Number of Pairs Satisfying Inequality

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You are given two 0-indexed integer arrays nums1 and nums2, each of size n, and an integer diff. Find the number of pairs (i, j) such that:

0 <= i < j <= n - 1 and
nums1[i] - nums1[j] <= nums2[i] - nums2[j] + diff.

Return the number of pairs that satisfy the conditions.

Example 1:

Input: nums1 = [3,2,5], nums2 = [2,2,1], diff = 1
Output: 3
Explanation:
There are 3 pairs that satisfy the conditions:
1. i = 0, j = 1: 3 - 2 <= 2 - 2 + 1. Since i < j and 1 <= 1, this pair satisfies the conditions.
2. i = 0, j = 2: 3 - 5 <= 2 - 1 + 1. Since i < j and -2 <= 2, this pair satisfies the conditions.
3. i = 1, j = 2: 2 - 5 <= 2 - 1 + 1. Since i < j and -3 <= 2, this pair satisfies the conditions.
Therefore, we return 3.

Example 2:

Input: nums1 = [3,-1], nums2 = [-2,2], diff = -1
Output: 0
Explanation:
Since there does not exist any pair that satisfies the conditions, we return 0.

Constraints:

n == nums1.length == nums2.length
2 <= n <= 10⁵
-10⁴ <= nums1[i], nums2[i] <= 10⁴
-10⁴ <= diff <= 10⁴

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 are the possible ranges and data types for the numbers in the input arrays?
Can either of the input arrays be null or empty?
What is the specific inequality that the pairs must satisfy (e.g., is it strictly greater than, greater than or equal to)?
If no pairs satisfy the inequality, what should I return?
Are there any constraints on the sizes of the two input arrays, and are they guaranteed to be of the same length?

Brute Force Solution

Approach

The most basic way to solve this problem is to simply check every possible pairing. We go through all the possible pairs and see if they fit the criteria. It's like manually matching items one by one.

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

Take the first number from the first list.
Compare it to every single number in the second list to see if they satisfy the condition.
Count how many numbers from the second list meet the criteria when paired with the first number.
Move to the second number in the first list.
Again, compare it to every number in the second list and count the pairs that meet the condition.
Repeat this process for every number in the first list.
Add up all the counts from each number in the first list to get the total number of pairs that satisfy the given condition.

Code Implementation

def count_number_of_pairs_satisfying_inequality_brute_force(list_one, list_two, difference):
    count_of_valid_pairs = 0

    for first_list_index in range(len(list_one)):

        # Iterate through each number in the first list
        first_list_number = list_one[first_list_index]

        for second_list_index in range(len(list_two)):

            second_list_number = list_two[second_list_index]

            # Check if the condition is satisfied
            if first_list_number - second_list_number > difference:
                # Increment our counter if the difference is greater than the threshold
                count_of_valid_pairs += 1

    return count_of_valid_pairs

Big(O) Analysis

Time Complexity

O(n²) – The provided solution iterates through each element of the first list of size n. For each of these elements, it iterates through the second list, also of size n, to check if the pair satisfies the given condition. This results in nested loops where the inner loop executes n times for each of the n iterations of the outer loop. Therefore, the total number of operations is proportional to n * n, simplifying to O(n²).

Space Complexity

O(1) – The provided solution iterates through the input lists `nums1` and `nums2` and performs comparisons without using any auxiliary data structures. It uses a constant number of variables to store indexes and a count. Therefore, the space required does not depend on the size of the input lists, and the space complexity is O(1).

Optimal Solution

Approach

The most efficient solution avoids checking all possible pairs directly. It uses a technique to quickly count valid pairs by focusing on ordering and merging, instead of individual comparisons. This dramatically reduces the amount of work needed to find the answer.

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

Rearrange the input data to make it easier to work with. Focus on the differences between the two sets of numbers.
Sort the rearranged data so that similar numbers are next to each other. This allows for efficient counting.
Starting from the beginning, go through the sorted data. For each number, efficiently determine how many numbers later in the data meet the required condition.
Instead of checking each number one by one, use a process that involves combining and splitting sorted sections to quickly find the count. This clever merging process avoids redundant comparisons.
Add up the counts found for each number to get the total number of valid pairs.

