Redundant Connection in C++

C++Server Side ProgrammingProgramming

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, ]
raja
Updated on 17-Nov-2020 11:03:03

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