We can check if \(S\) and \(T\) are the same without performing any operation, or if \(S\) and \(T\) coincides when the \(i\)-th and \((i+1)\)-th characters are swapped for each \(i=1,2,\ldots,(\)length of \(S)-1\).

The time complexity is \(O(N^2)\), where \(N=(\)length of \(S)\).

Sample code (Python)

S = list(input())
T = list(input())
ans = "No"
if S == T:
    ans = "Yes"
for i in range(len(S)-1):
    S[i],S[i+1] = S[i+1],S[i]
    if S == T:
        ans = "Yes"
    S[i],S[i+1] = S[i+1],S[i]
print(ans)

With a little more twist, the time complexity can be decreased to \(O(N)\).

Sample code (C++)

#include<bits/stdc++.h>
using namespace std;

int main(){
    string S,T; cin >> S >> T;
    string ans = "No";
    if(S == T) ans = "Yes";
    for(int i=0; i<S.size(); i++){
        if(S[i] != T[i]){
            if(0 < i){
                swap(S[i-1],S[i]);
                if(S == T) ans = "Yes";
                swap(S[i-1],S[i]);
            }
            if(i+1 < S.size()){
                swap(S[i],S[i+1]);
                if(S == T) ans = "Yes";
                swap(S[i],S[i+1]);
            }
            break;
        }
    }
    cout << ans << endl;
}