Redundant Connection in C++

Suppose we have one unrooted tree; this is one undirected graph with no cycles. The given input is a graph that started as a tree with N nodes (values of the nodes are distinct values ranging from 1 through N), with one additional edge added. The added edge has two different vertices chosen from 1 to N, and was not an edge that already existed.

The final graph is given as a 2D-array of edges. Each element of edges is a pair [u, v] where u < v, that represents an undirected edge connecting nodes u and v.

We have to find an edge that can be removed so that the resulting graph is a tree of N nodes. There may be multiple answers, we have to find the answer that occurs last in the given 2Darray. The answer edge [u, v] should be in the same format, with u < v.

So, if the input is like [[1,2], [2,3], [3,4], [1,4], [1,5]],

then the output will be [1,4]

To solve this, we will follow these steps −

• N := 1000

• Define an array parent of size: N+5.

• Define an array rank of size: N+5.

• Define a function getParent(), this will take n,

• if parent[n] is same as -1, then −

  • return n

• return parent[n] = getParent(parent[n])

• Define a function unionn(), this will take a, b,

• pa := getParent(a), pb := getParent(b)

• if pa is same as pb, then −

  • return false

• if rank[pa] > rank[pb], then −

  • rank[pa] := rank[pa] + rank[pb]

  • parent[pb] := pa

• Otherwise

  • rank[pb] := rank[pb] + rank[pa]

  • parent[pa] := pb

• return true

• From the main method, do the following −

• n := size of edges list

• for initialize i := 0, when i < n, update (increase i by 1), do −

  • parent[edges[i, 0]] := -1, parent[edges[i, 1]] := - 1

  • rank[edges[i, 0]] := -1, rank[edges[i, 1]] := 1

• Define an array ans

• for initialize i := 0, when i < n, update (increase i by 1), do −

  • u := edges[i, 0]

  • v := edges[i, 1]

  • if unionn(u, v) is zero, then −

    • ans := edges[i]

• return ans

Example 

Let us see the following implementation to get better understanding −

 Live Demo

#include <bits/stdc++.h> using namespace std; void print_vector(vector v){    cout << "[";    for(int i = 0; i<v.size(); i++){       cout << v[i] << ", ";    }    cout << "]"<<endl; } const int N = 1000; class Solution { public:    int parent[N + 5];    int rank[N + 5];    int getParent(int n){       if (parent[n] == -1)          return n;       return parent[n] = getParent(parent[n]);    }    bool unionn(int a, int b){       int pa = getParent(a);       int pb = getParent(b);       if (pa == pb)          return false;       if (rank[pa] > rank[pb]) {          rank[pa] += rank[pb];          parent[pb] = pa;       }       else {          rank[pb] += rank[pa];          parent[pa] = pb;       }       return true;    }    vector<int> findRedundantConnection(vector<vector<int>>& edges) {       int n = edges.size();       for (int i = 0; i < n; i++) {          parent[edges[i][0]] = parent[edges[i][1]] = -1;          rank[edges[i][0]] = rank[edges[i][1]] = 1;       }       vector<int> ans;       for (int i = 0; i < n; i++) {          int u = edges[i][0];          int v = edges[i][1];          if (!unionn(u, v)) {             ans = edges[i];          }       }       return ans;    } }; main(){    Solution ob;    vector<vector<int>> v = {{1,2}, {2,3}, {3,4}, {1,4}, {1,5}};    print_vector(ob.findRedundantConnection(v)); }

Input

{{1,2}, {2,3}, {3,4}, {1,4}, {1,5}}

Output

[1, 4, ]