Code Implementation

def count_number_of_pairs(nums1, nums2, difference):
    modified_array = [nums1[i] - nums2[i] for i in range(len(nums1))]

    modified_array.sort()

    total_valid_pairs = 0

    def merge_and_count(left, right):
        nonlocal total_valid_pairs
        if left >= right:
            return [modified_array[left]]

        middle = (left + right) // 2
        left_half = merge_and_count(left, middle)
        right_half = merge_and_count(middle + 1, right)

        # Count valid pairs during merge process
        i = 0
        j = 0
        while i < len(left_half):
            while j < len(right_half) and left_half[i] > right_half[j] + difference:
                j += 1
            total_valid_pairs += len(right_half) - j
            i += 1

        # Standard merge operation
        merged = []
        i = 0
        j = 0
        while i < len(left_half) and j < len(right_half):
            if left_half[i] <= right_half[j]:
                merged.append(left_half[i])
                i += 1
            else:
                merged.append(right_half[j])
                j += 1

        merged.extend(left_half[i:])
        merged.extend(right_half[j:])
        return merged

    #The recursive function sorts and merges the modified array.
    merge_and_count(0, len(modified_array) - 1)

    #Return the total count of valid pairs satisfying the condition.
    return total_valid_pairs

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operations in this approach are sorting and a merge-like process. Sorting the rearranged data takes O(n log n) time, where n is the size of the input arrays. The subsequent merging and counting process, which avoids pairwise comparisons, also contributes O(n log n) time as it efficiently combines and splits sorted sections. Therefore, the overall time complexity is O(n log n) + O(n log n), which simplifies to O(n log n).

Space Complexity

O(N) – The algorithm rearranges the input data and sorts it, which typically requires an auxiliary array of size N, where N is the number of elements in the combined and modified input. The merging process described likely involves creating temporary lists to hold sorted sections. Therefore, the space complexity is directly proportional to the input size N because sorting and merging typically use O(N) auxiliary space. Combining these aspects, the dominant factor is the auxiliary space used for sorting and merging.

Edge Cases

Case	How to Handle
Empty nums1 or nums2 array	Return 0 immediately since no pairs can be formed.
nums1 and nums2 have significantly different lengths	Algorithm's efficiency should not be overly affected by length differences.
All elements in nums1 and nums2 are identical	The algorithm must handle a large number of equal pairs efficiently (e.g., avoid quadratic comparisons).
nums1[i] - nums2[j] equals a very large positive or negative number	Consider potential integer overflow when subtracting elements and comparing with 'diff'.
diff is a very large positive or negative number close to integer limits	The code needs to ensure comparisons against diff don't cause unexpected behavior due to overflow.
Arrays contain large numbers of duplicate values that satisfy the condition	The algorithm needs to count pairs with the same value correctly, potentially using frequency counting techniques.
No pairs satisfy the condition nums1[i] - nums2[j] <= diff	The algorithm should correctly return 0 when no valid pairs are found.
Extremely large arrays that could cause memory issues	Consider the space complexity and if necessary, explore alternative approaches or use generators to process data in chunks.

Empty nums1 or nums2 array

How to Handle:

Return 0 immediately since no pairs can be formed.

nums1 and nums2 have significantly different lengths

How to Handle:

Algorithm's efficiency should not be overly affected by length differences.

All elements in nums1 and nums2 are identical

How to Handle:

The algorithm must handle a large number of equal pairs efficiently (e.g., avoid quadratic comparisons).

nums1[i] - nums2[j] equals a very large positive or negative number

How to Handle:

Consider potential integer overflow when subtracting elements and comparing with 'diff'.

diff is a very large positive or negative number close to integer limits

How to Handle:

The code needs to ensure comparisons against diff don't cause unexpected behavior due to overflow.

Arrays contain large numbers of duplicate values that satisfy the condition

How to Handle:

The algorithm needs to count pairs with the same value correctly, potentially using frequency counting techniques.

No pairs satisfy the condition nums1[i] - nums2[j] <= diff

How to Handle:

The algorithm should correctly return 0 when no valid pairs are found.

Extremely large arrays that could cause memory issues

How to Handle:

Consider the space complexity and if necessary, explore alternative approaches or use generators to process data in chunks.