Introduction

In this blog post, we will explore how to solve the LeetCode problem "3250" 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 an array of positive integers nums of length n. We call a pair of non-negative integer arrays (arr1, arr2) monotonic if: The lengths of both arrays are n. arr1 is monotonically non-decreasing, in other words, arr1[0] <= arr1[1] <= ... <= arr1[n - 1]. arr2 is monotonically non-increasing, in other words, arr2[0] >= arr2[1] >= ... >= arr2[n - 1]. arr1[i] + arr2[i] == nums[i] for all 0 <= i <= n - 1. Return the count of monotonic pairs. Since the answer may be very large, return it modulo 109 + 7. Example 1: Input: nums = [2,3,2] Output: 4 Explanation: The good pairs are: ([0, 1, 1], [2, 2, 1]) ([0, 1, 2], [2, 2, 0]) ([0, 2, 2], [2, 1, 0]) ([1, 2, 2], [1, 1, 0]) Example 2: Input: nums = [5,5,5,5] Output: 126 Constraints: 1 <= n == nums.length <= 2000 1 <= nums[i] <= 50

Explanation

Here's the breakdown of the solution:

• Dynamic Programming: We use dynamic programming to build up the count of monotonic pairs. dp[i][j] represents the number of valid arr1 prefixes of length i+1 such that arr1[i] = j. From this, we can derive arr2[i] = nums[i] - j.
• Monotonicity Enforcement: The core of the DP transition ensures monotonicity. For arr1, we only transition from dp[i-1][k] to dp[i][j] if k <= j. Similarly, for arr2, we check if nums[i-1] - k >= nums[i] - j. This guarantees that arr1 is non-decreasing and arr2 is non-increasing.

• Modulo Arithmetic: Since the counts can be very large, we apply the modulo operator % (10**9 + 7) throughout the calculation to prevent overflow.

• Time and Space Complexity: O(n m^2), where n is the length of nums and m is the maximum value in nums. Space complexity is O(n m).

Code

    def count_monotonic_pairs(nums):
    n = len(nums)
    MOD = 10**9 + 7
    max_num = max(nums)

    dp = [[0] * (max_num + 1) for _ in range(n)]

    # Initialize the first row of dp
    for j in range(nums[0] + 1):
        dp[0][j] = 1

    # Iterate through the remaining rows
    for i in range(1, n):
        for j in range(nums[i] + 1):
            for k in range(j + 1):
                if nums[i - 1] - k >= nums[i] - j:
                    dp[i][j] = (dp[i][j] + dp[i - 1][k]) % MOD

    # Sum the last row to get the total count
    total_count = 0
    for j in range(max_num + 1):
        total_count = (total_count + dp[n - 1][j]) % MOD

    return total_count