# Solving Leetcode Interviews in Seconds with AI: Find the Count of Monotonic Pairs I


	# 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](https://www.chatmagic.app), 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
	```python
	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
	```