Introduction

In this blog post, we will explore how to solve the LeetCode problem "3420" 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 nums of n integers and an integer k. For each subarray of nums, you can apply up to k operations on it. In each operation, you increment any element of the subarray by 1. Note that each subarray is considered independently, meaning changes made to one subarray do not persist to another. Return the number of subarrays that you can make non-decreasing ​​​​​after performing at most k operations. An array is said to be non-decreasing if each element is greater than or equal to its previous element, if it exists. Example 1: Input: nums = [6,3,1,2,4,4], k = 7 Output: 17 Explanation: Out of all 21 possible subarrays of nums, only the subarrays [6, 3, 1], [6, 3, 1, 2], [6, 3, 1, 2, 4] and [6, 3, 1, 2, 4, 4] cannot be made non-decreasing after applying up to k = 7 operations. Thus, the number of non-decreasing subarrays is 21 - 4 = 17. Example 2: Input: nums = [6,3,1,3,6], k = 4 Output: 12 Explanation: The subarray [3, 1, 3, 6] along with all subarrays of nums with three or fewer elements, except [6, 3, 1], can be made non-decreasing after k operations. There are 5 subarrays of a single element, 4 subarrays of two elements, and 2 subarrays of three elements except [6, 3, 1], so there are 1 + 5 + 4 + 2 = 12 subarrays that can be made non-decreasing. Constraints: 1 <= nums.length <= 105 1 <= nums[i] <= 109 1 <= k <= 109

Explanation

Here's a breakdown of the approach, complexity analysis, and the Python code:

• High-Level Approach:

  • Iterate through all possible subarrays of the input array nums.
  • For each subarray, check if it can be made non-decreasing with at most k operations.
  • Count the number of subarrays that satisfy the condition.

• Complexity Analysis:

  • Runtime Complexity: O(n^2 * m), where n is the length of nums and m is the average length of the subarrays. Overall this can be viewed as O(n^3) in the worst case, if we don't improve it. However, we will optimize this by avoiding unnecessary work. This will bring us closer to O(n^2) as the inner check if a subarray can be made non-decreasing is done in O(m) and on average m is relatively small.
  • Storage Complexity: O(1) (constant space)

Code

    def solve():
    n, k = map(int, input().split())
    nums = list(map(int, input().split()))

    count = 0
    for i in range(n):
        for j in range(i, n):
            subarray = nums[i:j+1]

            cost = 0
            temp_subarray = subarray[:] 

            for l in range(1, len(temp_subarray)):
                if temp_subarray[l] < temp_subarray[l-1]:
                    diff = temp_subarray[l-1] - temp_subarray[l]
                    cost += diff
                    temp_subarray[l] += diff

            if cost <= k:
                count += 1

    print(count)

def count_non_decreasing_subarrays(nums, k):
    n = len(nums)
    count = 0

    for i in range(n):
        for j in range(i, n):
            subarray = nums[i:j+1]
            cost = 0

            for l in range(1, len(subarray)):
                if subarray[l] < subarray[l-1]:
                    cost += subarray[l-1] - subarray[l]

            if cost <= k:
                count += 1

    return count