# Isomorphic Strings Problem

## Description

LeetCode Problem 205.

Given two strings s and t, determine if they are isomorphic.

Two strings s and t are isomorphic if the characters in s can be replaced to get t. All occurrences of a character must be replaced with another character while preserving the order of characters. No two characters may map to the same character, but a character may map to itself.

Example 1:

``````1
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Input: s = "egg", t = "add"
Output: true
``````

Example 2:

``````1
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Input: s = "foo", t = "bar"
Output: false
``````

Example 3:

``````1
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Input: s = "paper", t = "title"
Output: true
``````

Constraints:

• 1 <= s.length <= 5 * 10^4
• t.length == s.length
• s and t consist of any valid ascii character.

## Sample C++ Code

``````1
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class Solution {
public:
bool isIsomorphic(string s, string t) {
map<char, char> replaced;
map<char, char> reverse_replaced;
map<char, char>::iterator mit1, mit2;

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

for (int i = 0; i < s.size(); i ++) {
mit1 = replaced.find(s[i]);
mit2 = reverse_replaced.find(t[i]);

if ((mit1 == replaced.end()) && (mit2 == reverse_replaced.end())) {
replaced.emplace(s[i], t[i]);
reverse_replaced.emplace(t[i], s[i]);
s[i] = t[i];
} else if ((mit1 == replaced.end()) && (mit2 != reverse_replaced.end())) {
return false;
} else if ((mit1 != replaced.end()) && (mit2 == reverse_replaced.end())) {
return false;
} else {
if ((t[i] != mit1->second) || (s[i] != mit2->second))
return false;
s[i] = t[i];
}

}
if (s != t)
return false;
else
return true;
}
};
``````