclic graphs?

  • No: The problem is not about DAGs or dependencies between locks.

Is the problem related to shortest paths?

  • No: We're not finding paths through a graph in the traditional sense.

Does the problem involve connectivity?

  • No: We're not checking if components are connected.

Does the problem have small constraints?

  • Yes: With n locks, we have n! possible orderings to consider. The constraints are typically small enough to allow exploring different permutations.

DFS/backtracking?

  • Yes: We can use DFS with backtracking to explore all possible permutations of lock breaking orders and find the minimum time.

Conclusion: The flowchart suggests using DFS/backtracking to solve this problem. We can use DFS to:

  1. Try each unvisited lock at the current position
  2. Calculate the time needed (based on the lock's strength and current factor X)
  3. Recursively solve for the remaining locks
  4. Backtrack and try different orderings
  5. Track the minimum total time across all valid permutations

The solution provided actually uses a more sophisticated approach (Minimum Cost Flow), but the fundamental insight of needing to explore different orderings aligns with the DFS/backtracking pattern identified by the flowchart.

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Intuition

The key insight is recognizing that this is fundamentally an assignment problem. We have n locks and n positions in a sequence, and we need to find the optimal assignment that minimizes total time.

Let's think about what happens when we break locks in a certain order:

  • When breaking the 1st lock (any lock we choose), we have factor X = 1, so time = ⌈strength/1⌉
  • When breaking the 2nd lock, we have factor X = 2, so time = ⌈strength/2⌉
  • When breaking the 3rd lock, we have factor X = 3, so time = ⌈strength/3⌉
  • And so on...

This reveals an important pattern: stronger locks should be broken later when the factor X is higher, because dividing by a larger number reduces the time more significantly. Conversely, weaker locks should be broken early when the factor is small.

However, it's not as simple as just sorting the locks. Consider locks with strengths [10, 9, 8]:

  • If we break them in order [8, 9, 10]: times are ⌈8/1⌉ + ⌈9/2⌉ + ⌈10/3⌉ = 8 + 5 + 4 = 17
  • If we break them in order [9, 8, 10]: times are ⌈9/1⌉ + ⌈8/2⌉ + ⌈10/3⌉ = 9 + 4 + 4 = 17
  • If we break them in order [10, 8, 9]: times are ⌈10/1⌉ + ⌈8/2⌉ + ⌈9/3⌉ = 10 + 4 + 3 = 17

Multiple orderings can give the same result, and the optimal assignment depends on the specific values and how they divide by different factors.

This leads us to model the problem as a bipartite matching problem:

  • One set of nodes represents the locks (with their strengths)
  • Another set represents the positions (1st, 2nd, 3rd, etc.)
  • The edge weight between lock i and position j is the time needed: ⌈strength[i] / (j+1)⌉
  • We need to find a perfect matching that minimizes the total weight

Since we need to minimize the total cost of matching all locks to all positions, this is a Minimum Cost Maximum Flow problem. The solution uses a specialized algorithm (Min Cost Flow) to efficiently find the optimal assignment without having to check all n! permutations.

The formula (a[i] - 1) // (j + 1) + 1 in the code is equivalent to ⌈a[i] / (j+1)⌉, calculating the ceiling division which gives us the exact number of minutes needed for lock i at position j.

Learn more about Depth-First Search and Graph patterns.

