# 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 :

```                        ```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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