LeetCode #2068 — EASY

Check Whether Two Strings are Almost Equivalent

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

The Problem

Problem Statement

Two strings word1 and word2 are considered almost equivalent if the differences between the frequencies of each letter from 'a' to 'z' between word1 and word2 is at most 3.

Given two strings word1 and word2, each of length n, return true if word1 and word2 are almost equivalent, or false otherwise.

The frequency of a letter x is the number of times it occurs in the string.

Example 1:

Input: word1 = "aaaa", word2 = "bccb"
Output: false
Explanation: There are 4 'a's in "aaaa" but 0 'a's in "bccb".
The difference is 4, which is more than the allowed 3.

Example 2:

Input: word1 = "abcdeef", word2 = "abaaacc"
Output: true
Explanation: The differences between the frequencies of each letter in word1 and word2 are at most 3:
- 'a' appears 1 time in word1 and 4 times in word2. The difference is 3.
- 'b' appears 1 time in word1 and 1 time in word2. The difference is 0.
- 'c' appears 1 time in word1 and 2 times in word2. The difference is 1.
- 'd' appears 1 time in word1 and 0 times in word2. The difference is 1.
- 'e' appears 2 times in word1 and 0 times in word2. The difference is 2.
- 'f' appears 1 time in word1 and 0 times in word2. The difference is 1.

Example 3:

Input: word1 = "cccddabba", word2 = "babababab"
Output: true
Explanation: The differences between the frequencies of each letter in word1 and word2 are at most 3:
- 'a' appears 2 times in word1 and 4 times in word2. The difference is 2.
- 'b' appears 2 times in word1 and 5 times in word2. The difference is 3.
- 'c' appears 3 times in word1 and 0 times in word2. The difference is 3.
- 'd' appears 2 times in word1 and 0 times in word2. The difference is 2.

Constraints:

n == word1.length == word2.length
1 <= n <= 100
word1 and word2 consist only of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: Two strings word1 and word2 are considered almost equivalent if the differences between the frequencies of each letter from 'a' to 'z' between word1 and word2 is at most 3. Given two strings word1 and word2, each of length n, return true if word1 and word2 are almost equivalent, or false otherwise. The frequency of a letter x is the number of times it occurs in the string.

Baseline thinking

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

Pattern signal: Hash Map

Example 1

"aaaa"
"bccb"

Example 2

"abcdeef"
"abaaacc"

Example 3

"cccddabba"
"babababab"

Core Insight

What unlocks the optimal approach

What data structure can we use to count the frequency of each character?
Are there edge cases where a character is present in one string but not the other?

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 #2068: Check Whether Two Strings are Almost Equivalent
class Solution {
    public boolean checkAlmostEquivalent(String word1, String word2) {
        int[] cnt = new int[26];
        for (int i = 0; i < word1.length(); ++i) {
            ++cnt[word1.charAt(i) - 'a'];
        }
        for (int i = 0; i < word2.length(); ++i) {
            --cnt[word2.charAt(i) - 'a'];
        }
        for (int x : cnt) {
            if (Math.abs(x) > 3) {
                return false;
            }
        }
        return true;
    }
}

// Accepted solution for LeetCode #2068: Check Whether Two Strings are Almost Equivalent
func checkAlmostEquivalent(word1 string, word2 string) bool {
	cnt := [26]int{}
	for _, c := range word1 {
		cnt[c-'a']++
	}
	for _, c := range word2 {
		cnt[c-'a']--
	}
	for _, x := range cnt {
		if x > 3 || x < -3 {
			return false
		}
	}
	return true
}

# Accepted solution for LeetCode #2068: Check Whether Two Strings are Almost Equivalent
class Solution:
    def checkAlmostEquivalent(self, word1: str, word2: str) -> bool:
        cnt = Counter(word1)
        for c in word2:
            cnt[c] -= 1
        return all(abs(x) <= 3 for x in cnt.values())

// Accepted solution for LeetCode #2068: Check Whether Two Strings are Almost Equivalent
struct Solution;

use std::collections::HashMap;

impl Solution {
    fn check_almost_equivalent(word1: String, word2: String) -> bool {
        let mut hm1: HashMap<char, i32> = HashMap::new();
        let mut hm2: HashMap<char, i32> = HashMap::new();
        for c in word1.chars() {
            *hm1.entry(c).or_default() += 1;
        }
        for c in word2.chars() {
            *hm2.entry(c).or_default() += 1;
        }
        for c in 'a'..='z' {
            let x1 = *hm1.entry(c).or_default();
            let x2 = *hm2.entry(c).or_default();
            if (x1 - x2).abs() > 3 {
                return false;
            }
        }
        true
    }
}

#[test]
fn test() {
    let word1 = "aaaa".to_string();
    let word2 = "bccb".to_string();
    let res = false;
    assert_eq!(Solution::check_almost_equivalent(word1, word2), res);
    let word1 = "abcdeef".to_string();
    let word2 = "abaaacc".to_string();
    let res = true;
    assert_eq!(Solution::check_almost_equivalent(word1, word2), res);
    let word1 = "cccddabba".to_string();
    let word2 = "babababab".to_string();
    let res = true;
    assert_eq!(Solution::check_almost_equivalent(word1, word2), res);
}

// Accepted solution for LeetCode #2068: Check Whether Two Strings are Almost Equivalent
function checkAlmostEquivalent(word1: string, word2: string): boolean {
    const cnt: number[] = new Array(26).fill(0);
    for (const c of word1) {
        ++cnt[c.charCodeAt(0) - 97];
    }
    for (const c of word2) {
        --cnt[c.charCodeAt(0) - 97];
    }
    return cnt.every(x => Math.abs(x) <= 3);
}

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.