Solution Implementation

1from typing import List, Tuple, Optional, NamedTuple, cast 2from heapq import heappush, heappop 3 4 5class MCFGraph: 6 """Minimum Cost Flow Graph implementation using successive shortest path algorithm.""" 7 8 class Edge(NamedTuple): 9 """Represents an edge in the original graph with flow information.""" 10 src: int # Source vertex 11 dst: int # Destination vertex 12 cap: int # Original capacity 13 flow: int # Current flow through the edge 14 cost: int # Cost per unit flow 15 16 class _Edge: 17 """Internal edge representation for the residual graph.""" 18 def __init__(self, dst: int, cap: int, cost: int) -> None: 19 self.dst = dst # Destination vertex 20 self.cap = cap # Residual capacity 21 self.cost = cost # Cost per unit flow 22 self.rev: Optional['MCFGraph._Edge'] = None # Reverse edge reference 23 24 def __init__(self, n: int) -> None: 25 """Initialize a flow graph with n vertices.""" 26 self._n = n # Number of vertices 27 self._g: List[List[MCFGraph._Edge]] = [[] for _ in range(n)] # Adjacency list 28 self._edges: List[MCFGraph._Edge] = [] # List of original edges 29 30 def add_edge(self, src: int, dst: int, cap: int, cost: int) -> int: 31 """ 32 Add an edge from src to dst with given capacity and cost. 33 Returns the index of the added edge. 34 """ 35 assert 0 <= src < self._n 36 assert 0 <= dst < self._n 37 assert 0 <= cap 38 39 edge_id = len(self._edges) 40 41 # Create forward and reverse edges for residual graph 42 forward_edge = MCFGraph._Edge(dst, cap, cost) 43 reverse_edge = MCFGraph._Edge(src, 0, -cost) 44 45 # Link forward and reverse edges 46 forward_edge.rev = reverse_edge 47 reverse_edge.rev = forward_edge 48 49 # Add edges to adjacency list 50 self._g[src].append(forward_edge) 51 self._g[dst].append(reverse_edge) 52 53 # Store original edge for reference 54 self._edges.append(forward_edge) 55 56 return edge_id 57 58 def get_edge(self, i: int) -> Edge: 59 """Get edge information by its index.""" 60 assert 0 <= i < len(self._edges) 61 62 edge = self._edges[i] 63 reverse_edge = cast(MCFGraph._Edge, edge.rev) 64 65 # Calculate original capacity and current flow 66 original_capacity = edge.cap + reverse_edge.cap 67 current_flow = reverse_edge.cap 68 69 return MCFGraph.Edge( 70 src=reverse_edge.dst, 71 dst=edge.dst, 72 cap=original_capacity, 73 flow=current_flow, 74 cost=edge.cost 75 ) 76 77 def edges(self) -> List[Edge]: 78 """Get all edges in the graph.""" 79 return [self.get_edge(i) for i in range(len(self._edges))] 80 81 def flow(self, s: int, t: int, flow_limit: Optional[int] = None) -> Tuple[int, int]: 82 """ 83 Find maximum flow from s to t with minimum cost. 84 Returns (flow_amount, total_cost). 85 """ 86 return self.slope(s, t, flow_limit)[-1] 87 88 def slope( 89 self, s: int, t: int, flow_limit: Optional[int] = None 90 ) -> List[Tuple[int, int]]: 91 """ 92 Find the cost-flow curve from s to t. 93 Returns list of (flow, cost) pairs representing the piecewise linear function. 94 """ 95 assert 0 <= s < self._n 96 assert 0 <= t < self._n 97 assert s != t 98 99 # Set flow limit to sum of outgoing capacities from source if not specified 100 if flow_limit is None: 101 flow_limit = cast(int, sum(e.cap for e in self._g[s])) 102 103 # Dual variables for reduced cost 104 dual = [0] * self._n 105 # Previous vertex and edge in shortest path 106 prev: List[Optional[Tuple[int, MCFGraph._Edge]]] = [None] * self._n 107 108 def refine_dual() -> bool: 109 """ 110 Find shortest path from s to t using Dijkstra's algorithm 111 and update dual variables. Returns True if path exists. 112 """ 113 # Priority queue: (distance, vertex) 114 priority_queue = [(0, s)] 115 visited = [False] * self._n 116 dist: List[Optional[int]] = [None] * self._n 117 dist[s] = 0 118 119 while priority_queue: 120 dist_v, v = heappop(priority_queue) 121 122 if visited[v]: 123 continue 124 visited[v] = True 125 126 if v == t: 127 break 128 129 dual_v = dual[v] 130 131 # Check all outgoing edges 132 for edge in self._g[v]: 133 w = edge.dst 134 135 # Skip if already visited or no residual capacity 136 if visited[w] or edge.cap == 0: 137 continue 138 139 # Calculate reduced cost 140 reduced_cost = edge.cost - dual[w] + dual_v 141 new_dist = dist_v + reduced_cost 142 dist_w = dist[w] 143 144 # Update distance if shorter path found 145 if dist_w is None or new_dist < dist_w: 146 dist[w] = new_dist 147 prev[w] = v, edge 148 heappush(priority_queue, (new_dist, w)) 149 else: 150 # No path to t found 151 return False 152 153 # Update dual variables to maintain reduced costs 154 dist_t = dist[t] 155 for v in range(self._n): 156 if visited[v]: 157 dual[v] -= cast(int, dist_t) - cast(int, dist[v]) 158 159 return True 160 161 # Initialize flow and cost 162 total_flow = 0 163 total_cost = 0 164 prev_cost_per_flow: Optional[int] = None 165 result = [(total_flow, total_cost)] 166 167 # Main loop: find augmenting paths until flow limit reached 168 while total_flow < flow_limit: 169 # Find shortest augmenting path 170 if not refine_dual(): 171 break 172 173 # Find bottleneck capacity along the path 174 bottleneck = flow_limit - total_flow 175 v = t 176 while prev[v] is not None: 177 u, edge = cast(Tuple[int, MCFGraph._Edge], prev[v]) 178 bottleneck = min(bottleneck, edge.cap) 179 v = u 180 181 # Update residual capacities along the path 182 v = t 183 while prev[v] is not None: 184 u, edge = cast(Tuple[int, MCFGraph._Edge], prev[v]) 185 edge.cap -= bottleneck 186 assert edge.rev is not None 187 edge.rev.cap += bottleneck 188 v = u 189 190 # Update flow and cost 191 cost_per_unit = -dual[s] 192 total_flow += bottleneck 193 total_cost += bottleneck * cost_per_unit 194 195 # Merge with previous segment if same slope 196 if cost_per_unit == prev_cost_per_flow: 197 result.pop() 198 199 result.append((total_flow, total_cost)) 200 prev_cost_per_flow = cost_per_unit 201 202 return result 203 204 205class Solution: 206 def findMinimumTime(self, a: List[int]) -> int: 207 """ 208 Find minimum total time to process all tasks. 209 Each task i takes ceil(a[i]/(j+1)) time if assigned to worker j. 210 """ 211 n = len(a) 212 213 # Create bipartite graph with source and sink 214 source = n * 2 # Source node 215 sink = source + 1 # Sink node 216 graph = MCFGraph(sink + 1) 217 218 # Build bipartite matching graph 219 for i in range(n): 220 # Connect source to each task with capacity 1 221 graph.add_edge(source, i, 1, 0) 222 223 # Connect each worker to sink with capacity 1 224 graph.add_edge(i + n, sink, 1, 0) 225 226 # Connect each task to each worker with appropriate cost 227 for j in range(n): 228 # Cost is the time task i takes on worker j 229 time_cost = (a[i] - 1) // (j + 1) + 1 # Equivalent to ceil(a[i]/(j+1)) 230 graph.add_edge(i, j + n, 1, time_cost) 231 232 # Find minimum cost perfect matching 233 _, min_cost = graph.flow(source, sink, n) 234 return min_cost 235
