XI Samara Regional Intercollegiate Programming Contest Problem E. Substring Reverse

Two strings s and t of the same length are given. Determine whether it is possible to make t from s using
exactly one reverse of some its substring.
Input
The first line contains the string s, and the second — the string t. Both strings have the same length from
1 to 200000 characters and consist of lowercase Latin letters.
Output
Output «YES», if it is possible to reverse some substring of s to make s equal to t, and «NO», otherwise.
Examples
standard input standard output
abcdefg
abedcfg
YES
abcdefg
abdecfg
NO

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<string>
#include<bitset>
#include<ctime>
#include<deque>
typedef long long ll;
using namespace std;
typedef unsigned long long int ull;
#define maxn 500005
#define ms(x) memset(x,0,sizeof(x))
#define Inf 0x7fffffff
#define inf 0x3f3f3f3f
const long long int mod = 1e9 + 7;
#define pi acos(-1.0)
#define pii pair<int,int>
#define eps 1e-7
#define pll pair<ll,ll>



ll quickpow(ll a, ll b) {
    ll ans = 1;
    a = a % mod;
    while (b > 0) {
        if (b % 2)ans = ans * a;
        b = b / 2;
        a = a * a;
    }
    return ans;
}

int gcd(int a, int b) {
    return b == 0 ? a : gcd(b, a%b);
}

bool prime(int x) {
    if (x == 0 || x == 1)return false;
    for (int i = 2; i <= sqrt(x); i++) {
        if (x%i == 0)return false;
    }
    return true;
}


int main()
{
    ios::sync_with_stdio(false);
    string s, t;
    cin >> s >> t;
    int i, j;
    int len = s.length();
    int pos1, pos2;
    if (s == t) {
        int flag = 0;
        cout << "YES" << endl;
    }
    else {
        for (i = 0; i < len; i++) {
            if (s[i] != t[i]) {
                pos1 = i;
                break;
            }
        }
        for (i = len - 1; i >= 0; i--) {
            if (s[i] != t[i]) {
                pos2 = i;
                break;
            }
        }
        int pp1 = pos1, pp2 = pos2;
        int  flag = 0;
        while (1) {
            if (s[pos1] != t[pos2]) {
                flag = 1;
                break;
            }
            pos1++;
            pos2--;
            if (pos1 > pp2)break;
        }
        if (flag)cout << "NO" << endl;
        else cout << "YES" << endl;
    }
}