Introduction

In this blog post, we will explore how to solve the LeetCode problem "1420" 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

You are given three integers n, m and k. Consider the following algorithm to find the maximum element of an array of positive integers: You should build the array arr which has the following properties: arr has exactly n integers. 1 <= arr[i] <= m where (0 <= i < n). After applying the mentioned algorithm to arr, the value search_cost is equal to k. Return the number of ways to build the array arr under the mentioned conditions. As the answer may grow large, the answer must be computed modulo 109 + 7. Example 1: Input: n = 2, m = 3, k = 1 Output: 6 Explanation: The possible arrays are [1, 1], [2, 1], [2, 2], [3, 1], [3, 2] [3, 3] Example 2: Input: n = 5, m = 2, k = 3 Output: 0 Explanation: There are no possible arrays that satisfy the mentioned conditions. Example 3: Input: n = 9, m = 1, k = 1 Output: 1 Explanation: The only possible array is [1, 1, 1, 1, 1, 1, 1, 1, 1] Constraints: 1 <= n <= 50 1 <= m <= 100 0 <= k <= n

Explanation

Here's the solution:

• Dynamic Programming: Use a 3D DP table dp[i][j][l] to store the number of ways to build an array of length i with the maximum element being j and the search cost being l.
• Base Cases and Transitions: Initialize base cases and then iterate through the DP table, calculating the number of ways based on whether the current element is the new maximum or not.

• Modulo Arithmetic: Apply modulo operation % (10**9 + 7) throughout the calculation to avoid integer overflow.

• Runtime Complexity: O(n*m*k*m) can be optimized to O(n*m*k). Storage Complexity: O(n*m*k)

Code

    def numOfArrays(n: int, m: int, k: int) -> int:
    """
    Calculates the number of ways to build an array under the given conditions.

    Args:
        n: The number of integers in the array.
        m: The upper bound of the integers in the array.
        k: The required search cost.

    Returns:
        The number of ways to build the array modulo 10^9 + 7.
    """
    if k > m:
        return 0

    dp = [[[0] * (k + 1) for _ in range(m + 1)] for _ in range(n + 1)]
    prefix_sum = [[[0] * (k + 1) for _ in range(m + 2)] for _ in range(n + 1)]
    MOD = 10**9 + 7

    for j in range(1, m + 1):
        dp[1][j][1] = 1
        prefix_sum[1][j+1][1] = prefix_sum[1][j][1] + dp[1][j][1]

    for i in range(2, n + 1):
        for j in range(1, m + 1):
            for l in range(1, k + 1):
                # Case 1: arr[i] is not the largest element
                dp[i][j][l] = (dp[i][j][l] + (dp[i - 1][j][l] * j)) % MOD

                # Case 2: arr[i] is the largest element
                dp[i][j][l] = (dp[i][j][l] + prefix_sum[i-1][j][l-1]) % MOD
            prefix_sum[i][j+1] = [ (prefix_sum[i][j][x] + dp[i][j][x]) % MOD for x in range(k+1)]


    ans = 0
    for j in range(1, m + 1):
        ans = (ans + dp[n][j][k]) % MOD

    return ans