1import java.util.*; 2 3/** 4 * Minimum Cost Flow Graph implementation using the Successive Shortest Path algorithm. 5 * This class solves the minimum cost maximum flow problem on a directed graph. 6 */ 7class MCFGraph { 8 /** 9 * Represents an edge in the flow network with source, destination, capacity, flow, and cost. 10 */ 11 static class Edge { 12 int source; 13 int destination; 14 int capacity; 15 int flow; 16 int cost; 17 18 Edge(int source, int destination, int capacity, int flow, int cost) { 19 this.source = source; 20 this.destination = destination; 21 this.capacity = capacity; 22 this.flow = flow; 23 this.cost = cost; 24 } 25 } 26 27 /** 28 * Internal edge representation for the residual graph. 29 * Each edge maintains a reference to its reverse edge for efficient flow updates. 30 */ 31 static class InternalEdge { 32 int destination; 33 int capacity; 34 int cost; 35 InternalEdge reverseEdge; 36 37 InternalEdge(int destination, int capacity, int cost) { 38 this.destination = destination; 39 this.capacity = capacity; 40 this.cost = cost; 41 this.reverseEdge = null; 42 } 43 } 44 45 /** 46 * Helper class to store a pair of vertex and its incoming edge in the shortest path. 47 */ 48 static class VertexEdgePair { 49 int vertex; 50 InternalEdge edge; 51 52 VertexEdgePair(int vertex, InternalEdge edge) { 53 this.vertex = vertex; 54 this.edge = edge; 55 } 56 } 57 58 private int numberOfVertices; 59 private List<List<InternalEdge>> adjacencyList; 60 private List<InternalEdge> edgeList; 61 62 /** 63 * Constructs a minimum cost flow graph with n vertices. 64 * @param n the number of vertices in the graph 65 */ 66 public MCFGraph(int n) { 67 this.numberOfVertices = n; 68 this.adjacencyList = new ArrayList<>(); 69 this.edgeList = new ArrayList<>(); 70 71 // Initialize adjacency list for each vertex 72 for (int i = 0; i < n; i++) { 73 adjacencyList.add(new ArrayList<>()); 74 } 75 } 76 77 /** 78 * Adds a directed edge from source to destination with given capacity and cost. 79 * @param src source vertex 80 * @param dst destination vertex 81 * @param cap edge capacity 82 * @param cost cost per unit flow 83 * @return the index of the added edge 84 */ 85 public int addEdge(int src, int dst, int cap, int cost) { 86 // Validate input parameters 87 assert (0 <= src && src < numberOfVertices); 88 assert (0 <= dst && dst < numberOfVertices); 89 assert (0 <= cap); 90 91 int edgeIndex = edgeList.size(); 92 93 // Create forward and reverse edges for residual graph 94 InternalEdge forwardEdge = new InternalEdge(dst, cap, cost); 95 InternalEdge reverseEdge = new InternalEdge(src, 0, -cost); 96 97 // Link forward and reverse edges 98 forwardEdge.reverseEdge = reverseEdge; 99 reverseEdge.reverseEdge = forwardEdge; 100 101 // Add edges to adjacency list 102 adjacencyList.get(src).add(forwardEdge); 103 adjacencyList.get(dst).add(reverseEdge); 104 105 // Store the forward edge in edge list 106 edgeList.add(forwardEdge); 107 108 return edgeIndex; 109 } 110 111 /** 112 * Retrieves the edge at the given index. 113 * @param i edge index 114 * @return Edge object with current flow information 115 */ 116 public Edge getEdge(int i) { 117 assert (0 <= i && i < edgeList.size()); 118 119 InternalEdge forwardEdge = edgeList.get(i); 120 InternalEdge reverseEdge = forwardEdge.reverseEdge; 121 122 // Calculate original capacity and current flow 123 int originalCapacity = forwardEdge.capacity + reverseEdge.capacity; 124 int currentFlow = reverseEdge.capacity; 125 126 return new Edge(reverseEdge.destination, forwardEdge.destination, 127 originalCapacity, currentFlow, forwardEdge.cost); 128 } 129 130 /** 131 * Returns all edges in the graph. 132 * @return list of all edges with their current flow information 133 */ 134 public List<Edge> edges() { 135 List<Edge> result = new ArrayList<>(); 136 for (int i = 0; i < edgeList.size(); i++) { 137 result.add(getEdge(i)); 138 } 139 return result; 140 } 141 142 /** 143 * Computes the minimum cost flow from source to sink with optional flow limit. 144 * @param s source vertex 145 * @param t sink vertex 146 * @param flowLimit maximum flow limit (null for unlimited) 147 * @return array containing [total flow, total cost] 148 */ 149 public int[] flow(int s, int t, Integer flowLimit) { 150 List<int[]> slopeResult = slope(s, t, flowLimit); 151 return slopeResult.get(slopeResult.size() - 1); 152 } 153 154 /** 155 * Computes the slope function of minimum cost flow. 156 * Returns a sequence of [flow, cost] pairs representing the piecewise linear cost function. 157 * @param s source vertex 158 * @param t sink vertex 159 * @param flowLimit maximum flow limit (null for unlimited) 160 * @return list of [flow, cost] pairs 161 */ 162 public List<int[]> slope(int s, int t, Integer flowLimit) { 163 // Validate input parameters 164 assert (0 <= s && s < numberOfVertices); 165 assert (0 <= t && t < numberOfVertices); 166 assert (s != t); 167 168 // Set flow limit to sum of outgoing capacities from source if not specified 169 if (flowLimit == null) { 170 flowLimit = adjacencyList.get(s).stream() 171 .mapToInt(e -> e.capacity) 172 .sum(); 173 } 174 175 // Initialize dual variables and predecessor array 176 int[] dualVariables = new int[numberOfVertices]; 177 VertexEdgePair[] predecessors = new VertexEdgePair[numberOfVertices]; 178 179 // Initialize result with zero flow and cost 180 List<int[]> result = new ArrayList<>(); 181 result.add(new int[] {0, 0}); 182 183 // Main loop: find augmenting paths and update flow 184 while (true) { 185 // Find shortest path using reduced costs 186 if (!findShortestPath(s, t, dualVariables, predecessors)) { 187 break; 188 } 189 190 // Calculate bottleneck capacity along the path 191 int bottleneckFlow = flowLimit; 192 int currentVertex = t; 193 while (predecessors[currentVertex] != null) { 194 VertexEdgePair pair = predecessors[currentVertex]; 195 int previousVertex = pair.vertex; 196 InternalEdge edge = pair.edge; 197 bottleneckFlow = Math.min(bottleneckFlow, edge.capacity); 198 currentVertex = previousVertex; 199 } 200 201 // Update flow along the path 202 currentVertex = t; 203 while (predecessors[currentVertex] != null) { 204 VertexEdgePair pair = predecessors[currentVertex]; 205 int previousVertex = pair.vertex; 206 InternalEdge edge = pair.edge; 207 edge.capacity -= bottleneckFlow; 208 edge.reverseEdge.capacity += bottleneckFlow; 209 currentVertex = previousVertex; 210 } 211 212 // Calculate cost and update result 213 int unitCost = -dualVariables[s]; 214 int[] previousResult = result.get(result.size() - 1); 215 result.add(new int[] { 216 previousResult[0] + bottleneckFlow, 217 previousResult[1] + bottleneckFlow * unitCost 218 }); 219 220 // Remove redundant point if cost remains the same 221 if (result.size() >= 2) { 222 int[] lastPoint = result.get(result.size() - 1); 223 int[] secondLastPoint = result.get(result.size() - 2); 224 if (unitCost == (secondLastPoint[1] / secondLastPoint[0])) { 225 result.remove(result.size() - 2); 226 } 227 } 228 } 229 230 return result; 231 } 232 233 /** 234 * Finds the shortest path from source to sink using Dijkstra's algorithm with reduced costs. 235 * Updates dual variables to maintain optimality conditions. 236 * @param s source vertex 237 * @param t sink vertex 238 * @param dualVariables array of dual variables 239 * @param predecessors array to store path information 240 * @return true if a path exists, false otherwise 241 */ 242 private boolean findShortestPath(int s, int t, int[] dualVariables, VertexEdgePair[] predecessors) { 243 // Initialize priority queue for Dijkstra's algorithm 244 PriorityQueue<int[]> priorityQueue = new PriorityQueue<>( 245 Comparator.comparingInt(a -> a[0]) 246 ); 247 priorityQueue.add(new int[] {0, s}); 248 249 // Initialize tracking arrays 250 boolean[] visited = new boolean[numberOfVertices]; 251 Integer[] distances = new Integer[numberOfVertices]; 252 Arrays.fill(distances, null); 253 distances[s] = 0; 254 255 // Dijkstra's algorithm with reduced costs 256 while (!priorityQueue.isEmpty()) { 257 int[] current = priorityQueue.poll(); 258 int currentDistance = current[0]; 259 int currentVertex = current[1]; 260 261 // Skip if already visited 262 if (visited[currentVertex]) continue; 263 visited[currentVertex] = true; 264 265 // Stop if we reached the sink 266 if (currentVertex == t) break; 267 268 // Explore neighbors 269 int currentDual = dualVariables[currentVertex]; 270 for (InternalEdge edge : adjacencyList.get(currentVertex)) { 271 int neighbor = edge.destination; 272 273 // Skip if visited or no capacity 274 if (visited[neighbor] || edge.capacity == 0) continue; 275 276 // Calculate reduced cost and new distance 277 int reducedCost = edge.cost - dualVariables[neighbor] + currentDual; 278 int newDistance = currentDistance + reducedCost; 279 Integer neighborDistance = distances[neighbor]; 280 281 // Update if we found a shorter path 282 if (neighborDistance == null || newDistance < neighborDistance) { 283 distances[neighbor] = newDistance; 284 predecessors[neighbor] = new VertexEdgePair(currentVertex, edge); 285 priorityQueue.add(new int[] {newDistance, neighbor}); 286 } 287 } 288 } 289 290 // Check if sink is reachable 291 if (!visited[t]) return false; 292 293 // Update dual variables to maintain optimality 294 int sinkDistance = distances[t]; 295 for (int vertex = 0; vertex < numberOfVertices; vertex++) { 296 if (visited[vertex]) { 297 dualVariables[vertex] -= sinkDistance - distances[vertex]; 298 } 299 } 300 301 return true; 302 } 303} 304 305/** 306 * Solution class for the minimum time problem using minimum cost flow. 307 */ 308class Solution { 309 /** 310 * Finds the minimum time to process all tasks with given strengths. 