String Isomorphism - Amazon Top Interview Questions


Problem Statement :


Given lowercase alphabet strings s, and t return whether you can create a 1-to-1 mapping for each letter in s to another letter (could be the same letter) such that s could be mapped to t, with the ordering of characters preserved.

Constraints

n ≤ 100,000 where n is the length of s
m ≤ 100,000 where m is the length of t


Example 1

Input
s = "coco"
t = "kaka"


Output
True


Explanation
We can create this mapping:

"c" -> "k"
"o" -> "a"


Example 2

Input
s = "cat"
t = "foo"

Output
False


Explanation
We can't transform both "a" and "t" into "o" since it has to be a 1-to-1 mapping.




Example 3
Input
s = "hello"
t = "hello"

Output
True


Explanation
The mapping can just map each letter to itself.


Solution :



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                        Solution in C++ :

bool solve(string s, string t) {
    unordered_map<char, char> st, ts;

    if (s.size() != t.size()) return 0;

    for (int i = 0; i < s.size(); i++) {
        // check the s -> t mapping
        if (not st.count(s[i]) or st[s[i]] == t[i])
            // this is the first time I've seen this char in s,
            // or second time with the same mapped value as last time
            st[s[i]] = t[i];
        else
            return false;

        // check the t -> s mapping (same logic as above)
        if (not ts.count(t[i]) or ts[t[i]] == s[i])
            ts[t[i]] = s[i];
        else
            return false;
    }

    return true;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public boolean solve(String s, String t) {
        if (s == t)
            return true;
        if (s.length() != t.length())
            return false;
        for (int i = 0; i < s.length(); i++) {
            if (s.indexOf(s.charAt(i)) != t.indexOf(t.charAt(i)))
                return false;
        }
        return true;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, s, t):
        ds, dt = {}, {}
        for i in range(len(s)):
            if s[i] not in ds:
                ds[s[i]] = t[i]
            else:
                if ds[s[i]] != t[i]:
                    return False
            if t[i] not in dt:
                dt[t[i]] = s[i]
            else:
                if dt[t[i]] != s[i]:
                    return False
        return True


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