Jobs

Interviews

Discussions

Reverse Pairs

Hard

Google

2 views

Topics:

ArraysRecursion

Given an integer array nums, return the number of reverse pairs in the array. A reverse pair is a pair (i, j) where:

0 <= i < j < nums.length and
nums[i] > 2 * nums[j].

For example:

nums = [1,3,2,3,1]

Output: 2

Explanation: The reverse pairs are:

(1, 4) --> nums[1] = 3, nums[4] = 1, 3 > 2 * 1

(3, 4) --> nums[3] = 3, nums[4] = 1, 3 > 2 * 1
nums = [2,4,3,5,1]

Output: 3

Explanation: The reverse pairs are:

(1, 4) --> nums[1] = 4, nums[4] = 1, 4 > 2 * 1

(2, 4) --> nums[2] = 3, nums[4] = 1, 3 > 2 * 1

(3, 4) --> nums[3] = 5, nums[4] = 1, 5 > 2 * 1

How would you efficiently count the number of reverse pairs in the array? Discuss the time and space complexity of your solution. Consider any edge cases.

Reverse Pairs

Hard

Google

2 views

Topics:

ArraysRecursion

Given an integer array nums, return the number of reverse pairs in the array. A reverse pair is a pair (i, j) where:

0 <= i < j < nums.length and
nums[i] > 2 * nums[j].

For example:

nums = [1,3,2,3,1]

Output: 2

Explanation: The reverse pairs are:

(1, 4) --> nums[1] = 3, nums[4] = 1, 3 > 2 * 1

(3, 4) --> nums[3] = 3, nums[4] = 1, 3 > 2 * 1
nums = [2,4,3,5,1]

Output: 3

Explanation: The reverse pairs are:

(1, 4) --> nums[1] = 4, nums[4] = 1, 4 > 2 * 1

(2, 4) --> nums[2] = 3, nums[4] = 1, 3 > 2 * 1

(3, 4) --> nums[3] = 5, nums[4] = 1, 5 > 2 * 1

How would you efficiently count the number of reverse pairs in the array? Discuss the time and space complexity of your solution. Consider any edge cases.

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 range of integer values within the input array? Can I expect negative numbers, zero, or only positive integers?
What should I return if no reverse pairs exist in the input array? Should I return 0, -1, or throw an exception?
Are duplicate numbers allowed in the input array, and if so, how should they be handled when identifying reverse pairs?
Is the input array guaranteed to be non-empty? What should I do if the input array is null or empty?
To clarify the definition of a 'reverse pair': is it strictly `nums[i] > 2 * nums[j]` where `i < j`, or could it also be `nums[i] >= 2 * nums[j]`?

Brute Force Solution

Approach

The basic idea is to check every single pair of numbers to see if they meet the 'reverse pair' condition. We will look at all possible combinations, comparing each number with every other number that comes after it.

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

Start with the very first number in the collection.
Compare this number with every single number that comes after it in the collection.
For each comparison, check if the first number is more than twice as big as the second number.
If it is, we've found a reverse pair, so make a note of it.
Move to the second number in the collection and repeat the process, comparing it to every number after it.
Keep doing this for each number in the collection, until you've compared every possible pair.
Finally, count up all the times you found a reverse pair. That's your answer.

Code Implementation

def reverse_pairs_brute_force(numbers):
    reverse_pairs_count = 0

    # Iterate through each number in the list
    for first_index in range(len(numbers)):

        # Compare the current number with all subsequent numbers
        for second_index in range(first_index + 1, len(numbers)):

            # Check the reverse pair condition.
            # We must use float to avoid integer division issues
            if numbers[first_index] > 2 * numbers[second_index]:
                reverse_pairs_count += 1

    return reverse_pairs_count

Big(O) Analysis

Time Complexity

O(n²) – The provided solution iterates through each of the n elements in the input array. For each element, it compares it with every subsequent element in the array. This means for the first element, n-1 comparisons are made, for the second, n-2 comparisons are made, and so on. The total number of comparisons approximates to n * (n-1) / 2. This simplifies to O(n²).

Space Complexity

O(1) – The provided plain English explanation outlines a brute-force approach. It iterates through pairs of numbers using nested loops but does not utilize any auxiliary data structures beyond a counter for reverse pairs. Only a few constant space variables are used to store indices and the count of reverse pairs. Therefore, the algorithm's space complexity is independent of the input size N and remains constant.