311 * Uses minimum cost flow to model the assignment problem. 312 * @param strength array of task strengths 313 * @return minimum total time 314 */ 315 public int findMinimumTime(int[] strength) { 316 int n = strength.length; 317 318 // Create super source and super sink vertices 319 int superSource = n * 2; 320 int superSink = superSource + 1; 321 322 // Initialize minimum cost flow graph 323 MCFGraph graph = new MCFGraph(superSink + 1); 324 325 // Build the bipartite graph 326 for (int i = 0; i < n; i++) { 327 // Connect source to left side (tasks) 328 graph.addEdge(superSource, i, 1, 0); 329 330 // Connect right side (positions) to sink 331 graph.addEdge(i + n, superSink, 1, 0); 332 333 // Connect each task to each position with appropriate cost 334 for (int j = 0; j < n; j++) { 335 // Cost represents time to complete task i at position j 336 int timeCost = (strength[i] - 1) / (j + 1) + 1; 337 graph.addEdge(i, j + n, 1, timeCost); 338 } 339 } 340 341 // Find minimum cost flow and return the total cost 342 return graph.flow(superSource, superSink, n)[1]; 343 } 344} 345
1#include <vector> 2#include <queue> 3#include <algorithm> 4#include <limits> 5#include <cassert> 6 7using namespace std; 8 9/** 10 * Minimum Cost Flow Graph implementation using the Successive Shortest Path algorithm. 11 * This class solves the minimum cost maximum flow problem on a directed graph. 12 */ 13class MCFGraph { 14public: 15 /** 16 * Represents an edge in the flow network with source, destination, capacity, flow, and cost. 17 */ 18 struct Edge { 19 int source; 20 int destination; 21 int capacity; 22 int flow; 23 int cost; 24 25 Edge(int src, int dst, int cap, int flw, int cst) 26 : source(src), destination(dst), capacity(cap), flow(flw), cost(cst) {} 27 }; 28 29private: 30 /** 31 * Internal edge representation for the residual graph. 32 * Each edge maintains a reference to its reverse edge for efficient flow updates. 33 */ 34 struct InternalEdge { 35 int destination; 36 int capacity; 37 int cost; 38 InternalEdge* reverse_edge; 39 40 InternalEdge(int dst, int cap, int cst) 41 : destination(dst), capacity(cap), cost(cst), reverse_edge(nullptr) {} 42 }; 43 44 /** 45 * Helper class to store a pair of vertex and its incoming edge in the shortest path. 46 */ 47 struct VertexEdgePair { 48 int vertex; 49 InternalEdge* edge; 50 51 VertexEdgePair() : vertex(-1), edge(nullptr) {} 52 VertexEdgePair(int v, InternalEdge* e) : vertex(v), edge(e) {} 53 }; 54 55 int number_of_vertices; 56 vector<vector<InternalEdge*>> adjacency_list; 57 vector<InternalEdge*> edge_list; 58 59public: 60 /** 61 * Constructs a minimum cost flow graph with n vertices. 62 * @param n the number of vertices in the graph 63 */ 64 MCFGraph(int n) : number_of_vertices(n), adjacency_list(n) {} 65 66 /** 67 * Destructor to clean up dynamically allocated edges 68 */ 69 ~MCFGraph() { 70 for (auto& edges : adjacency_list) { 71 for (auto* edge : edges) { 72 delete edge; 73 } 74 } 75 } 76 77 /** 78 * Adds a directed edge from source to destination with given capacity and cost. 79 * @param src source vertex 80 * @param dst destination vertex 81 * @param cap edge capacity 82 * @param cost cost per unit flow 83 * @return the index of the added edge 84 */ 85 int addEdge(int src, int dst, int cap, int cost) { 86 // Validate input parameters 87 assert(0 <= src && src < number_of_vertices); 88 assert(0 <= dst && dst < number_of_vertices); 89 assert(0 <= cap); 90 91 int edge_index = edge_list.size(); 92 93 // Create forward and reverse edges for residual graph 94 InternalEdge* forward_edge = new InternalEdge(dst, cap, cost); 95 InternalEdge* reverse_edge = new InternalEdge(src, 0, -cost); 96 97 // Link forward and reverse edges 98 forward_edge->reverse_edge = reverse_edge; 99 reverse_edge->reverse_edge = forward_edge; 100 101 // Add edges to adjacency list 102 adjacency_list[src].push_back(forward_edge); 103 adjacency_list[dst].push_back(reverse_edge); 104 105 // Store the forward edge in edge list 106 edge_list.push_back(forward_edge); 107 108 return edge_index; 109 } 110 111 /** 112 * Retrieves the edge at the given index. 113 * @param i edge index 114 * @return Edge object with current flow information 115 */ 116 Edge getEdge(int i) { 117 assert(0 <= i && i < edge_list.size()); 118 119 InternalEdge* forward_edge = edge_list[i]; 120 InternalEdge* reverse_edge = forward_edge->reverse_edge; 121 122 // Calculate original capacity and current flow 123 int original_capacity = forward_edge->capacity + reverse_edge->capacity; 124 int current_flow = reverse_edge->capacity; 125 126 return Edge(reverse_edge->destination, forward_edge->destination, 127 original_capacity, current_flow, forward_edge->cost); 128 } 129 130 /** 131 * Returns all edges in the graph. 132 * @return vector of all edges with their current flow information 133 */ 134 vector<Edge> edges() { 135 vector<Edge> result; 136 for (int i = 0; i < edge_list.size(); i++) { 137 result.push_back(getEdge(i)); 138 } 139 return result; 140 } 141 142 /** 143 * Computes the minimum cost flow from source to sink with optional flow limit. 144 * @param s source vertex 145 * @param t sink vertex 146 * @param flow_limit maximum flow limit (-1 for unlimited) 147 * @return pair containing [total flow, total cost] 148 */ 149 pair<int, int> flow(int s, int t, int flow_limit = -1) { 150 vector<pair<int, int>> slope_result = slope(s, t, flow_limit); 151 return slope_result.back(); 152 } 153 154 /** 155 * Computes the slope function of minimum cost flow. 156 * Returns a sequence of [flow, cost] pairs representing the piecewise linear cost function. 157 * @param s source vertex 158 * @param t sink vertex 159 * @param flow_limit maximum flow limit (-1 for unlimited) 160 * @return vector of [flow, cost] pairs 161 */ 162 vector<pair<int, int>> slope(int s, int t, int flow_limit = -1) { 163 // Validate input parameters 164 assert(0 <= s && s < number_of_vertices); 165 assert(0 <= t && t < number_of_vertices); 166 assert(s != t); 167 168 // Set flow limit to sum of outgoing capacities from source if not specified 169 if (flow_limit == -1) { 170 flow_limit = 0; 171 for (auto* edge : adjacency_list[s]) { 172 flow_limit += edge->capacity; 173 } 174 } 175 176 // Initialize dual variables and predecessor array 177 vector<int> dual_variables(number_of_vertices, 0); 178 vector<VertexEdgePair> predecessors(number_of_vertices); 179 180 // Initialize result with zero flow and cost 181 vector<pair<int, int>> result; 182 result.push_back({0, 0}); 183 184 // Main loop: find augmenting paths and update flow 185 while (flow_limit > 0) { 186 // Find shortest path using reduced costs 187 if (!findShortestPath(s, t, dual_variables, predecessors)) { 188 break; 189 } 190 191 // Calculate bottleneck capacity along the path 192 int bottleneck_flow = flow_limit; 193 int current_vertex = t; 194 while (predecessors[current_vertex].edge != nullptr) { 195 VertexEdgePair& pair = predecessors[current_vertex]; 196 int previous_vertex = pair.vertex; 197 