LeetCode #205 — EASY

Isomorphic Strings

Build confidence with an intuition-first walkthrough focused on hash map fundamentals.

The Problem

Problem Statement

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

Two strings s and t 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

Explanation:

The strings s and t can be made identical by:

Mapping 'e' to 'a'.
Mapping 'g' to 'd'.

Example 2:

Input: s = "f11", t = "b23"

Output: false

Explanation:

The strings s and t can not be made identical as '1' needs to be mapped to both '2' and '3'.

Example 3:

Input: s = "paper", t = "title"

Output: true

Constraints:

1 <= s.length <= 5 * 10⁴
t.length == s.length
s and t consist of any valid ascii character.

Step 01

Brute Force Baseline

Problem summary: Given two strings s and t, determine if they are isomorphic. Two strings s and t 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map

Example 1

"egg"
"add"

Example 2

"foo"
"bar"

Example 3

"paper"
"title"

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #205: Isomorphic Strings
class Solution {
    public boolean isIsomorphic(String s, String t) {
        Map<Character, Character> d1 = new HashMap<>();
        Map<Character, Character> d2 = new HashMap<>();
        int n = s.length();
        for (int i = 0; i < n; ++i) {
            char a = s.charAt(i), b = t.charAt(i);
            if (d1.containsKey(a) && d1.get(a) != b) {
                return false;
            }
            if (d2.containsKey(b) && d2.get(b) != a) {
                return false;
            }
            d1.put(a, b);
            d2.put(b, a);
        }
        return true;
    }
}

// Accepted solution for LeetCode #205: Isomorphic Strings
func isIsomorphic(s string, t string) bool {
	d1 := [256]int{}
	d2 := [256]int{}
	for i := range s {
		if d1[s[i]] != d2[t[i]] {
			return false
		}
		d1[s[i]] = i + 1
		d2[t[i]] = i + 1
	}
	return true
}

# Accepted solution for LeetCode #205: Isomorphic Strings
class Solution:
    def isIsomorphic(self, s: str, t: str) -> bool:
        d1 = {}
        d2 = {}
        for a, b in zip(s, t):
            if (a in d1 and d1[a] != b) or (b in d2 and d2[b] != a):
                return False
            d1[a] = b
            d2[b] = a
        return True

// Accepted solution for LeetCode #205: Isomorphic Strings
use std::collections::HashMap;
impl Solution {
    fn help(s: &[u8], t: &[u8]) -> bool {
        let mut map = HashMap::new();
        for i in 0..s.len() {
            if map.contains_key(&s[i]) {
                if map.get(&s[i]).unwrap() != &t[i] {
                    return false;
                }
            } else {
                map.insert(s[i], t[i]);
            }
        }
        true
    }

    pub fn is_isomorphic(s: String, t: String) -> bool {
        let (s, t) = (s.as_bytes(), t.as_bytes());
        Self::help(s, t) && Self::help(t, s)
    }
}

// Accepted solution for LeetCode #205: Isomorphic Strings
function isIsomorphic(s: string, t: string): boolean {
    const d1: number[] = new Array(256).fill(0);
    const d2: number[] = new Array(256).fill(0);
    for (let i = 0; i < s.length; ++i) {
        const a = s.charCodeAt(i);
        const b = t.charCodeAt(i);
        if (d1[a] !== d2[b]) {
            return false;
        }
        d1[a] = i + 1;
        d2[b] = i + 1;
    }
    return true;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(C)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan the rest of the array looking for a match. Two nested loops give n × (n−1)/2 comparisons = O(n²). No extra space since we only use loop indices.

HASH MAP

O(n) time

O(n) space

One pass through the input, performing O(1) hash map lookups and insertions at each step. The hash map may store up to n entries in the worst case. This is the classic space-for-time tradeoff: O(n) extra memory eliminates an inner loop.

Shortcut: Need to check “have I seen X before?” → hash map → O(n) time, O(n) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.