Optimal Solution

Approach

The straightforward way to solve this problem is slow. A much faster approach relies on splitting the data into smaller pieces that are easier to manage, then merging those pieces back together while carefully counting the special pairs as we go.

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

Divide the whole group of numbers into two smaller subgroups.
Keep dividing each subgroup into even smaller subgroups until you have subgroups of just one number each.
Now, start merging the subgroups back together in a specific order.
During the merging process, count the number of special pairs between the two subgroups you're merging. A pair is special if a number in the first subgroup is more than twice as big as a number in the second subgroup.
The key here is to efficiently count these special pairs during the merge instead of checking every possible pair. Sorting the sub groups before the merge helps.
Continue merging the subgroups and counting the special pairs at each stage. Adding up all the special pairs found during each merge gives the total number of special pairs in the whole group of numbers.

Code Implementation

def reverse_pairs(numbers):
    def merge_and_count(left, right):
        count = 0
        index_left = 0
        index_right = 0
        sorted_array = []

        # Count reverse pairs while merging.
        while index_left < len(left) and index_right < len(right):
            if left[index_left] > 2 * right[index_right]:
                count += len(left) - index_left
                index_right += 1
            else:
                index_left += 1

        index_left = 0
        index_right = 0

        # Standard merge operation.
        while index_left < len(left) and index_right < len(right):
            if left[index_left] <= right[index_right]:
                sorted_array.append(left[index_left])
                index_left += 1
            else:
                sorted_array.append(right[index_right])
                index_right += 1

        sorted_array.extend(left[index_left:])
        sorted_array.extend(right[index_right:])

        return count, sorted_array

    def merge_sort(sub_array):
        array_length = len(sub_array)
        if array_length <= 1:
            return 0, sub_array

        midpoint = array_length // 2
        
        # Recursively divide the array.
        left_count, left_array = merge_sort(sub_array[:midpoint])
        right_count, right_array = merge_sort(sub_array[midpoint:])

        # Accumulate counts from subproblems and merge.
        merge_count, sorted_array = merge_and_count(left_array, right_array)
        return left_count + right_count + merge_count, sorted_array

    # Start the merge sort process to count reverse pairs.
    total_count, _ = merge_sort(numbers)
    return total_count

Big(O) Analysis

Time Complexity

O(n log n) – The algorithm employs a divide-and-conquer strategy similar to merge sort. Dividing the array into subproblems takes O(log n) steps. During each merge step, we count reverse pairs. While a naive pair-checking approach would be O(n^2), the suggested method of sorting sub arrays then incrementing a pointer as we find matching pairs means that on average, we look at each of the 'n' elements only a fixed number of times for comparisons during a merge. Since the merge process also takes O(n) time, and we perform it at each level of the recursion (log n levels), the total time complexity becomes O(n log n).

Space Complexity

O(N) – The algorithm uses a divide and conquer approach, repeatedly splitting the input array of size N into smaller subarrays. During the merging steps, a temporary array is created to store the merged sorted subarray, which can be of size up to N in the worst case. This auxiliary array is used in each merge step at each level of the recursion/iteration, so the space complexity is primarily determined by this temporary array which holds N elements. Therefore, the space complexity is O(N).

Edge Cases

Case	How to Handle
Empty input array	Return 0 as there are no reverse pairs.
Array with one element	Return 0 as a reverse pair requires at least two elements.
Array with all elements equal to zero	Handle the integer division by 2 to avoid potential errors.
Array with a large number of reverse pairs (e.g., sorted in descending order)	Ensure the algorithm's time complexity doesn't degrade significantly in the worst-case scenario.
Array with very large numbers (close to Integer.MAX_VALUE or Integer.MIN_VALUE)	Watch out for potential integer overflows when multiplying or dividing by 2.
Array with negative numbers	The comparison 'nums[i] > 2 * nums[j]' should handle negative numbers correctly according to the problem description.
Array is already sorted in ascending order	The algorithm should still function correctly and efficiently even if no reverse pairs exist.
Array with duplicate values that form reverse pairs.	The algorithm should correctly count all valid reverse pairs, even if some numbers appear multiple times.