InternalEdge* edge = pair.edge; 198 bottleneck_flow = min(bottleneck_flow, edge->capacity); 199 current_vertex = previous_vertex; 200 } 201 202 // Update flow along the path 203 current_vertex = t; 204 while (predecessors[current_vertex].edge != nullptr) { 205 VertexEdgePair& pair = predecessors[current_vertex]; 206 int previous_vertex = pair.vertex; 207 InternalEdge* edge = pair.edge; 208 edge->capacity -= bottleneck_flow; 209 edge->reverse_edge->capacity += bottleneck_flow; 210 current_vertex = previous_vertex; 211 } 212 213 // Calculate cost and update result 214 int unit_cost = -dual_variables[s]; 215 pair<int, int> previous_result = result.back(); 216 result.push_back({ 217 previous_result.first + bottleneck_flow, 218 previous_result.second + bottleneck_flow * unit_cost 219 }); 220 221 // Update flow limit 222 flow_limit -= bottleneck_flow; 223 224 // Remove redundant point if cost remains the same 225 if (result.size() >= 2) { 226 pair<int, int>& last_point = result.back(); 227 pair<int, int>& second_last_point = result[result.size() - 2]; 228 if (result.size() >= 3) { 229 pair<int, int>& third_last_point = result[result.size() - 3]; 230 // Check if the slope is the same 231 long long slope1 = (long long)(second_last_point.second - third_last_point.second); 232 long long slope2 = (long long)(last_point.second - second_last_point.second); 233 long long flow1 = second_last_point.first - third_last_point.first; 234 long long flow2 = last_point.first - second_last_point.first; 235 if (slope1 * flow2 == slope2 * flow1) { 236 result.erase(result.end() - 2); 237 } 238 } 239 } 240 } 241 242 return result; 243 } 244 245private: 246 /** 247 * Finds the shortest path from source to sink using Dijkstra's algorithm with reduced costs. 248 * Updates dual variables to maintain optimality conditions. 249 * @param s source vertex 250 * @param t sink vertex 251 * @param dual_variables array of dual variables 252 * @param predecessors array to store path information 253 * @return true if a path exists, false otherwise 254 */ 255 bool findShortestPath(int s, int t, vector<int>& dual_variables, vector<VertexEdgePair>& predecessors) { 256 // Initialize priority queue for Dijkstra's algorithm 257 priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> priority_queue; 258 priority_queue.push({0, s}); 259 260 // Initialize tracking arrays 261 vector<bool> visited(number_of_vertices, false); 262 vector<int> distances(number_of_vertices, numeric_limits<int>::max()); 263 distances[s] = 0; 264 265 // Reset predecessors 266 fill(predecessors.begin(), predecessors.end(), VertexEdgePair()); 267 268 // Dijkstra's algorithm with reduced costs 269 while (!priority_queue.empty()) { 270 auto [current_distance, current_vertex] = priority_queue.top(); 271 priority_queue.pop(); 272 273 // Skip if already visited 274 if (visited[current_vertex]) continue; 275 visited[current_vertex] = true; 276 277 // Stop if we reached the sink 278 if (current_vertex == t) break; 279 280 // Explore neighbors 281 int current_dual = dual_variables[current_vertex]; 282 for (InternalEdge* edge : adjacency_list[current_vertex]) { 283 int neighbor = edge->destination; 284 285 // Skip if visited or no capacity 286 if (visited[neighbor] || edge->capacity == 0) continue; 287 288 // Calculate reduced cost and new distance 289 int reduced_cost = edge->cost - dual_variables[neighbor] + current_dual; 290 int new_distance = current_distance + reduced_cost; 291 292 // Update if we found a shorter path 293 if (new_distance < distances[neighbor]) { 294 distances[neighbor] = new_distance; 295 predecessors[neighbor] = VertexEdgePair(current_vertex, edge); 296 priority_queue.push({new_distance, neighbor}); 297 } 298 } 299 } 300 301 // Check if sink is reachable 302 if (!visited[t]) return false; 303 304 // Update dual variables to maintain optimality 305 int sink_distance = distances[t]; 306 for (int vertex = 0; vertex < number_of_vertices; vertex++) { 307 if (visited[vertex]) { 308 dual_variables[vertex] -= sink_distance - distances[vertex]; 309 } 310 } 311 312 return true; 313 } 314}; 315 316/** 317 * Solution class for the minimum time problem using minimum cost flow. 318 */ 319class Solution { 320public: 321 /** 322 * Finds the minimum time to process all tasks with given strengths. 323 * Uses minimum cost flow to model the assignment problem. 324 * @param strength array of task strengths 325 * @return minimum total time 326 */ 327 int findMinimumTime(vector<int>& strength) { 328 int n = strength.size(); 329 330 // Create super source and super sink vertices 331 int super_source = n * 2; 332 int super_sink = super_source + 1; 333 334 // Initialize minimum cost flow graph 335 MCFGraph graph(super_sink + 1); 336 337 // Build the bipartite graph 338 for (int i = 0; i < n; i++) { 339 // Connect source to left side (tasks) 340 graph.addEdge(super_source, i, 1, 0); 341 342 // Connect right side (positions) to sink 343 graph.addEdge(i + n, super_sink, 1, 0); 344 345 // Connect each task to each position with appropriate cost 346 for (int j = 0; j < n; j++) { 347 // Cost represents time to complete task i at position j 348 int time_cost = (strength[i] - 1) / (j + 1) + 1; 349 graph.addEdge(i, j + n, 1, time_cost); 350 } 351 } 352 353 // Find minimum cost flow and return the total cost 354 return graph.flow(super_source, super_sink, n).second; 355 } 356}; 357
1/** 2 * Represents an edge in the flow network with source, destination, capacity, flow, and cost. 3 */ 4interface Edge { 5 source: number; 6 destination: number; 7 capacity: number; 8 flow: number; 9 cost: number; 10} 11 12/** 13 * Internal edge representation for the residual graph. 14 * Each edge maintains a reference to its reverse edge for efficient flow updates. 15 */ 16class InternalEdge { 17 destination: number; 18 capacity: number; 19 cost: number; 20 reverseEdge: InternalEdge | null; 21 22 constructor(destination: number, capacity: number, cost: number) { 23 this.destination = destination; 24 this.capacity = capacity; 25 this.cost = cost; 26 this.reverseEdge = null; 27 } 28} 29 30/** 31 * Helper class to store a pair of vertex and its incoming edge in the shortest path. 32 */ 33interface VertexEdgePair { 34 vertex: number; 35 edge: InternalEdge; 36} 37 38// Global variables for the MCF graph 39let numberOfVertices: number; 40let adjacencyList: InternalEdge[][]; 41let edgeList: InternalEdge[]; 42 43/** 44 * Initializes a minimum cost flow graph with n vertices. 45 * @param n - The number of vertices in the graph 46 */ 47function initializeMCFGraph(n: number): void { 48 numberOfVertices = n; 49 adjacencyList = []; 50 edgeList = []; 51 52 // Initialize adjacency list for each vertex 53 for (let i = 0; i < n; i++) { 54 adjacencyList.push([]); 55 } 56} 57 58/** 59 * Adds a directed edge from source to destination with given capacity and cost. 60 * @param src - Source vertex 61 * @param dst - Destination vertex 62 * @param cap - Edge capacity 63 * @param cost - Cost per unit flow 64 * @returns The index of the added edge 65 */ 66function addEdge(src: number, dst: number, cap: number, cost: number): number { 67 // Validate input parameters 68 if (src < 0 || src >= numberOfVertices || dst < 0 || dst >= numberOfVertices || cap < 0) { 69 throw new Error("Invalid edge parameters"); 70 } 71 72 const edgeIndex = edgeList.length; 73 74 // Create forward and reverse edges for residual graph 75 const forwardEdge = new InternalEdge(dst, cap, cost); 76 const reverseEdge = new InternalEdge(src, 0, -cost); 77 78 // Link forward and reverse edges 79 forwardEdge.reverseEdge = reverseEdge; 80 reverseEdge.reverseEdge = forwardEdge; 81 82 // Add edges to adjacency list 83 adjacencyList[src].push(forwardEdge); 84 adjacencyList[dst].push(reverseEdge); 85 86 // Store the forward edge in edge list 87 edgeList.push(forwardEdge); 88 89 return edgeIndex; 90} 91 92/** 93 * Retrieves the edge at the given index. 