0200 - 0299
0205 - Isomorphic Strings (Easy)
Problem Link
https://leetcode.com/problems/isomorphic-strings/
Problem Statement
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:
Input: s = "egg", t = "add"
Output: trueExample 2:
Input: s = "foo", t = "bar"
Output: falseExample 3:
Input: s = "paper", t = "title"
Output: trueConstraints:
1 <= s.length <= 5 * 10^4t.length == s.lengthsandtconsist of any valid ascii character.
Approach 1: Hash Map
class Solution {
public:
bool isIsomorphic(string s, string t) {
// m1 is used to map s[i] to t[i]
// m2 is used to map t[i] to s[i]
// example 1:
// m1['e'] --> 'a'
// m1['g'] --> 'd'
// m2['a'] --> 'e'
// m2['d'] --> 'g'
unordered_map<char, char> m1, m2;
for (int i = 0; i < s.size(); i++) {
// if s[i] isn't in m1 and t[i] isn't in m2
if (!m1.count(s[i]) && !m2.count(t[i])) {
// then we can map s[i] to t[i]
m1[s[i]] = t[i];
// and t[i] to s[i]
m2[t[i]] = s[i];
} else {
// if they are they are isomorphic,
// m1[s[i]] should map to t[i], and m2[t[i]] should map to s[i]
// we return false here if this condition is not satisfied
if (m1[s[i]] != t[i] || m2[t[i]] != s[i]) {
return false;
}
}
}
// they are isomorphic
return true;
}
};