1506C - Double-ended Strings - CodeForces Solution

brute force implementation strings *1000

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Python Code: 

for _ in range(int(input())):
    a = input()     b = input()     s = set()
    if len(b) > len(a):
        b, a = a, b
   
    for i in range(len(b)):
            for j in range(i+1, len(b)+1):
                if b[i:j] in a:
                    s.add(len(b[i:j]))
    print((len(a)-max(s))+(len(b)-max(s)) if len(s) > 0 else len(a)+len(b))

C++ Code: 

#include <bits/stdc++.h>

#define ll long long int
#define PI 3.141592653589793238462643383279502884197
#define endl "\n"
using namespace std;


int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);

    int t = 1;
    cin>>t;
    while (t--) {
        string a,b; cin>>a>>b;
        if(a==b)
        {
            cout<<0<<endl;
            continue;
        }
        string found="";
        for (int i = 0; i <b.size(); ++i) {
            string search_for="";
            for (int j = i; j <b.size(); ++j) {
                search_for+=b[j];

                if(a.find(search_for)!=-1)
                {
                    if(found.size()<=search_for.size())
                    {
                        found=search_for;
                    }
                }
            }

        }
        if(found=="")
        {
            cout<<a.size()+b.size()<<endl;
        }
        else
        {
            cout<<a.size()+b.size() -(found.size()*2)<<endl;
        }



    }
    return 0;
}


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