94 * @param i - Edge index 95 * @returns Edge object with current flow information 96 */ 97function getEdge(i: number): Edge { 98 if (i < 0 || i >= edgeList.length) { 99 throw new Error("Invalid edge index"); 100 } 101 102 const forwardEdge = edgeList[i]; 103 const reverseEdge = forwardEdge.reverseEdge!; 104 105 // Calculate original capacity and current flow 106 const originalCapacity = forwardEdge.capacity + reverseEdge.capacity; 107 const currentFlow = reverseEdge.capacity; 108 109 return { 110 source: reverseEdge.destination, 111 destination: forwardEdge.destination, 112 capacity: originalCapacity, 113 flow: currentFlow, 114 cost: forwardEdge.cost 115 }; 116} 117 118/** 119 * Returns all edges in the graph. 120 * @returns List of all edges with their current flow information 121 */ 122function edges(): Edge[] { 123 const result: Edge[] = []; 124 for (let i = 0; i < edgeList.length; i++) { 125 result.push(getEdge(i)); 126 } 127 return result; 128} 129 130/** 131 * Finds the shortest path from source to sink using Dijkstra's algorithm with reduced costs. 132 * Updates dual variables to maintain optimality conditions. 133 * @param s - Source vertex 134 * @param t - Sink vertex 135 * @param dualVariables - Array of dual variables 136 * @param predecessors - Array to store path information 137 * @returns True if a path exists, false otherwise 138 */ 139function findShortestPath( 140 s: number, 141 t: number, 142 dualVariables: number[], 143 predecessors: (VertexEdgePair | null)[] 144): boolean { 145 // Initialize priority queue for Dijkstra's algorithm 146 const priorityQueue: Array<[number, number]> = [[0, s]]; 147 148 // Initialize tracking arrays 149 const visited: boolean[] = new Array(numberOfVertices).fill(false); 150 const distances: (number | null)[] = new Array(numberOfVertices).fill(null); 151 distances[s] = 0; 152 153 // Dijkstra's algorithm with reduced costs 154 while (priorityQueue.length > 0) { 155 // Sort to get minimum distance element (simulating priority queue) 156 priorityQueue.sort((a, b) => a[0] - b[0]); 157 const [currentDistance, currentVertex] = priorityQueue.shift()!; 158 159 // Skip if already visited 160 if (visited[currentVertex]) continue; 161 visited[currentVertex] = true; 162 163 // Stop if we reached the sink 164 if (currentVertex === t) break; 165 166 // Explore neighbors 167 const currentDual = dualVariables[currentVertex]; 168 for (const edge of adjacencyList[currentVertex]) { 169 const neighbor = edge.destination; 170 171 // Skip if visited or no capacity 172 if (visited[neighbor] || edge.capacity === 0) continue; 173 174 // Calculate reduced cost and new distance 175 const reducedCost = edge.cost - dualVariables[neighbor] + currentDual; 176 const newDistance = currentDistance + reducedCost; 177 const neighborDistance = distances[neighbor]; 178 179 // Update if we found a shorter path 180 if (neighborDistance === null || newDistance < neighborDistance) { 181 distances[neighbor] = newDistance; 182 predecessors[neighbor] = { vertex: currentVertex, edge: edge }; 183 priorityQueue.push([newDistance, neighbor]); 184 } 185 } 186 } 187 188 // Check if sink is reachable 189 if (!visited[t]) return false; 190 191 // Update dual variables to maintain optimality 192 const sinkDistance = distances[t]!; 193 for (let vertex = 0; vertex < numberOfVertices; vertex++) { 194 if (visited[vertex]) { 195 dualVariables[vertex] -= sinkDistance - distances[vertex]!; 196 } 197 } 198 199 return true; 200} 201 202/** 203 * Computes the slope function of minimum cost flow. 204 * Returns a sequence of [flow, cost] pairs representing the piecewise linear cost function. 205 * @param s - Source vertex 206 * @param t - Sink vertex 207 * @param flowLimit - Maximum flow limit (null for unlimited) 208 * @returns List of [flow, cost] pairs 209 */ 210function slope(s: number, t: number, flowLimit: number | null): number[][] { 211 // Validate input parameters 212 if (s < 0 || s >= numberOfVertices || t < 0 || t >= numberOfVertices || s === t) { 213 throw new Error("Invalid source or sink vertex"); 214 } 215 216 // Set flow limit to sum of outgoing capacities from source if not specified 217 if (flowLimit === null) { 218 flowLimit = adjacencyList[s].reduce((sum, edge) => sum + edge.capacity, 0); 219 } 220 221 // Initialize dual variables and predecessor array 222 const dualVariables: number[] = new Array(numberOfVertices).fill(0); 223 const predecessors: (VertexEdgePair | null)[] = new Array(numberOfVertices).fill(null); 224 225 // Initialize result with zero flow and cost 226 const result: number[][] = [[0, 0]]; 227 228 // Main loop: find augmenting paths and update flow 229 while (true) { 230 // Clear predecessors for new iteration 231 predecessors.fill(null); 232 233 // Find shortest path using reduced costs 234 if (!findShortestPath(s, t, dualVariables, predecessors)) { 235 break; 236 } 237 238 // Calculate bottleneck capacity along the path 239 let bottleneckFlow = flowLimit; 240 let currentVertex = t; 241 while (predecessors[currentVertex] !== null) { 242 const pair = predecessors[currentVertex]!; 243 const edge = pair.edge; 244 bottleneckFlow = Math.min(bottleneckFlow, edge.capacity); 245 currentVertex = pair.vertex; 246 } 247 248 // Update flow along the path 249 currentVertex = t; 250 while (predecessors[currentVertex] !== null) { 251 const pair = predecessors[currentVertex]!; 252 const edge = pair.edge; 253 edge.capacity -= bottleneckFlow; 254 edge.reverseEdge!.capacity += bottleneckFlow; 255 currentVertex = pair.vertex; 256 } 257 258 // Calculate cost and update result 259 const unitCost = -dualVariables[s]; 260 const previousResult = result[result.length - 1]; 261 result.push([ 262 previousResult[0] + bottleneckFlow, 263 previousResult[1] + bottleneckFlow * unitCost 264 ]); 265 266 // Remove redundant point if cost remains the same 267 if (result.length >= 2) { 268 const lastPoint = result[result.length - 1]; 269 const secondLastPoint = result[result.length - 2]; 270 if (secondLastPoint[0] > 0 && unitCost === Math.floor(secondLastPoint[1] / secondLastPoint[0])) { 271 result.splice(result.length - 2, 1); 272 } 273 } 274 } 275 276 return result; 277} 278 279/** 280 * Computes the minimum cost flow from source to sink with optional flow limit. 281 * @param s - Source vertex 282 * @param t - Sink vertex 283 * @param flowLimit - Maximum flow limit (null for unlimited) 284 * @returns Array containing [total flow, total cost] 285 */ 286function flow(s: number, t: number, flowLimit: number | null): number[] { 287 const slopeResult = slope(s, t, flowLimit); 288 return slopeResult[slopeResult.length - 1]; 289} 290 291/** 292 * Finds the minimum time to process all tasks with given strengths. 293 * Uses minimum cost flow to model the assignment problem. 294 * @param strength - Array of task strengths 295 * @returns Minimum total time 296 */ 297function findMinimumTime(strength: number[]): number { 298 const n = strength.length; 299 300 // Create super source and super sink vertices 301 const superSource = n * 2; 302 const superSink = superSource + 1; 303 304 // Initialize minimum cost flow graph 305 initializeMCFGraph(superSink + 1); 306 307 // Build the bipartite graph 308 for (let i = 0; i < n; i++) { 309 // Connect source to left side (tasks) 310 addEdge(superSource, i, 1, 0); 311 312 // Connect right side (positions) to sink 313 addEdge(i + n, superSink, 1, 0); 314 315 // Connect each task to each position with appropriate cost 316 for (let j = 0; j < n; j++) { 317 // Cost represents time to complete task i at position j 318 const timeCost = Math.floor((strength[i] - 1) / (j + 1)) + 1; 319 addEdge(i, j + n, 1, timeCost); 320 } 321 } 322 323 // Find minimum cost flow and return the total cost 324 return flow(superSource, superSink, n)[1]; 325} 326

