Introduction

In this blog post, we will explore how to solve the LeetCode problem "629" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

Here's a breakdown of the solution:

• Dynamic Programming: The problem is solved using dynamic programming. dp[i][j] stores the number of arrays of length i with exactly j inverse pairs.
• State Transition: When adding the number i to an array of length i-1, it can be placed in any of the i positions. Placing it at the end adds 0 inversions, placing it in the second to last position adds 1 inversion, and so on. This allows us to derive dp[i][j] from the previous row dp[i-1][].

• Prefix Sum Optimization: A naive implementation of the state transition would be O(n) leading to O(n^2k) overall. The state transition inherently involves summing a range of values from the previous row. We use prefix sums to compute these sums in O(1) time, reducing overall complexity.

• Runtime Complexity: O(n*k), Storage Complexity: O(n*k)

Code

    def kInversePairs(n: int, k: int) -> int:
    """
    Calculates the number of different arrays consisting of numbers from 1 to n
    such that there are exactly k inverse pairs.

    Args:
        n: The number of elements in the array (1 to n).
        k: The desired number of inverse pairs.

    Returns:
        The number of different arrays modulo 10^9 + 7.
    """

    MOD = 10**9 + 7

    # dp[i][j] stores the number of arrays of length i with j inverse pairs
    dp = [[0] * (k + 1) for _ in range(n + 1)]

    # Base case: an array of length 1 has 0 inverse pairs
    dp[1][0] = 1

    # Iterate through array lengths from 2 to n
    for i in range(2, n + 1):
        # Prefix sum array for the previous row
        prefix_sum = [0] * (k + 2)  # Extra element for easier calculations
        for l in range(k + 1):
            prefix_sum[l + 1] = (prefix_sum[l] + dp[i - 1][l]) % MOD

        # Iterate through the possible number of inverse pairs (0 to k)
        for j in range(k + 1):
            # Calculate dp[i][j] using prefix sums
            upper_bound = j + 1
            lower_bound = max(0, j - i + 1)
            dp[i][j] = (prefix_sum[upper_bound] - prefix_sum[lower_bound]) % MOD
            if dp[i][j] < 0:
                dp[i][j] += MOD  # Handle potential negative modulo

    return dp[n][k]