Isomorphic Strings - Hash Table - Easy - LeetCode
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Isomorphic Strings - Hash Table - Easy - LeetCode

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  • 1Isomorphic strings allow character replacements while preserving order, with no two characters mapping to the same character.
  • 2The provided solution uses two hash maps to track character mappings between the two strings.
  • 3The algorithm has a time complexity of O(n) and a space complexity of O(n).

AI-generated summary · May not capture all nuances

Key Insight
AskGif

"Isomorphic strings allow character replacements while preserving order, with no two characters mapping to the same character."

Isomorphic Strings - Hash Table - Easy - LeetCode

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

Two strings 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 Note: You may assume both s and t have the same length.

public class Solution {
 public bool IsIsomorphic(string s, string t) {
 
 if(s.Length != t.Length){
 return false;
 }
 
 var map1 = new Dictionary<char,char>(); 
 var map2 = new Dictionary<char,char>(); 
 for(int i=0;i<s.Length;i++){
 if(map1.ContainsKey(s[i])){
 map1[s[i]] = t[i];
 }
 else{
 map1.Add(s[i],t[i]);
 } 
 
 if(map2.ContainsKey(t[i])){
 map2[t[i]] = s[i];
 }
 else{
 map2.Add(t[i],s[i]);
 } 
 }
 
 for(int i=0;i<s.Length;i++){
 if(map1[s[i]]!=t[i]){
 return false;
 }
 
 if(map2[t[i]]!=s[i]){
 return false;
 }
 }
 
 return true;
 }
}

Time Complexity: O(n)

Space Complexity: O(n)

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sumitc91

Published on 27 September 2020 · 1 min read · 171 words

Part of AskGif Blog · coding

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