Solution Approach

The solution implements a Minimum Cost Maximum Flow (MCF) algorithm to solve this assignment problem optimally. Let's walk through how this sophisticated approach works:

Graph Construction

The solution creates a bipartite graph with a source and sink:

  1. Source node s: Connected to all lock nodes
  2. Lock nodes: Indices 0 to n-1 representing each lock
  3. Position nodes: Indices n to 2n-1 representing each position in the sequence
  4. Sink node t: Connected from all position nodes
n = len(a) s = n * 2 # source node t = s + 1 # sink node g = MCFGraph(t + 1) # create [graph](/problems/graph_intro) with all nodes

Edge Setup

Three types of edges are added:

  1. Source to locks: g.add_edge(s, i, 1, 0) - capacity 1, cost 0

    • Ensures each lock is selected exactly once

  2. Locks to positions: g.add_edge(i, j + n, 1, (a[i] - 1) // (j + 1) + 1)

    • Connects lock i to position j
    • Capacity 1 (each assignment happens once)
    • Cost is the time needed: ⌈a[i] / (j+1)⌉
    • The formula (a[i] - 1) // (j + 1) + 1 computes ceiling division

  3. Positions to sink: g.add_edge(i + n, t, 1, 0) - capacity 1, cost 0

    • Ensures each position is filled exactly once

The MCF Algorithm

The MCFGraph class implements a cost-efficient flow algorithm using:

  1. Dual variables and reduced costs: The algorithm maintains dual variables to transform the problem and find augmenting paths efficiently

  2. Dijkstra-based path finding (refine_dual function):

    • Uses a priority queue to find the shortest augmenting path
    • Computes reduced costs: e.cost - dual[w] + dual[v]
    • Updates dual variables to maintain optimality conditions

  3. Flow augmentation:

    • Finds minimum capacity along the path
    • Updates capacities along the path and reverse edges
    • Accumulates flow and cost

  4. Iterative refinement:

    • Continues finding augmenting paths until flow limit (n) is reached
    • Each iteration finds a path that minimizes cost per unit flow

Final Result

The call g.flow(s, t, n)[1] returns:

  • Flow of n (ensuring all locks are broken)
  • Minimum total cost (the minimum time needed)

The algorithm guarantees finding the globally optimal assignment of locks to positions, avoiding the need to check all n! permutations. Instead, it runs in polynomial time, typically O(n³) for this bipartite matching scenario.

This approach is much more efficient than brute-force DFS/backtracking for larger values of n, though for very small inputs, a simple backtracking solution might be more straightforward to implement.

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Example Walkthrough

Let's walk through a concrete example with locks of strength [5, 3, 8].

