Isomorphic Strings

Published at16 Mar 2026

String Easy 12 views

Like0

Two strings are isomorphic when the character pattern in one string can be matched to the character pattern in the other string with a consistent one-to-one mapping.

That means if a character in the first string appears again later, it must map to the same character again in the second string. It also means two different characters from the first string are not allowed to collapse into the same character in the second string.

For example, "egg" and "add" are isomorphic because e -> a and g -> d stay consistent across the whole string. But "foo" and "bar" are not isomorphic because the repeated o would need to map inconsistently.

So the task is to check whether one stable one-to-one character mapping can explain every position in both strings.

Example Input & Output

Example 1

Input

s = "egg", t = "add"

Output

true

Explanation

e->a and g->d is consistent.

Example 2

Input

s = "foo", t = "bar"

Output

false

Explanation

o cannot map to both a and r.

Example 3

Input

s = "paper", t = "title"

Output

true

Explanation

A consistent one-to-one mapping exists.

Algorithm Flow

Recommendation Algorithm Flow for Isomorphic Strings

Solution Approach

Check if two strings s and t are isomorphic, meaning characters in s can be replaced to get t while preserving the character mapping. Each character must map to exactly one other character, and no two characters can map to the same character. Use two hash maps tracking s-to-t and t-to-s mappings.

function isIsomorphic(s, t) {
  var sMap = {}, tMap = {};
  for (var i = 0; i < s.length; i++) {
    if (!sMap[s[i]] && !tMap[t[i]]) { sMap[s[i]] = t[i]; tMap[t[i]] = s[i]; }
    else if (sMap[s[i]] !== t[i] || tMap[t[i]] !== s[i]) return false;
  }
  return true;
}

Both direction maps ensure the mapping is a bijection. When a new character pair is encountered, both maps are updated. Any conflict with an existing mapping returns false.

Time complexity is O(n), space complexity is O(k).

Best Answers
These are the verified best solutions for this Challenge. You can use them as a reference to improve your solution.

java

import java.util.*;
class Solution {
    public boolean isomorphic_strings(String s, String t) {
        if (s.length() != t.length()) return false;
        Map<Character, Character> st = new HashMap<>();
        Map<Character, Character> ts = new HashMap<>();
        for (int i = 0; i < s.length(); i++) {
            char a = s.charAt(i), b = t.charAt(i);
            if (st.containsKey(a) && st.get(a) != b) return false;
            if (ts.containsKey(b) && ts.get(b) != a) return false;
            st.put(a, b);
            ts.put(b, a);
        }
        return true;
    }
}

javascript

function isomorphic_strings(s, t) {
    if (s.length !== t.length) return false;
    const st = {}, ts = {};
    for (let i = 0; i < s.length; i++) {
        const a = s[i], b = t[i];
        if (st[a] !== undefined && st[a] !== b) return false;
        if (ts[b] !== undefined && ts[b] !== a) return false;
        st[a] = b;
        ts[b] = a;
    }
    return true;
}

php

<?php
function isomorphic_strings($s, $t) {
    if (strlen($s) !== strlen($t)) return false;
    $st = []; $ts = [];
    for ($i = 0; $i < strlen($s); $i++) {
        $a = $s[$i]; $b = $t[$i];
        if (isset($st[$a]) && $st[$a] !== $b) return false;
        if (isset($ts[$b]) && $ts[$b] !== $a) return false;
        $st[$a] = $b;
        $ts[$b] = $a;
    }
    return true;
}
?>

python

def isomorphic_strings(s: str, t: str) -> bool:
    if len(s) != len(t):
        return False
    st, ts = {}, {}
    for a, b in zip(s, t):
        if a in st and st[a] != b:
            return False
        if b in ts and ts[b] != a:
            return False
        st[a] = b
        ts[b] = a
    return True

rust

use std::collections::HashMap;
fn isomorphic_strings(s: String, t: String) -> bool {
    if s.len() != t.len() { return false; }
    let mut st: HashMap<char, char> = HashMap::new();
    let mut ts: HashMap<char, char> = HashMap::new();
    let sa: Vec<char> = s.chars().collect();
    let ta: Vec<char> = t.chars().collect();
    for i in 0..sa.len() {
        let a = sa[i];
        let b = ta[i];
        if let Some(v) = st.get(&a) { if *v != b { return false; } }
        if let Some(v) = ts.get(&b) { if *v != a { return false; } }
        st.insert(a, b);
        ts.insert(b, a);
    }
    true
}

typescript

function isomorphic_strings(s: string, t: string): boolean {
    if (s.length !== t.length) return false;
    var st: { [key: string]: string } = {};
    var ts: { [key: string]: string } = {};
    for (var i = 0; i < s.length; i++) {
        var a = s.charAt(i);
        var b = t.charAt(i);
        if (st[a] !== undefined && st[a] !== b) return false;
        if (ts[b] !== undefined && ts[b] !== a) return false;
        st[a] = b;
        ts[b] = a;
    }
    return true;
}

Related String Challenges

Comments (0)

Join the Discussion

Share your thoughts, ask questions, or help others with this Challenge.

Isomorphic Strings

Example Input & Output

Algorithm Flow

Solution Approach

Best AnswersThese are the verified best solutions for this Challenge. You can use them as a reference to improve your solution.

Solutions Coming Soon

Related String Challenges

Longest Substring Without Repeating Characters

Maximum Vowels In Substring Length K

Length of Last Word

Longest Word Length

Maximum Repeating Substring

Zigzag Conversion

Comments (0)

Join the Discussion

Best Answers
These are the verified best solutions for this Challenge. You can use them as a reference to improve your solution.