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K Inverse Pairs Array

Hard

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Topics:

Dynamic Programming

For an integer array nums, an inverse pair is a pair of integers [i, j] where 0 <= i < j < nums.length and nums[i] > nums[j].

Given two integers n and k, return the number of different arrays consisting of numbers from 1 to n such that there are exactly k inverse pairs. Since the answer can be huge, return it modulo 10⁹ + 7.

Example 1:

Input: n = 3, k = 0
Output: 1
Explanation: Only the array [1,2,3] which consists of numbers from 1 to 3 has exactly 0 inverse pairs.

Example 2:

Input: n = 3, k = 1
Output: 2
Explanation: The array [1,3,2] and [2,1,3] have exactly 1 inverse pair.

Constraints:

1 <= n <= 1000
0 <= k <= 1000

Try coding it in LeetCode

K Inverse Pairs Array

Hard

Asked by:

12 views

Topics:

Dynamic Programming

For an integer array nums, an inverse pair is a pair of integers [i, j] where 0 <= i < j < nums.length and nums[i] > nums[j].

Example 1:

Input: n = 3, k = 0
Output: 1
Explanation: Only the array [1,2,3] which consists of numbers from 1 to 3 has exactly 0 inverse pairs.

Example 2:

Input: n = 3, k = 1
Output: 2
Explanation: The array [1,3,2] and [2,1,3] have exactly 1 inverse pair.

Constraints:

1 <= n <= 1000
0 <= k <= 1000

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 maximum possible values for `n` and `k`? Specifically, what is the upper bound for these inputs?
Can `n` or `k` ever be zero?
If it's impossible to construct an array with exactly `k` inverse pairs, what should I return? Should I return 0, -1, or throw an exception?
Are `n` distinct positive integers required to be consecutive, starting from 1, or can they be any distinct positive integers?
Could you provide a small example input and the expected output, to ensure I understand the problem correctly?

Brute Force Solution

Approach

The brute force approach to this problem involves generating every single possible arrangement of numbers from 1 to n. Then, for each arrangement, we count the number of 'inverse pairs'. Finally, we count how many of these arrangements have exactly k inverse pairs.

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

First, imagine you have all the numbers from 1 up to a certain number, say 3. You need to arrange them in a specific order.
Start by listing all the different ways you can arrange those numbers. For example, with 1, 2, and 3, you could have 1-2-3, 1-3-2, 2-1-3, 2-3-1, 3-1-2, or 3-2-1.
For each of these arrangements, count how many pairs of numbers are in the wrong order. A pair is in the wrong order if the bigger number comes before the smaller number. For example, in 3-1-2, the pairs 3-1 and 3-2 are in the wrong order, so there are 2 inverse pairs.
Finally, once you've gone through every arrangement and counted the number of inverse pairs for each, count how many of those arrangements have the exact number of inverse pairs that you are looking for. That count is your answer.

Code Implementation

def k_inverse_pairs_brute_force(number_limit, target_inversions):

    def count_inversions(permutation):
        inversion_count = 0
        for i in range(len(permutation)):
            for j in range(i + 1, len(permutation)):
                if permutation[i] > permutation[j]:
                    inversion_count += 1
        return inversion_count

    import itertools
    numbers = list(range(1, number_limit + 1))
    permutations = list(itertools.permutations(numbers))

    # Iterate through all permutations and count valid results
    valid_permutation_count = 0

    for permutation in permutations:
        # Count inversions for the current permutation
        inversions = count_inversions(permutation)

        # Check if the current permutation has the target number of inversions
        if inversions == target_inversions:

            valid_permutation_count += 1

    return valid_permutation_count

Big(O) Analysis

Time Complexity

O(n! * n^2) – The algorithm generates all possible permutations of numbers from 1 to n. There are n! such permutations. For each permutation, we iterate through the array to count the inverse pairs. Counting inverse pairs requires comparing each element to every other element, which takes O(n^2) time. Therefore, the overall time complexity is O(n! * n^2), as we perform the inverse pair counting for each of the n! permutations.

