Count Almost Equal Pairs I

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ArraysStringsTwo Pointers

You are given an array nums consisting of positive integers.

We call two integers x and y in this problem almost equal if both integers can become equal after performing the following operation at most once:

Choose either x or y and swap any two digits within the chosen number.

Return the number of indices i and j in nums where i < j such that nums[i] and nums[j] are almost equal.

Note that it is allowed for an integer to have leading zeros after performing an operation.

Example 1:

Input: nums = [3,12,30,17,21]

Output: 2

Explanation:

The almost equal pairs of elements are:

3 and 30. By swapping 3 and 0 in 30, you get 3.
12 and 21. By swapping 1 and 2 in 12, you get 21.

Example 2:

Input: nums = [1,1,1,1,1]

Output: 10

Explanation:

Every two elements in the array are almost equal.

Example 3:

Input: nums = [123,231]

Output: 0

Explanation:

We cannot swap any two digits of 123 or 231 to reach the other.

Constraints:

2 <= nums.length <= 100
1 <= nums[i] <= 10⁶

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum possible size of the input array?
Can the input array contain negative numbers, zero, or floating-point numbers?
What exactly is meant by 'almost equal'? Is it defined as |nums[i] - nums[j]| <= difference for a given difference value?
Are duplicate numbers allowed in the input array, and if so, should pairs like (nums[i], nums[i]) with i != j be considered?
If no almost equal pairs exist, what value should be returned?

Brute Force Solution

Approach

The brute force approach to counting almost equal pairs involves comparing every number in the list to every other number. We'll check each pair to see if they are almost equal, according to the problem's definition.

Here's how the algorithm would work step-by-step:

Take the first number in the list.
Compare that first number to every other number in the list, one at a time.
For each comparison, check if the two numbers are considered 'almost equal' according to the rules.
If they are almost equal, count that pair.
After comparing the first number to all other numbers, move to the second number in the list.
Repeat the process of comparing this second number to all the remaining numbers in the list.
Continue this process for each number in the list, making sure you don't double-count pairs (if number A has already been compared to number B, you don't need to compare number B to number A again).
At the end, you'll have a count of all the 'almost equal' pairs in the list.

Code Implementation

def count_almost_equal_pairs(numbers):
    number_of_almost_equal_pairs = 0

    # Iterate through each number in the list
    for first_index in range(len(numbers)): 
        # Avoid redundant comparisons
        for second_index in range(first_index + 1, len(numbers)): 
            # Checking for the 'almost equal' condition
            if abs(numbers[first_index] - numbers[second_index]) <= 1:
                number_of_almost_equal_pairs += 1
    
    return number_of_almost_equal_pairs

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through the input array of size n. For each element, it compares it with the remaining elements to check for almost equal pairs. In the worst case, the outer loop iterates n times, and the inner comparison iterates approximately n/2 times on average. Therefore, the number of operations scales roughly as n * (n/2), which simplifies to O(n²).

Space Complexity

O(1) – The brute force approach, as described, iterates through the input list using nested loops and compares elements directly. It does not create any auxiliary data structures like arrays, hash maps, or other collections to store intermediate results or track visited elements. The algorithm only uses a constant number of variables for loop indices and potentially a counter for the almost equal pairs. Therefore, the auxiliary space complexity is O(1), indicating constant space usage regardless of the input size N.

Optimal Solution

Approach

To efficiently find almost equal pairs, we avoid checking every possible pair. Instead, we sort the data and then count pairs that are close together in the sorted order, because almost equal values will be near each other.

Here's how the algorithm would work step-by-step:

First, organize the list of numbers from smallest to largest. This puts similar numbers next to each other.
Then, go through the sorted list. For each number, look at the numbers that come after it.
If a number is very close to the first number (the difference is at most 2), count it as a pair.
Repeat this process for each number in the sorted list. Don't recount the same pair.
The total number of pairs you count is the answer.

Code Implementation

def count_almost_equal_pairs(numbers):
    numbers.sort()
    pair_count = 0

    for i in range(len(numbers)):
        # Iterate through the sorted list.

        for j in range(i + 1, len(numbers)):
            # Iterate through the rest of the list to find pairs.

            if abs(numbers[i] - numbers[j]) <= 2:
                # Count pairs if the absolute difference is within the limit.

                pair_count += 1

    return pair_count

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operation in this approach is initially sorting the input array of size n, which takes O(n log n) time. After sorting, we iterate through each of the n elements and compare each element to the elements that come after it. In the worst-case scenario, for each element, we might compare it to all other elements after it, which is at most n-1 comparisons. However, since the sorting step dominates the runtime and the pair checking is only done with subsequent elements in the array, this contributes an additional O(n) to the algorithm. Therefore, the overall time complexity is dominated by the initial sorting, resulting in O(n log n).

Space Complexity

O(1) – The algorithm first sorts the input array. If the sorting is done in-place (like heapsort), it requires no extra space beyond a few temporary variables for swapping elements. After sorting, the algorithm iterates through the sorted array, comparing each element with subsequent elements. This process only requires a fixed number of variables to keep track of indices and the count of almost equal pairs. Thus, the auxiliary space used is constant and independent of the input size N.

Edge Cases

Case	How to Handle
Empty input array	Return 0, as there are no pairs to count.
Array with one element	Return 0, as a pair requires at least two elements.
Array with all identical elements	The solution should correctly count all pairs whose absolute difference is less than or equal to the threshold.
Large array with integer overflow potential in the count.	Use a data type that can accommodate large counts (e.g., long) to prevent overflow.
Array with large numbers that, when differenced, could cause overflow.	Check for possible overflow when calculating the absolute difference between two numbers.
Array with negative numbers	The absolute difference calculation should handle negative numbers correctly.
k = 0 (threshold is zero)	The solution should count pairs with identical values.
Very large k value (threshold)	The solution's efficiency should not significantly degrade with very large k values.

Empty input array

How to Handle:

Return 0, as there are no pairs to count.

Array with one element

How to Handle:

Return 0, as a pair requires at least two elements.

Array with all identical elements

How to Handle:

The solution should correctly count all pairs whose absolute difference is less than or equal to the threshold.

Large array with integer overflow potential in the count.

How to Handle:

Use a data type that can accommodate large counts (e.g., long) to prevent overflow.

Array with large numbers that, when differenced, could cause overflow.

How to Handle:

Check for possible overflow when calculating the absolute difference between two numbers.

Array with negative numbers

How to Handle:

The absolute difference calculation should handle negative numbers correctly.

k = 0 (threshold is zero)

How to Handle:

The solution should count pairs with identical values.

Very large k value (threshold)

How to Handle:

The solution's efficiency should not significantly degrade with very large k values.