How do you solve Count Almost Equal Pairs II on LeetCode?

Count Almost Equal Pairs II (LeetCode #3267) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Understanding how to manipulate digits can lead to different representations of numbers.

What is the brute force solution for Count Almost Equal Pairs II?

The brute force approach for Count Almost Equal Pairs II is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check every pair of numbers and see if they can be made equal by performing up to two digit swaps. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Count Almost Equal Pairs II?

Count Almost Equal Pairs II uses the following concepts: Array, Hash Table, Sorting, Counting, Enumeration. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Almost Equal Pairs II?

Always start with a brute-force solution to understand the problem deeply. Think about how you can reduce the number of operations by using data structures like hash maps. Practice explaining your thought process out loud as you code.

What are common mistakes when solving Count Almost Equal Pairs II?

Not considering leading zeros when swapping digits. Failing to optimize the brute-force approach by recognizing patterns in digit occurrences.

#3267

Count Almost Equal Pairs II

Hard

ArrayHash TableSortingCountingEnumerationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(n)

💡

Intuition

Time O(n²)Space O(n)

Instead of generating all possible swaps, we can use a frequency map to count how many times each digit appears in each number. This allows us to quickly determine if two numbers can be made equal with up to two swaps.

⚙️

Algorithm

3 steps

1Step 1: Create a frequency map for each number in the array, counting occurrences of each digit.
2Step 2: Store these frequency maps in a list.
3Step 3: Compare each pair of frequency maps to see if they can be made equal with up to two swaps.

solution.py22 lines

1from collections import Counter
2
3def countAlmostEqualPairs(nums):
4    def digit_count(num):
5        return Counter(str(num))
6
7    count = 0
8    n = len(nums)
9    freq_maps = [digit_count(num) for num in nums]
10    for i in range(n):
11        for j in range(i + 1, n):
12            if can_be_equal(freq_maps[i], freq_maps[j]):
13                count += 1
14    return count
15
16
17def can_be_equal(map1, map2):
18    diff = 0
19    for digit in set(map1.keys()).union(map2.keys()):
20        diff += abs(map1[digit] - map2[digit])
21    return diff <= 4  # Up to 2 swaps means 4 differences
22

ℹ

Complexity note: The complexity remains O(n²) due to the pairwise comparison, but we reduce the overhead of generating swaps by using frequency maps, making it more efficient in practice.

1Understanding how to manipulate digits can lead to different representations of numbers.
2Using frequency maps allows for efficient comparisons without generating all possible swaps.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.