1005B - Delete from the Left - CodeForces Solution


brute force implementation strings *900

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Python Code:

s= input()
t=input()
w=0
while True:
    i = len(s) - w - 1
    j = len(t)-w-1
    if i>=0 and j>=0 and t[j]==s[i]:
        w += 1
    else:
        break

print(len(s)+len(t)-2*w)

C++ Code:

#include<iostream>
#include<queue>
#include<stack>
#include<deque>
#include<algorithm>
#include<vector>
#include<cmath>
#include<string>
#include<iomanip>
#include<map>
#include<set>
#include<bitset>
#define ll long long
#define T int t;  cin>>t; while(t--)
#define pb  push_back
#define vv vector
#define ss string
#define so(vec) sort(vec.begin(), vec.end())
#define srt(vec) sort(vec.rbegin(), vec.rend())
#define s(s) sort(s.begin(), s.end())
#define all(vec) vec.begin(), vec.end()
#define sor(arr,n) sort(arr, arr+n)
#define yes cout<<"YES\n"
#define no cout<<"NO\n"
#define  Mod 1000000007
const double PI = 3.14159265358979323846;
using namespace std;
void  HaiDO0da() {
    ios_base::sync_with_stdio(false);
    cout.tie(NULL);
    cin.tie(NULL);
}
int main()
{
  HaiDO0da();
ss x,y;    cin>>x >>y;
int  cnt=0,b=x.size()+y.size(),c=min(x.size(),y.size());
reverse(x.begin(),x.end());
reverse(y.begin(),y.end());
for(int i=0;i<c;i++){
if(x[i]==y[i]) cnt++;
else break;
}
//cout<<cnt;
cout<<b-(cnt*2);
}




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