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 range of integer values within the input array? Can I expect negative numbers, zero, or only positive integers?
What should I return if no reverse pairs exist in the input array? Should I return 0, -1, or throw an exception?
Are duplicate numbers allowed in the input array, and if so, how should they be handled when identifying reverse pairs?
Is the input array guaranteed to be non-empty? What should I do if the input array is null or empty?
To clarify the definition of a 'reverse pair': is it strictly `nums[i] > 2 * nums[j]` where `i < j`, or could it also be `nums[i] >= 2 * nums[j]`?

Brute Force Solution

Approach

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

Start with the very first number in the collection.
Compare this number with every single number that comes after it in the collection.
For each comparison, check if the first number is more than twice as big as the second number.
If it is, we've found a reverse pair, so make a note of it.
Move to the second number in the collection and repeat the process, comparing it to every number after it.
Keep doing this for each number in the collection, until you've compared every possible pair.
Finally, count up all the times you found a reverse pair. That's your answer.

Code Implementation

def reverse_pairs_brute_force(numbers):
    reverse_pairs_count = 0

    # Iterate through each number in the list
    for first_index in range(len(numbers)):

        # Compare the current number with all subsequent numbers
        for second_index in range(first_index + 1, len(numbers)):

            # Check the reverse pair condition.
            # We must use float to avoid integer division issues
            if numbers[first_index] > 2 * numbers[second_index]:
                reverse_pairs_count += 1

    return reverse_pairs_count

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Divide the whole group of numbers into two smaller subgroups.
Keep dividing each subgroup into even smaller subgroups until you have subgroups of just one number each.
Now, start merging the subgroups back together in a specific order.
During the merging process, count the number of special pairs between the two subgroups you're merging. A pair is special if a number in the first subgroup is more than twice as big as a number in the second subgroup.
The key here is to efficiently count these special pairs during the merge instead of checking every possible pair. Sorting the sub groups before the merge helps.
Continue merging the subgroups and counting the special pairs at each stage. Adding up all the special pairs found during each merge gives the total number of special pairs in the whole group of numbers.

Code Implementation

def reverse_pairs(numbers):
    def merge_and_count(left, right):
        count = 0
        index_left = 0
        index_right = 0
        sorted_array = []

        # Count reverse pairs while merging.
        while index_left < len(left) and index_right < len(right):
            if left[index_left] > 2 * right[index_right]:
                count += len(left) - index_left
                index_right += 1
            else:
                index_left += 1

        index_left = 0
        index_right = 0

        # Standard merge operation.
        while index_left < len(left) and index_right < len(right):
            if left[index_left] <= right[index_right]:
                sorted_array.append(left[index_left])
                index_left += 1
            else:
                sorted_array.append(right[index_right])
                index_right += 1

        sorted_array.extend(left[index_left:])
        sorted_array.extend(right[index_right:])

        return count, sorted_array

    def merge_sort(sub_array):
        array_length = len(sub_array)
        if array_length <= 1:
            return 0, sub_array

        midpoint = array_length // 2
        
        # Recursively divide the array.
        left_count, left_array = merge_sort(sub_array[:midpoint])
        right_count, right_array = merge_sort(sub_array[midpoint:])

        # Accumulate counts from subproblems and merge.
        merge_count, sorted_array = merge_and_count(left_array, right_array)
        return left_count + right_count + merge_count, sorted_array

    # Start the merge sort process to count reverse pairs.
    total_count, _ = merge_sort(numbers)
    return total_count

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty input array	Return 0 as there are no reverse pairs.
Array with one element	Return 0 as a reverse pair requires at least two elements.
Array with all elements equal to zero	Handle the integer division by 2 to avoid potential errors.
Array with a large number of reverse pairs (e.g., sorted in descending order)	Ensure the algorithm's time complexity doesn't degrade significantly in the worst-case scenario.
Array with very large numbers (close to Integer.MAX_VALUE or Integer.MIN_VALUE)	Watch out for potential integer overflows when multiplying or dividing by 2.
Array with negative numbers	The comparison 'nums[i] > 2 * nums[j]' should handle negative numbers correctly according to the problem description.
Array is already sorted in ascending order	The algorithm should still function correctly and efficiently even if no reverse pairs exist.
Array with duplicate values that form reverse pairs.	The algorithm should correctly count all valid reverse pairs, even if some numbers appear multiple times.