Leetcode Daily Challenge Solutions

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Today's 02-04-24 Problem Link

205. Isomorphic Strings

Intuition

To check if two strings are isomorphic, we need to establish a one-to-one mapping between characters in one string to characters in the other string. This mapping must preserve the order of characters and ensure that no two characters map to the same character, while allowing a character to map to itself.

Approach

Initialization : Initialized two hash maps (sToTMap and tToSMap) to store mappings from characters in string s to characters in string t, and vice versa.

Iteration : Iterated through each character in the strings s and t.

Mapping Check :

• For each character pair (sChar, tChar), checked if sChar is already mapped to a character in t and if tChar is already mapped to a character in s.
• If a mapping existed and the mapped characters are not equal to each other, returned false.
• If a mapping does not exist, established the mapping by adding the characters to their respective maps.

Completion : If the loop completes without returning false, returned true, indicating that the strings are isomorphic.

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Have a look at the code , still have any confusion then please let me know in the comments Keep Solving.:)

Complexity

• Time complexity : $O(n)$

$n$ : length of the strings s and t

• Space complexity : $O(min(m, n))$

$m$ : size of the character set

Code

class Solution {
    public boolean isIsomorphic(String s, String t) {

        // Initializing two hash maps to store mappings
        Map<Character, Character> sToTMap = new HashMap<>();
        Map<Character, Character> tToSMap = new HashMap<>();
        
        // Iterating through each character in the strings s and t
        for (int i = 0; i < s.length(); i++) {
            char sChar = s.charAt(i);
            char tChar = t.charAt(i);
            
            // Checking if sChar is already mapped to a character in t
            if (sToTMap.containsKey(sChar)) {
                // If the mapped character is not equal to tChar, returning false
                if (sToTMap.get(sChar) != tChar) {
                    return false;
                }
            } else {
                // If sChar is not mapped, mapping it to tChar
                sToTMap.put(sChar, tChar);
            }
            
            // Checking if tChar is already mapped to a character in s
            if (tToSMap.containsKey(tChar)) {
                // If the mapped character is not equal to sChar, returning false
                if (tToSMap.get(tChar) != sChar) {
                    return false;
                }
            } else {
                // If tChar is not mapped, mapping it to sChar
                tToSMap.put(tChar, sChar);
            }
        }
        
        // If the loop completes without returning false, returning true
        return true;
    }
}