205. Isomorphic Strings

LeetCode problem link (opens new window)

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

Hint: You may assume both s and t have the same length.

Approach

The prompt does not specify that the strings are lowercase letters only, so using an array is not suitable. We'll use a map to create a mapping.

Use two maps to save the mapping: from s[i] to t[j] and from t[j] to s[i]. If at any point the mapping does not match, return false.

The C++ code is as follows:

class Solution {
public:
    bool isIsomorphic(string s, string t) {
        unordered_map<char, char> map1;
        unordered_map<char, char> map2;
        for (int i = 0, j = 0; i < s.size(); i++, j++) {
            if (map1.find(s[i]) == map1.end()) { // map1 saves s[i] to t[j] mapping
                map1[s[i]] = t[j];
            }
            if (map2.find(t[j]) == map2.end()) { // map2 saves t[j] to s[i] mapping
                map2[t[j]] = s[i];
            }
            // If the mapping does not match, return false immediately
            if (map1[s[i]] != t[j] || map2[t[j]] != s[i]) {
                return false;
            }
        }
        return true;
    }
};

Other Language Versions

Java

class Solution {
    public boolean isIsomorphic(String s, String t) {
        Map<Character, Character> map1 = new HashMap<>();
        Map<Character, Character> map2 = new HashMap<>();
        for (int i = 0, j = 0; i < s.length(); i++, j++) {
            if (!map1.containsKey(s.charAt(i))) {
                map1.put(s.charAt(i), t.charAt(j)); // map1 saves the mapping from s[i] to t[j]
            }
            if (!map2.containsKey(t.charAt(j))) {
                map2.put(t.charAt(j), s.charAt(i)); // map2 saves the mapping from t[j] to s[i]
            }
            // If mapping cannot be done, return false
            if (map1.get(s.charAt(i)) != t.charAt(j) || map2.get(t.charAt(j)) != s.charAt(i)) {
                return false;
            }
        }
        return true;
    }
}

Python

class Solution:
    def isIsomorphic(self, s: str, t: str) -> bool:
        default_dict1 = defaultdict(str)
        default_dict2 = defaultdict(str)
    
        if len(s) != len(t): return false

        for i in range(len(s)):
            if not default_dict1[s[i]]:
                default_dict1[s[i]] = t[i]
            
            if not default_dict2[t[i]]:
                default_dict2[t[i]] = s[i]

            if default_dict1[s[i]] != t[i] or default_dict2[t[i]] != s[i]:
                return False
            
        return True

Go

func isIsomorphic(s string, t string) bool {
	map1 := make(map[byte]byte)
	map2 := make(map[byte]byte)
	for i := range s {
		if _, ok := map1[s[i]]; !ok {
			map1[s[i]] = t[i] // map1 saves the mapping from s[i] to t[j]
		}
		if _, ok := map2[t[i]]; !ok {
			map2[t[i]] = s[i] // map2 saves the mapping from t[i] to s[j]
		}
		// If mapping cannot be done, return false
		if (map1[s[i]] != t[i]) || (map2[t[i]] != s[i]) {
			return false
		}
	}
	return true
}

JavaScript

var isIsomorphic = function(s, t) {
    let len = s.length;
    if(len === 0) return true;
    let maps = new Map();
    let mapt = new Map();
    for(let i = 0, j = 0; i < len; i++, j++){
        if(!maps.has(s[i])){
            maps.set(s[i],t[j]);// maps saves the mapping from s[i] to t[j]
        } 
        if(!mapt.has(t[j])){
            mapt.set(t[j],s[i]);// mapt saves the mapping from t[j] to s[i]
        }
        // If mapping cannot be done, return false
        if(maps.get(s[i]) !== t[j] || mapt.get(t[j]) !== s[i]){
            return false;
        }
    };
    return true;
};

TypeScript

function isIsomorphic(s: string, t: string): boolean {
    const helperMap1: Map<string, string> = new Map();
    const helperMap2: Map<string, string> = new Map();
    for (let i = 0, length = s.length; i < length; i++) {
        let temp1: string | undefined = helperMap1.get(s[i]);
        let temp2: string | undefined = helperMap2.get(t[i]);
        if (temp1 === undefined && temp2 === undefined) {
            helperMap1.set(s[i], t[i]);
            helperMap2.set(t[i], s[i]);
        } else if (temp1 !== t[i] || temp2 !== s[i]) {
            return false;
        }
    }
    return true;
};