Check Whether Two Strings are Almost Equivalent

Check Whether Two Strings are Almost Equivalent - Problem

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.

Input & Output

Example 1 — Basic Almost Equivalent

$ Input: word1 = "aabc", word2 = "aabb"

› Output: true

💡 Note: Character frequencies: 'a' appears 2 times in both (diff=0), 'b' appears 1 time in word1 and 2 times in word2 (diff=1), 'c' appears 1 time in word1 and 0 times in word2 (diff=1). All differences ≤ 3.

Example 2 — Not Almost Equivalent

$ Input: word1 = "abc", word2 = "bcd"

› Output: true

💡 Note: Character 'a' appears 1 time in word1 and 0 times in word2 (diff=1), 'b' appears 1 time in both (diff=0), 'c' appears 1 time in both (diff=0), 'd' appears 0 times in word1 and 1 time in word2 (diff=1). All differences ≤ 3, so the strings are almost equivalent.

Example 3 — Large Difference

$ Input: word1 = "aaaa", word2 = "bbbb"

› Output: false

💡 Note: Character 'a' appears 4 times in word1 and 0 times in word2 (diff=4 > 3), 'b' appears 0 times in word1 and 4 times in word2 (diff=4 > 3). Since differences exceed 3, strings are not almost equivalent.

Constraints

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

Visualization

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Asked in

M Meta 15 G Google 12 a Amazon 8

The key insight is to track frequency differences between characters in both strings. The optimal approach uses a single difference array where we increment for word1 characters and decrement for word2 characters, then check if any absolute difference exceeds 3. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force - Count Each Character

⏱️ Time: O(n) Space: O(1)

For each character from 'a' to 'z', iterate through both strings to count occurrences, then compare the difference. This approach checks all 26 characters individually.

Frequency Count with Arrays

⏱️ Time: O(n) Space: O(1)

Create two frequency arrays of size 26 to count occurrences of each character in both strings. Then iterate through the arrays to check if any difference exceeds 3.

Single Pass Difference Tracking

⏱️ Time: O(n) Space: O(1)

Instead of building complete frequency arrays, maintain only the differences between character counts. Process both strings simultaneously to update difference counts and check the constraint in real-time.

Brute Force - Count Each Character — Algorithm Steps

Step 1: For each character 'a' to 'z'
Step 2: Count occurrences in word1 and word2
Step 3: Check if absolute difference ≤ 3

Visualization

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Step-by-Step Walkthrough

For each char

Iterate through 'a' to 'z'

Count in both

Count occurrences in word1 and word2

Check diff

Verify difference ≤ 3

Code -

solution.c — C

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int solution(char* word1, char* word2) {
    // Check each character from 'a' to 'z'
    for (int i = 0; i < 26; i++) {
        char ch = 'a' + i;
        int count1 = 0, count2 = 0;
        
        for (int j = 0; word1[j]; j++) {
            if (word1[j] == ch) count1++;
        }
        
        for (int j = 0; word2[j]; j++) {
            if (word2[j] == ch) count2++;
        }
        
        if (abs(count1 - count2) > 3) {
            return 0;
        }
    }
    return 1;
}

int main() {
    char word1[1000], word2[1000];
    fgets(word1, sizeof(word1), stdin);
    fgets(word2, sizeof(word2), stdin);
    
    // Remove newlines
    word1[strcspn(word1, "\n")] = 0;
    word2[strcspn(word2, "\n")] = 0;
    
    int result = solution(word1, word2);
    printf(result ? "true\n" : "false\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

26 iterations × O(n) counting = O(26n) = O(n)

✓ Linear Growth

Space Complexity

O(1)

Only using constant variables, no data structures

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Count Each Character — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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