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, 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

    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