Problem Description

You are given two strings word1 and word2, both of the same length n. Your task is to determine if these two strings are "almost equivalent."

Two strings are considered almost equivalent if, for every letter from 'a' to 'z', the absolute difference between how many times that letter appears in word1 versus word2 is at most 3.

For example:

• If the letter 'a' appears 5 times in word1 and 2 times in word2, the difference is |5 - 2| = 3, which is acceptable.
• If the letter 'b' appears 7 times in word1 and 3 times in word2, the difference is |7 - 3| = 4, which exceeds 3, so the strings would not be almost equivalent.

The function should return true if the strings are almost equivalent, and false otherwise.

The solution uses a Counter to track the frequency differences. It first counts all characters in word1, then subtracts the counts for characters in word2. This gives us the difference in frequencies for each character. Finally, it checks if all absolute differences are at most 3.

Intuition

The key insight is that we need to track the frequency difference for each letter between the two strings. Instead of counting frequencies separately for both strings and then comparing them, we can use a single counter to directly compute the differences.

Think of it like a balance sheet: we add counts for characters from word1 (positive contributions) and subtract counts for characters from word2 (negative contributions). After processing both strings, each value in our counter represents the net difference in frequency for that character.

For instance, if 'a' appears 5 times in word1 and 2 times in word2, our counter would show cnt['a'] = 5 - 2 = 3. This direct difference calculation is more efficient than maintaining two separate frequency maps.

The beauty of this approach is that after processing both strings, we simply need to check if any character has an absolute frequency difference greater than 3. If all differences are within the range [-3, 3], the strings are almost equivalent.

Using Python's Counter class makes this particularly elegant - we can initialize it with word1, then decrement counts for each character in word2 using subtraction. The final check all(abs(x) <= 3 for x in cnt.values()) ensures every frequency difference satisfies our constraint.

Solution Approach

The solution uses a counting approach with a single pass through both strings to determine if they are almost equivalent.

Step 1: Initialize the Counter We create a Counter object initialized with word1. This automatically counts the frequency of each character in the first string. For example, if word1 = "aaab", the counter would be {'a': 3, 'b': 1}.

Step 2: Process the Second String We iterate through each character c in word2 and decrement its count in our counter:

for c in word2:
    cnt[c] -= 1

This subtraction effectively computes the frequency difference. If a character appears in word2 but not in word1, the counter will have a negative value for that character.

Step 3: Check the Differences After processing both strings, each value in cnt represents the frequency difference for that character between word1 and word2. We use:

all(abs(x) <= 3 for x in cnt.values())

This checks if every frequency difference has an absolute value of at most 3.

Example Walkthrough:

• word1 = "aaaa", word2 = "bbb"
• After initializing: cnt = {'a': 4}
• After processing word2: cnt = {'a': 4, 'b': -3}
• Check: |4| = 4 > 3, so return false

The time complexity is O(n) where n is the length of the strings, and space complexity is O(1) since we only store at most 26 letters in the counter.

Example Walkthrough

Let's walk through the solution with word1 = "abcde" and word2 = "aabcd".

Step 1: Initialize Counter with word1

• Process each character in word1 = "abcde"
• Counter becomes: {'a': 1, 'b': 1, 'c': 1, 'd': 1, 'e': 1}

Step 2: Subtract counts for word2

• Process each character in word2 = "aabcd":
  • 'a': cnt['a'] = 1 - 1 = 0
  • 'a': cnt['a'] = 0 - 1 = -1
  • 'b': cnt['b'] = 1 - 1 = 0
  • 'c': cnt['c'] = 1 - 1 = 0
  • 'd': cnt['d'] = 1 - 1 = 0
• Counter becomes: {'a': -1, 'b': 0, 'c': 0, 'd': 0, 'e': 1}

Step 3: Check all differences

• Check each value in the counter:
  • |−1| = 1 ≤ 3 ✓
  • |0| = 0 ≤ 3 ✓
  • |0| = 0 ≤ 3 ✓
  • |0| = 0 ≤ 3 ✓
  • |1| = 1 ≤ 3 ✓
• All differences are at most 3, so return true

The key insight: the counter directly tracks the net difference for each character. Positive values mean the character appears more in word1, negative values mean it appears more in word2, and the absolute value gives us the actual difference we need to check.

Solution Implementation

Time and Space Complexity

The time complexity is O(n + m) where n is the length of word1 and m is the length of word2. The Counter(word1) operation takes O(n) time to count all characters in word1. The loop iterating through word2 takes O(m) time. Finally, checking all values in the counter takes O(C) time where C is the size of the character set (at most 26 for lowercase English letters). Since C is constant and typically n and m are of similar magnitude, this can be simplified to O(n) when considering both strings have comparable lengths.

The space complexity is O(C) where C is the size of the character set. The Counter object stores at most C = 26 key-value pairs (one for each lowercase English letter that appears in either string). This is constant space with respect to the input size, as the number of distinct characters is bounded by the alphabet size regardless of the string lengths.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

Pitfall 1: Forgetting to Handle Characters Present in Only One String

A common mistake is assuming that both strings contain the same set of characters. The Counter approach handles this elegantly, but if implementing manually with a dictionary, you might forget to account for characters that appear in word2 but not in word1.

Incorrect approach:

def checkAlmostEquivalent(self, word1: str, word2: str) -> bool:
    freq1 = {}
    for c in word1:
        freq1[c] = freq1.get(c, 0) + 1
  
    # Wrong: This only checks characters that exist in word1
    for c in freq1:
        count2 = word2.count(c)
        if abs(freq1[c] - count2) > 3:
            return False
    return True

This fails when word2 has characters not in word1. For example, with word1 = "aaa" and word2 = "bbbb", it would incorrectly return True because it never checks the frequency of 'b'.

Solution: Always check all 26 lowercase letters or use the Counter subtraction approach which automatically handles missing keys.

Pitfall 2: Using Two Separate Counters Without Proper Comparison

Another mistake is creating two separate counters and not properly comparing all characters from both.

Incorrect approach:

def checkAlmostEquivalent(self, word1: str, word2: str) -> bool:
    cnt1 = Counter(word1)
    cnt2 = Counter(word2)
  
    # Wrong: Only checks keys in cnt1
    for char in cnt1:
        if abs(cnt1[char] - cnt2.get(char, 0)) > 3:
            return False
    return True

Solution: Ensure you check all unique characters from both strings:

def checkAlmostEquivalent(self, word1: str, word2: str) -> bool:
    cnt1 = Counter(word1)
    cnt2 = Counter(word2)
  
    # Check all characters from both counters
    all_chars = set(cnt1.keys()) | set(cnt2.keys())
    for char in all_chars:
        if abs(cnt1.get(char, 0) - cnt2.get(char, 0)) > 3:
            return False
    return True

The original solution avoids this pitfall by using Counter subtraction, which automatically handles all characters from both strings.