Last updated on October 5th, 2024 at 04:20 pm
Here, We see Isomorphic Strings LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.
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Topics
Hash Table, Tree
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Level of Question
Medium
Isomorphic Strings LeetCode Solution
Table of Contents
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: true
Example 2:
Input: s = “foo”, t = “bar”
Output: false
Example 3:
Input: s = “paper”, t = “title”
Output: true
1. Isomorphic Strings Leetcode Solution C++
class Solution { public: bool isIsomorphic(string s, string t) { unordered_map<char, char> mp, mp2; for (int i=0; i<s.length(); ++i) { if (mp[s[i]] && mp[s[i]]!=t[i]) return false; if (mp2[t[i]] && mp2[t[i]]!=s[i]) return false; mp[s[i]]=t[i]; mp2[t[i]]=s[i]; } return true; } };
2. Isomorphic Strings Leetcode Solution Java
class Solution { public boolean isIsomorphic(String s, String t) { int map1[]=new int[200]; int map2[]=new int[200]; if(s.length()!=t.length()) return false; for(int i=0;i<s.length();i++) { if(map1[s.charAt(i)]!=map2[t.charAt(i)]) return false; map1[s.charAt(i)]=i+1; map2[t.charAt(i)]=i+1; } return true; } }
3. Isomorphic Strings Leetcode Solution JavaScript
var isIsomorphic = function(s, t) { if(s.length != t.length) return false; const map1 = [256]; const map2 = [256]; for(let idx = 0; idx < s.length; idx++){ if(map1[s.charAt(idx)] != map2[t.charAt(idx)]) return false; map1[s.charAt(idx)] = idx + 1; map2[t.charAt(idx)] = idx + 1; } return true; };
4. Isomorphic Strings Leetcode Solution Python
class Solution(object): def isIsomorphic(self, s, t): map1 = [] map2 = [] for idx in s: map1.append(s.index(idx)) for idx in t: map2.append(t.index(idx)) if map1 == map2: return True return False