Suppose we have two lowercase strings s and t. Now consider an operation where we replace all occurrences of a character in s with another character. If we can perform this operation any number of times, we have to check whether s can be converted to t or not.
So, if the input is like s = "eye" t = "pip", then the output will be True, as we can replace "e"s with "p"s then "y" by "i".
To solve this, we will follow these steps −
Define one map m1 and another map m2
n := size of s
for initialize i := 0, when i < n, update (increase i by 1), do −
if s[i] is in m1, then −
if m1[s[i]] is same as t[i], then −
go for the next iteration
return false
Otherwise
m1[s[i]] := t[i]
return true
Let us see the following implementation to get a better understanding −
#include <bits/stdc++.h> using namespace std; bool solve(string s, string t) { map m1, m2; int n = s.size(); for(int i = 0; i <n; i++){ if(m1.count(s[i])){ if(m1[s[i]] == t[i]) continue; return false; } else{ m1[s[i]] = t[i]; } } return true; } int main(){ string s = "eye", t = "pip"; cout << solve(s, t); }
"eye","pip"
1