**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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