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


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3251" 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] <= 1000  

	# Explanation
	Here's the breakdown of the solution:

*   **Dynamic Programming:** We use dynamic programming to efficiently calculate the number of monotonic pairs. `dp[i][j]` stores the number of valid `arr1` prefixes of length `i+1` such that `arr1[i] == j`.
*   **State Transition:** The key is to calculate `dp[i][j]` based on the previous row `dp[i-1]`. Since `arr1` is monotonically non-decreasing, we sum up the `dp[i-1][k]` for all `k <= j`.
*   **Prefix Sum Optimization:** To optimize the state transition, we use prefix sums. `prefix_sum[i][j]` stores the sum of `dp[i][k]` for all `k <= j`. This allows us to calculate the sum in O(1) time.

*   **Runtime Complexity:** O(n\*m), where n is the length of `nums` and m is the maximum value in `nums`.
*   **Storage Complexity:** O(n\*m)

	
	# Code
	```python
	def count_monotonic_pairs(nums):
    n = len(nums)
    MOD = 10**9 + 7
    max_val = max(nums)

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

    # Base case: i = 0
    for j in range(nums[0] + 1):
        dp[0][j] = 1
        prefix_sum[0][j] = (prefix_sum[0][j-1] + dp[0][j]) % MOD if j > 0 else dp[0][j]

    # Iterate through the rest of the rows
    for i in range(1, n):
        for j in range(max_val + 1):
            if j <= nums[i]:
                dp[i][j] = prefix_sum[i-1][j]
            else:
                dp[i][j] = 0
            prefix_sum[i][j] = (prefix_sum[i][j-1] + dp[i][j]) % MOD if j > 0 else dp[i][j]

    # Sum up the last row
    result = 0
    for j in range(max_val + 1):
        result = (result + dp[n-1][j]) % MOD

    return result
	```