[NEW] Add solution for Building Teams

[vingoel26]

🎯 Problem Information

• Problem Name: Building Teams
• CSES Link: https://cses.fi/problemset/task/1668
• Category: Graphs
• Estimated Difficulty: Medium

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💡 Proposed Approach

Describe the algorithm/approach you plan to use:

• The problem can be solved using Graph Coloring (Bipartite Check).

• Each student is represented as a node, and each friendship as an undirected edge.

• We must divide pupils into two teams such that no two friends are in the same team.

• This can be checked using BFS or DFS traversal with two colors:

  • Assign color 1 to a node.
  • Assign color 2 to all its neighbors.
  • If any adjacent nodes have the same color → print "IMPOSSIBLE".

• If traversal finishes successfully for all components, print the color of each node.

• Time Complexity: O(N + M)

• Space Complexity: O(N + M)

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📝 Implementation Notes

Any special considerations or optimizations needed:

• Requires fast I/O
• Needs modular arithmetic
• Large input constraints
• Edge cases to consider:
  • Smallest input (n = 1)
  • Disconnected graphs
  • Odd cycles (no valid bipartite division)
  • Large inputs (n = 10^5, m = 2×10^5)

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🏷️ Labels

Please add appropriate labels:

• hacktoberfest
• good first issue
• help wanted

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👤 Assignee

• I would like to work on this issue
• I'm available for guidance/review

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📎 Additional Context

This code implements a Graph Coloring approach using BFS traversal to check bipartiteness.
If a valid two-coloring is possible, the nodes are divided into two teams (1 and 2).
If any conflict arises (same color on both ends of an edge), the answer is "IMPOSSIBLE".

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🧠 Code Implementation

#include <bits/stdc++.h>
using namespace std;

/*
========================================
Author:         Vinayak Goel
May the WA avoid you
========================================
*/

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int n, m;
    cin >> n >> m;
    
    vector<vector<int>> adj(n + 1);
    for (int i = 0; i < m; ++i) {
        int a, b;
        cin >> a >> b;
        adj[a].push_back(b);
        adj[b].push_back(a);
    }
    
    vector<int> color(n + 1, 0);
    
    for (int i = 1; i <= n; ++i) {
        if (color[i] != 0) continue;
        queue<int> q;
        q.push(i);
        color[i] = 1;
        
        while (!q.empty()) {
            int u = q.front();
            q.pop();
            for (int v : adj[u]) {
                if (color[v] == 0) {
                    color[v] = 3 - color[u];
                    q.push(v);
                } else if (color[v] == color[u]) {
                    cout << "IMPOSSIBLE\n";
                    return 0;
                }
            }
        }
    }
    
    for (int i = 1; i <= n; ++i)
        cout << color[i] << " ";
    cout << "\n";
    
    return 0;
}

[vingoel26]

@ks-iitjmu please assign me this issue