Space Complexity

O(n!) – The brute force approach generates all permutations of numbers from 1 to n. Storing each permutation requires an array of size n. Since there are n! possible permutations, the space needed to store all permutations is proportional to n! . Therefore, the space complexity is O(n!).

Optimal Solution

Approach

Instead of trying all possible arrangements of numbers, we use a clever trick to build up the solution step by step. We avoid recomputing the same information multiple times by building on previous calculations.

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

Consider small arrangements of numbers and the number of inversions they create.
Notice a pattern: to create a certain number of inversions, we can take a smaller arrangement and cleverly insert a new number to increase the inversions.
Store the number of ways to make each possible number of inversions for each arrangement size. We are interested in the one that has the needed number of inversions.
When we want to find the number of ways for an array of one size, we use the previous smaller sized array to calculate the ways we can make inversions for our new array
We add up how many ways can create the total amount of inversions needed, in the correct array size, using information from the previous array size

Code Implementation

def k_inverse_pairs(number, k_inversions):
    modulo = 10**9 + 7
    # This stores how many ways to get inversions for array size
    dp_table = [[0] * (k_inversions + 1) for _ in range(number + 1)]

    dp_table[0][0] = 1

    for array_size in range(1, number + 1):
        dp_table[array_size][0] = 1

        for inversion_count in range(1, k_inversions + 1):
            # Base inversions on the previous array size
            dp_table[array_size][inversion_count] = (dp_table[array_size][inversion_count - 1] + dp_table[array_size - 1][inversion_count]) % modulo

            # Subtract to avoid double counting
            if inversion_count >= array_size:
                dp_table[array_size][inversion_count] = (dp_table[array_size][inversion_count] - dp_table[array_size - 1][inversion_count - array_size]) % modulo

    # The final result is the number of ways to get the target inversions
    return dp_table[number][k_inversions]

Big(O) Analysis

Time Complexity

O(n*k) – The algorithm uses dynamic programming to build a table representing the number of ways to form k inversions using arrays of size n. The outer loop iterates up to n (the size of the array). The inner loop iterates up to k (the target number of inversions). Inside the inner loop, we potentially iterate up to k again to calculate the cumulative sum of previous values from the smaller array size, adding to the number of ways. Therefore, the dominant cost is proportional to n * k, making the overall time complexity O(n*k).

Space Complexity

O(N*K) – The solution utilizes dynamic programming to store the number of ways to create each possible number of inversions for each arrangement size. This involves creating a table (or equivalent data structure) where the rows represent the array size (up to N) and the columns represent the number of inversions (up to K). Therefore, the space required is proportional to the product of N and K, where N is the size of the array and K is the target number of inversions. This table stores intermediate results, avoiding redundant calculations. Thus the auxiliary space complexity is O(N*K).

Edge Cases

Case	How to Handle
n = 0, k = 0	Return 1 since an empty array has 0 inverse pairs.
n = 1, k = 0	Return 1 since a single element array has 0 inverse pairs.
k < 0	Return 0 since the number of inverse pairs cannot be negative.
k > n * (n - 1) / 2	Return 0 since the maximum number of inverse pairs for an array of size n is n*(n-1)/2.
Integer overflow in calculations (e.g., large n and k)	Use modulo operator (10^9 + 7) during intermediate calculations to prevent overflow.
n is large, causing recursion depth issues if a recursive approach is used	Implement a dynamic programming approach to avoid stack overflow errors.
k = 0 and n > 0	Return 1, as there is only one sorted array with 0 inverse pairs.
k = n * (n - 1) / 2	Return 1, as there is only one reverse-sorted array with the maximum number of inverse pairs.

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 maximum possible values for `n` and `k`? Specifically, what is the upper bound for these inputs?
Can `n` or `k` ever be zero?
If it's impossible to construct an array with exactly `k` inverse pairs, what should I return? Should I return 0, -1, or throw an exception?
Are `n` distinct positive integers required to be consecutive, starting from 1, or can they be any distinct positive integers?
Could you provide a small example input and the expected output, to ensure I understand the problem correctly?

