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

Count Almost Equal Pairs I (LeetCode #3265) can be solved using Brute Force. The optimal approach is Brute Force with undefined time complexity and undefined space complexity. Key insight: We will check every possible pair of numbers in the array to see if they can be made equal by swapping two digits in one of the numbers. This is straightforward but can be slow for larger arrays.

What is the time complexity of Count Almost Equal Pairs I?

The optimal solution for Count Almost Equal Pairs I runs in undefined time and uses undefined space. undefined

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

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

#3265

Count Almost Equal Pairs I

Medium

ArrayHash TableSortingCountingEnumeration

LeetCode ↗

Approaches

💡

Intuition

Time Space

We will check every possible pair of numbers in the array to see if they can be made equal by swapping two digits in one of the numbers. This is straightforward but can be slow for larger arrays.

⚙️

Algorithm

5 steps

1Step 1: Initialize a counter to 0 for counting almost equal pairs.
2Step 2: Iterate through each pair of numbers (i, j) where i < j.
3Step 3: For each pair, check if swapping any two digits in either number can make them equal.
4Step 4: If they can be made equal, increment the counter.
5Step 5: Return the counter after checking all pairs.

solution.py23 lines

1# Full working Python code
2
3def can_be_equal(x, y):
4    x_str, y_str = str(x), str(y)
5    for i in range(len(x_str)):
6        for j in range(i + 1, len(x_str)):
7            swapped_x = list(x_str)
8            swapped_x[i], swapped_x[j] = swapped_x[j], swapped_x[i]
9            if ''.join(swapped_x) == y_str:
10                return True
11    return False
12
13def count_almost_equal_pairs(nums):
14    count = 0
15    n = len(nums)
16    for i in range(n):
17        for j in range(i + 1, n):
18            if can_be_equal(nums[i], nums[j]) or can_be_equal(nums[j], nums[i]):
19                count += 1
20    return count
21
22# Example usage:
23print(count_almost_equal_pairs([3, 12, 30, 17, 21]))  # Output: 2

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