Problem statement

Two strings are isomorphic if the characters in s can be replaced to get t, preserving order, with every occurrence of a character mapping to the same character and no two characters mapping to the same character.

Given s and t, return true if they are isomorphic.

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

Constraints

• 1 <= s.length <= 30
• t.length == s.length
• Lowercase English letters.

Hints for Isomorphic Strings

1. Hint 1

   Keep a mapping in BOTH directions — one map alone misses collisions.

2. Hint 2

   Check consistency at every position before recording the pair.

How to solve Isomorphic Strings

Isomorphism is a bijection between characters, so one mapping table is not enough — you need both directions. s[i] -> t[i] must be consistent, and so must t[i] -> s[i]; without the second check, "ab" and "aa" would wrongly pass.

Approach

1. Keep two tables: forward, from characters of s to characters of t, and backward, the reverse.
2. For each position, look up both mappings.
3. If either has been set before and disagrees with the current pair, return false.
4. Otherwise record both directions and continue.
5. Surviving the scan means the strings are isomorphic.

Why it works

A valid replacement scheme must be a function (each character of s maps to exactly one character of t) and injective (no two characters of s collapse onto the same character of t). The forward table enforces the first property and the backward table the second — together they make the mapping a bijection on the characters actually used, which is precisely the definition.

Complexity

• Time — O(n)
• Space — O(1)

Pitfalls

• One map alone is the classic bug: it accepts "ab" → "aa", which collapses two characters into one.
• The strings are guaranteed to be the same length here, but a length check is the natural first guard in general.
• Use a sentinel that cannot be a real character (such as 0) to mean "unmapped".
• A character legitimately maps to itself; that is not a special case.

Reference solution

Python

def isIsomorphic(s: str, t: str) -> bool:
    if len(s) != len(t):
        return False
    m1 = {}
    m2 = {}
    for a, b in zip(s, t):
        if a in m1 and m1[a] != b:
            return False
        if b in m2 and m2[b] != a:
            return False
        m1[a] = b
        m2[b] = a
    return True

Java

public static boolean isIsomorphic(String s, String t) {
    int[] forward = new int[128];
    int[] backward = new int[128];
    Arrays.fill(forward, -1);
    Arrays.fill(backward, -1);
    for (int i = 0; i < s.length(); i++) {
        int a = s.charAt(i), b = t.charAt(i);
        if (forward[a] == -1 && backward[b] == -1) {
            forward[a] = b;
            backward[b] = a;
        } else if (forward[a] != b || backward[b] != a) {
            return false;
        }
    }
    return true;
}

C++

bool isIsomorphic(string s, string t) {
    vector<int> forward(128, -1), backward(128, -1);
    for (int i = 0; i < (int) s.size(); i++) {
        int a = (unsigned char) s[i], b = (unsigned char) t[i];
        if (forward[a] == -1 && backward[b] == -1) {
            forward[a] = b;
            backward[b] = a;
        } else if (forward[a] != b || backward[b] != a) {
            return false;
        }
    }
    return true;
}

JavaScript

var isIsomorphic = function(s, t) {
    if (s.length !== t.length) return false;
    const m1 = new Map(), m2 = new Map();
    for (let i = 0; i < s.length; i++) {
        const a = s[i], b = t[i];
        if (m1.has(a) && m1.get(a) !== b) return false;
        if (m2.has(b) && m2.get(b) !== a) return false;
        m1.set(a, b);
        m2.set(b, a);
    }
    return true;
};

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