Brute Force Solution

Approach

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

First, imagine you have all the numbers from 1 up to a certain number, say 3. You need to arrange them in a specific order.
Start by listing all the different ways you can arrange those numbers. For example, with 1, 2, and 3, you could have 1-2-3, 1-3-2, 2-1-3, 2-3-1, 3-1-2, or 3-2-1.
For each of these arrangements, count how many pairs of numbers are in the wrong order. A pair is in the wrong order if the bigger number comes before the smaller number. For example, in 3-1-2, the pairs 3-1 and 3-2 are in the wrong order, so there are 2 inverse pairs.
Finally, once you've gone through every arrangement and counted the number of inverse pairs for each, count how many of those arrangements have the exact number of inverse pairs that you are looking for. That count is your answer.

Code Implementation

def k_inverse_pairs_brute_force(number_limit, target_inversions):

    def count_inversions(permutation):
        inversion_count = 0
        for i in range(len(permutation)):
            for j in range(i + 1, len(permutation)):
                if permutation[i] > permutation[j]:
                    inversion_count += 1
        return inversion_count

    import itertools
    numbers = list(range(1, number_limit + 1))
    permutations = list(itertools.permutations(numbers))

    # Iterate through all permutations and count valid results
    valid_permutation_count = 0

    for permutation in permutations:
        # Count inversions for the current permutation
        inversions = count_inversions(permutation)

        # Check if the current permutation has the target number of inversions
        if inversions == target_inversions:

            valid_permutation_count += 1

    return valid_permutation_count

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Consider small arrangements of numbers and the number of inversions they create.
Notice a pattern: to create a certain number of inversions, we can take a smaller arrangement and cleverly insert a new number to increase the inversions.
Store the number of ways to make each possible number of inversions for each arrangement size. We are interested in the one that has the needed number of inversions.
When we want to find the number of ways for an array of one size, we use the previous smaller sized array to calculate the ways we can make inversions for our new array
We add up how many ways can create the total amount of inversions needed, in the correct array size, using information from the previous array size

Code Implementation

def k_inverse_pairs(number, k_inversions):
    modulo = 10**9 + 7
    # This stores how many ways to get inversions for array size
    dp_table = [[0] * (k_inversions + 1) for _ in range(number + 1)]

    dp_table[0][0] = 1

    for array_size in range(1, number + 1):
        dp_table[array_size][0] = 1

        for inversion_count in range(1, k_inversions + 1):
            # Base inversions on the previous array size
            dp_table[array_size][inversion_count] = (dp_table[array_size][inversion_count - 1] + dp_table[array_size - 1][inversion_count]) % modulo

            # Subtract to avoid double counting
            if inversion_count >= array_size:
                dp_table[array_size][inversion_count] = (dp_table[array_size][inversion_count] - dp_table[array_size - 1][inversion_count - array_size]) % modulo

    # The final result is the number of ways to get the target inversions
    return dp_table[number][k_inversions]

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
n = 0, k = 0	Return 1 since an empty array has 0 inverse pairs.
n = 1, k = 0	Return 1 since a single element array has 0 inverse pairs.
k < 0	Return 0 since the number of inverse pairs cannot be negative.
k > n * (n - 1) / 2	Return 0 since the maximum number of inverse pairs for an array of size n is n*(n-1)/2.
Integer overflow in calculations (e.g., large n and k)	Use modulo operator (10^9 + 7) during intermediate calculations to prevent overflow.
n is large, causing recursion depth issues if a recursive approach is used	Implement a dynamic programming approach to avoid stack overflow errors.
k = 0 and n > 0	Return 1, as there is only one sorted array with 0 inverse pairs.
k = n * (n - 1) / 2	Return 1, as there is only one reverse-sorted array with the maximum number of inverse pairs.