HackerEarth Minimize nodes problem solution

In this HackerEarth Minimize nodes problem solution, You are given N strings. Each string is given in the form of an array cnt of size 26 where the ith element in array denotes the count of the ith character of lowercase English alphabets in the string.

You have to select exactly 4 strings and insert the strings into a Trie. You are allowed to shuffle the characters in each string.

HackerEarth Minimize nodes problem solution.

#include<bits/stdc++.h>
using namespace std;
int solve(vector<int> a)
{
    int ans = 0;
    for(int i = 0; i < a.size() ; i++)
        ans += a[i];
    return ans;
}
int solve(vector<int> a, vector<int> b)
{
    int ans = 0;
    for(int i = 0; i < a.size() ; i++){
        int mn = min(a[i], b[i]);
        ans += mn;
        a[i] -= mn;
        b[i] -= mn;
    }
    return ans + solve(a) + solve(b);
}
int solve(vector<int> a, vector<int> b, vector<int> c)
{
    int ans = 0;
    for(int i = 0; i < a.size() ; i++){
        int mn = min({a[i], b[i], c[i]});
        ans += mn;
        a[i] -= mn;
        b[i] -= mn;
        c[i] -= mn;
    }
    int val = solve(a) + solve(b, c);
    val = min(val, solve(b) + solve(a, c));
    val = min(val, solve(c) + solve(a, b));
    return ans + val;
}
int solve(vector<int> a, vector<int> b, vector<int> c, vector<int> d)
{
    int ans = 0;
    for(int i = 0; i < a.size() ; i++){
        int mn = min({a[i], b[i], c[i], d[i]});
        ans += mn;
        a[i] -= mn;
        b[i] -= mn;
        c[i] -= mn;
        d[i] -= mn;
    }
    int val = solve(a) + solve(b, c, d);
    val = min(val, solve(b) + solve(a, c, d));
    val = min(val, solve(c) + solve(a, b, d));
    val = min(val, solve(d) + solve(a, b, c));
    val = min(val, solve(a, b) + solve(c, d));
    val = min(val, solve(a, c) + solve(b, d));
    val = min(val, solve(a, d) + solve(b, c));
    return ans + val;
}
long long solve (int N, vector<vector<int> > cnt) {
    // Write your code here
    int i, j, k, l;
    int ans = 1e9;
    assert(4 <= N && N <= 20);
    for(i = 0; i < N ; i++)
        for(j = 0; j < 26 ; j++)
            assert(0 <= cnt[i][j] && cnt[i][j] <= 100000);
    for(i = 0; i < N ; i++)
        for(j = i+1; j < N ; j++)
            for(k = j + 1 ; k < N ; k++)
                for(l = k + 1 ; l < N ; l++){
                    ans = min(ans, solve(cnt[i], cnt[j], cnt[k], cnt[l]));
                }
    return (1 + ans);
}

int main() {

    ios::sync_with_stdio(0);
    cin.tie(0);
    int T;
    cin >> T;
    assert(1 <= T && T <= 5);
    for(int t_i = 0; t_i < T; t_i++)
    {
        int N;
        cin >> N;
        vector<vector<int> > cnt(N, vector<int>(26));
        for (int i_cnt = 0; i_cnt < N; i_cnt++)
        {
            for(int j_cnt = 0; j_cnt < 26; j_cnt++)
            {
                cin >> cnt[i_cnt][j_cnt];
            }
        }

        long long out_;
        out_ = solve(N, cnt);
        cout << out_;
        cout << "\n";
    }
}