Codechef SEPT17 CookOff - Subsequence Equality Problem Code: LIKECS01

Problem: (Link: https://www.codechef.com/problems/LIKECS01 )⌗

Chef Tobby is playing a rapid fire with Bhuvan. He gives Bhuvan a string S and each time, Bhuvan has to guess whether there exists 2 equal subsequences in the string or not.
Bhuvan got a perfect score in the game with Chef Tobby. However, Chef Tobby has now asked Bhuvan to write a program that will do this automatically given a string S. Bhuvan is an intelligent man but he does not know how to write a code. Can you help him?
Find two different subsequences such that they are equal in their value, more formally, find two sequences of indices (a₁, a₂, …, a_k-1, a_k) and (b₁, b₂, …, b_k-1, b_k) such that:

1. 1≤ a_i, b_i ≤ |S|
2. a_i < a_i+1 for all valid i
3. b_i < b_i+1 for all valid i
4. S_ai = S_bi for all valid i
5. there exist at least one i such that a_i is not equal to b_i

Input section⌗

The first line contains T, the number of test cases.
Each of the next T lines contain one string S each.
Input will only consist of lowercase english characters

Output section⌗

For each test case, output “yes” or “no” (without quotes) as the solution to the problem.

Input constraints⌗

1 ≤ T ≤ 1000
1 ≤ length of S ≤ 100

Sample Input⌗

4
likecs
venivedivici
bhuvan
codechef

Sample Output⌗

no
yes
no
yes

Explanation⌗

In test case 2, one of the possible equal subsequence is “vi” and “vi”. (one at position {0, 3} and other at {4, 7}, assuming 0-based indexing).
In test case 4, one of the possible equal subsequence is “ce” and “ce”. (one at position {0, 3} and other at {4, 6}, assuming 0-based indexing).

Solution :⌗

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#include <iostream>
#include <set>
using namespace std;
int main()
{
    int t;
    cin >> t;
    while (t--) {
        set<char> st;
        string s;
        cin >> s;
        for (int i = 0; i < s.length(); i++)
            st.insert(s[i]);
        if (st.size() == s.length())
            cout << "no" << endl;
        else
            cout << "yes" << endl;
    }

    return 0;
}