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2912. Number of Ways to Reach Destination in the Grid

Description

You are given two integers n and m which represent the size of a 1-indexed grid. You are also given an integer k, a 1-indexed integer array source and a 1-indexed integer array dest, where source and dest are in the form [x, y] representing a cell on the given grid.

You can move through the grid in the following way:

  • You can go from cell [x1, y1] to cell [x2, y2] if either x1 == x2 or y1 == y2.
  • Note that you can't move to the cell you are already in e.g. x1 == x2 and y1 == y2.

Return the number of ways you can reach dest from source by moving through the grid exactly k times.

Since the answer may be very large, return it modulo 109 + 7.

 

Example 1:

 Input: n = 3, m = 2, k = 2, source = [1,1], dest = [2,2] Output: 2 Explanation: There are 2 possible sequences of reaching [2,2] from [1,1]: - [1,1] -> [1,2] -> [2,2] - [1,1] -> [2,1] -> [2,2]

Example 2:

 Input: n = 3, m = 4, k = 3, source = [1,2], dest = [2,3] Output: 9 Explanation: There are 9 possible sequences of reaching [2,3] from [1,2]: - [1,2] -> [1,1] -> [1,3] -> [2,3] - [1,2] -> [1,1] -> [2,1] -> [2,3] - [1,2] -> [1,3] -> [3,3] -> [2,3] - [1,2] -> [1,4] -> [1,3] -> [2,3] - [1,2] -> [1,4] -> [2,4] -> [2,3] - [1,2] -> [2,2] -> [2,1] -> [2,3] - [1,2] -> [2,2] -> [2,4] -> [2,3] - [1,2] -> [3,2] -> [2,2] -> [2,3] - [1,2] -> [3,2] -> [3,3] -> [2,3]

 

Constraints:

  • 2 <= n, m <= 109
  • 1 <= k <= 105
  • source.length == dest.length == 2
  • 1 <= source[1], dest[1] <= n
  • 1 <= source[2], dest[2] <= m

Solutions

Solution 1: Dynamic Programming

We define the following states:

  • $f[0]$ represents the number of ways to move from source to source itself;
  • $f[1]$ represents the number of ways to move from source to another row in the same column;
  • $f[2]$ represents the number of ways to move from source to another column in the same row;
  • $f[3]$ represents the number of ways to move from source to another row and another column.

Initially, $f[0] = 1$, and the other states are all $0$.

For each state, we can calculate the current state based on the previous state, as follows:

\[\begin{aligned} g[0] &= (n - 1) \times f[1] + (m - 1) \times f[2] \\ g[1] &= f[0] + (n - 2) \times f[1] + (m - 1) \times f[3] \\ g[2] &= f[0] + (m - 2) \times f[2] + (n - 1) \times f[3] \\ g[3] &= f[1] + f[2] + (n - 2) \times f[3] + (m - 2) \times f[3] \end{aligned}\]

We loop $k$ times, and finally check whether source and dest are in the same row or column, and return the corresponding state.

The time complexity is $O(k)$, where $k$ is the number of moves. The space complexity is $O(1)$.

  • class Solution { public int numberOfWays(int n, int m, int k, int[] source, int[] dest) { final int mod = 1000000007; long[] f = new long[4]; f[0] = 1; while (k-- > 0) { long[] g = new long[4]; g[0] = ((n - 1) * f[1] + (m - 1) * f[2]) % mod; g[1] = (f[0] + (n - 2) * f[1] + (m - 1) * f[3]) % mod; g[2] = (f[0] + (m - 2) * f[2] + (n - 1) * f[3]) % mod; g[3] = (f[1] + f[2] + (n - 2) * f[3] + (m - 2) * f[3]) % mod; f = g; } if (source[0] == dest[0]) { return source[1] == dest[1] ? (int) f[0] : (int) f[2]; } return source[1] == dest[1] ? (int) f[1] : (int) f[3]; } } 
  • class Solution { public: int numberOfWays(int n, int m, int k, vector<int>& source, vector<int>& dest) { const int mod = 1e9 + 7; vector<long long> f(4); f[0] = 1; while (k--) { vector<long long> g(4); g[0] = ((n - 1) * f[1] + (m - 1) * f[2]) % mod; g[1] = (f[0] + (n - 2) * f[1] + (m - 1) * f[3]) % mod; g[2] = (f[0] + (m - 2) * f[2] + (n - 1) * f[3]) % mod; g[3] = (f[1] + f[2] + (n - 2) * f[3] + (m - 2) * f[3]) % mod; f = move(g); } if (source[0] == dest[0]) { return source[1] == dest[1] ? f[0] : f[2]; } return source[1] == dest[1] ? f[1] : f[3]; } }; 
  • class Solution: def numberOfWays( self, n: int, m: int, k: int, source: List[int], dest: List[int] ) -> int: mod = 10**9 + 7 a, b, c, d = 1, 0, 0, 0 for _ in range(k): aa = ((n - 1) * b + (m - 1) * c) % mod bb = (a + (n - 2) * b + (m - 1) * d) % mod cc = (a + (m - 2) * c + (n - 1) * d) % mod dd = (b + c + (n - 2) * d + (m - 2) * d) % mod a, b, c, d = aa, bb, cc, dd if source[0] == dest[0]: return a if source[1] == dest[1] else c return b if source[1] == dest[1] else d 
  • func numberOfWays(n int, m int, k int, source []int, dest []int) int { const mod int = 1e9 + 7 f := []int{1, 0, 0, 0} for i := 0; i < k; i++ { g := make([]int, 4) g[0] = ((n-1)*f[1] + (m-1)*f[2]) % mod g[1] = (f[0] + (n-2)*f[1] + (m-1)*f[3]) % mod g[2] = (f[0] + (m-2)*f[2] + (n-1)*f[3]) % mod g[3] = (f[1] + f[2] + (n-2)*f[3] + (m-2)*f[3]) % mod f = g } if source[0] == dest[0] { if source[1] == dest[1] { return f[0] } return f[2] } if source[1] == dest[1] { return f[1] } return f[3] }

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