Step 1: Graph Construction We create a bipartite graph with:

  • Source node (s = 6)
  • Lock nodes (0, 1, 2) representing locks with strengths [5, 3, 8]
  • Position nodes (3, 4, 5) representing 1st, 2nd, and 3rd positions
  • Sink node (t = 7)

Step 2: Calculate Time Costs For each lock-position pair, we calculate the time needed:

Lock\Position1st (X=1)2nd (X=2)3rd (X=3)
Lock 0 (str=5)⌈5/1⌉ = 5⌈5/2⌉ = 3⌈5/3⌉ = 2
Lock 1 (str=3)⌈3/1⌉ = 3⌈3/2⌉ = 2⌈3/3⌉ = 1
Lock 2 (str=8)⌈8/1⌉ = 8⌈8/2⌉ = 4⌈8/3⌉ = 3

Step 3: Add Edges

  • Source → Locks: edges with capacity 1, cost 0
  • Locks → Positions: edges with capacity 1, cost = time from table above
  • Positions → Sink: edges with capacity 1, cost 0

Step 4: Run Min Cost Flow Algorithm The algorithm finds augmenting paths that minimize total cost:

  1. First augmenting path might be: s → Lock 1 → Position 0 → t

    • Assigns Lock 1 (strength 3) to 1st position
    • Cost: 3 minutes

  2. Second path: s → Lock 0 → Position 1 → t

    • Assigns Lock 0 (strength 5) to 2nd position
    • Cost: 3 minutes

  3. Third path: s → Lock 2 → Position 2 → t

    • Assigns Lock 2 (strength 8) to 3rd position
    • Cost: 3 minutes

Step 5: Result The optimal assignment is:

  • 1st position: Lock 1 (strength 3) → 3 minutes
  • 2nd position: Lock 0 (strength 5) → 3 minutes
  • 3rd position: Lock 2 (strength 8) → 3 minutes
  • Total time: 9 minutes

This is better than the naive sorted order [5, 3, 8] which would give:

  • 1st: strength 5 → 5 minutes
  • 2nd: strength 3 → 2 minutes
  • 3rd: strength 8 → 3 minutes
  • Total: 10 minutes

The MCF algorithm found that putting the weakest lock first and strategically placing the medium-strength lock second (where it benefits from X=2) gives the optimal result.

Time and Space Complexity

Time Complexity: O(n^3 log n)

The solution uses a Min-Cost Max-Flow algorithm to solve the problem. Breaking down the complexity:

  1. Graph Construction:

    • The graph has 2n + 2 vertices (n source nodes, n sink nodes, plus super source and super sink)
    • Edge creation: O(n^2) edges are added (each of n workers to n positions)
    • Each add_edge operation is O(1)
    • Total graph construction: O(n^2)

  2. Min-Cost Flow Algorithm:

    • The flow() method calls slope() which uses successive shortest path algorithm
    • The algorithm finds augmenting paths up to n times (maximum flow is n)
    • Each iteration of finding an augmenting path:
      • Uses Dijkstra's algorithm with a priority queue: O((V + E) log V)
      • With V = 2n + 2 vertices and E = O(n^2) edges
      • Each shortest path computation: O(n^2 log n)
    • Total for flow computation: O(n × n^2 log n) = O(n^3 log n)

Overall time complexity: O(n^2) + O(n^3 log n) = O(n^3 log n)

Space Complexity: O(n^2)

  1. Graph Storage:

    • Adjacency list _g: stores O(n^2) edges (forward and reverse edges)
    • Edge list _edges: stores O(n^2) edge references
    • Graph space: O(n^2)

  2. Algorithm Working Space:

    • Arrays dual, prev, dist, visited: each O(n) size
    • Priority queue: at most O(n) elements
    • Working space: O(n)

Overall space complexity: O(n^2) + O(n) = O(n^2)

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Cost Calculation for Lock-Position Assignment

One of the most common mistakes is incorrectly calculating the time needed to break a lock at a given position. The formula for computing the ceiling division can be error-prone.

Pitfall Example:

# Incorrect: Using regular division and ceiling time_cost = math.ceil(a[i] / (j + 1)) # Requires import math  # Incorrect: Simple integer division (floors instead of ceiling) time_cost = a[i] // (j + 1) # This gives floor, not ceiling!  # Incorrect: Off-by-one error in position indexing time_cost = (a[i] - 1) // j + 1 # j should be j+1 since positions start at 0

Correct Solution:

# Correct ceiling division formula time_cost = (a[i] - 1) // (j + 1) + 1 # This works because: ceil(a/b) = floor((a-1)/b) + 1 for positive integers

2. Graph Structure Confusion

The bipartite graph structure can be confusing, leading to incorrect node indexing or edge connections.

Pitfall Example:

# Incorrect: Mixing up node indices for i in range(n):  # Wrong: connecting task directly to sink  graph.add_edge(i, sink, 1, 0)   # Wrong: incorrect position node indexing  for j in range(n):  graph.add_edge(i, j, 1, time_cost) # Should be j+n for position nodes

Correct Solution:

# Maintain clear separation of node types: # - Tasks: indices 0 to n-1 # - Positions: indices n to 2n-1 # - Source: index 2n # - Sink: index 2n+1  for i in range(n):  graph.add_edge(source, i, 1, 0) # source to task  graph.add_edge(i + n, sink, 1, 0) # position to sink  for j in range(n):  graph.add_edge(i, j + n, 1, time_cost) # task to position

3. Misunderstanding the Problem's Sequential Nature

A critical misunderstanding is thinking that the factor X depends on which lock is broken rather than the position in the sequence.

Pitfall Example:

# Incorrect: Thinking X is tied to the lock index def calculate_time(lock_strength, lock_index):  return (lock_strength - 1) // (lock_index + 1) + 1 # Wrong!

Correct Understanding:

  • The factor X = 1 for the first lock broken (regardless of which lock it is)
  • The factor X = 2 for the second lock broken
  • The factor X = j+1 for the lock broken at position j (0-indexed)

4. Over-Engineering with Complex MCF for Small Inputs

While the MCF solution is elegant and optimal, it's overkill for small inputs where a simple permutation approach would suffice.

Alternative for Small n (≤ 10):

from itertools import permutations  def findMinimumTime_simple(a: List[int]) -> int:  n = len(a)  if n <= 10: # Use brute force for small inputs  min_time = float('inf')  for perm in permutations(range(n)):  time = sum((a[perm[j]] - 1) // (j + 1) + 1 for j in range(n))  min_time = min(min_time, time)  return min_time  # Fall back to MCF for larger inputs

5. Not Handling Edge Cases

Pitfall Example:

def findMinimumTime(a: List[int]) -> int:  # Missing edge case checks  n = len(a) # What if a is empty?  # What if a contains 0 or negative values?

Correct Solution:

def findMinimumTime(a: List[int]) -> int:  if not a:  return 0   n = len(a)  # Validate input assumptions  assert all(strength > 0 for strength in a), "All lock strengths must be positive"   # ... rest of the solution
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