Check if every vertex triplet in graph contains two vertices connected to third vertex

Given an undirected graph with N vertices and K edges, the task is to check if for every combination of three vertices in the graph, there exists two vertices which are connected to third vertex. In other words, for every vertex triplet (a, b, c), if there exists a path between a and c, then there should also exist a path between b and c.


Input: N = 4, K = 3
Edges: 1 -> 2, 2 -> 3, 3 -> 4
Output: YES
Since the whole graph is connected, the above condition will always be valid.

Input: N =  5 and K = 3
Edges: 1 -> 3, 3 -> 4, 2 -> 5.
Output: NO
If we consider the triplet (1, 2, 3) then there is a path between vertices 1 and 3 but there is no path between vertices 2 and 3.

Approach: Follow the steps below to solve the problem –

  • Traverse the graph by DFS Traversal technique from any component and maintain two variables to store the component minimum and component maximum.
  • Store every component maximum and minimum in a vector.
  • Now, if any two components have an intersection in their minimum and maximum values interval, then there will exist a valid (a < b < c) triplet. Hence, both of the components should be connected. Otherwise, the graph is not valid.

Below is the implementation of the above approach






// C++ program of the
// above approach
#include <bits/stdc++.h>
using namespace std;
// Function to add edge into
// the graph
void addEdge(vector<int> adj[],
             int u, int v)
void DFSUtil(int u, vector<int> adj[],
             vector<bool>& visited,
             int& componentMin,
             int& componentMax)
    visited[u] = true;
    // Finding the maximum and
    // minimum values in each component
    componentMax = max(componentMax, u);
    componentMin = min(componentMin, u);
    for (int i = 0; i < adj[u].size(); i++)
        if (visited[adj[u][i]] == false)
            DFSUtil(adj[u][i], adj, visited,
                    componentMin, componentMax);
// Function for checking whether
// the given graph is valid or not
bool isValid(vector<pair<int, int> >& v)
    int MAX = -1;
    bool ans = 0;
    // Checking for intersecting intervals
    for (auto i : v) {
        // If intersection is found
        if (i.first <= MAX) {
            // Graph is not valid
            ans = 1;
        MAX = max(MAX, i.second);
    return (ans == 0 ? 1 : 0);
// Function for the DFS Traversal
void DFS(vector<int> adj[], int V)
    std::vector<pair<int, int> > v;
    // Traversing for every vertex
    vector<bool> visited(V, false);
    for (int u = 1; u <= V; u++) {
        if (visited[u] == false) {
            int componentMax = u;
            int componentMin = u;
            DFSUtil(u, adj, visited,
                    componentMin, componentMax);
            // Storing maximum and minimum
            // values of each component
            v.push_back({ componentMin,
                          componentMax });
    bool check = isValid(v);
    if (check)
        cout << "Yes";
        cout << "No";
// Driver code
int main()
    int N = 4, K = 3;
    vector<int> adj[N + 1];
    addEdge(adj, 1, 2);
    addEdge(adj, 2, 3);
    addEdge(adj, 3, 4);
    DFS(adj, N);
    return 0;




Time Complexity: O(N + E)
Auxiliary Space:  O(N) 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up

Recommended Posts:

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using or mail your article to See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.