LeetCode 3420: Count Non-Decreasing Subarrays After K Operations Solution

Master LeetCode problem 3420 (Count Non-Decreasing Subarrays After K Operations), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3420. Count Non-Decreasing Subarrays After K Operations

Hard

Array Stack Segment Tree Queue Sliding Window Monotonic Stack Monotonic Queue

Problem Explanation

Explanation

To solve this problem, we can use a sliding window approach. We iterate over all subarrays and for each subarray, we find the number of operations needed to make it non-decreasing. If this number is less than or equal to k, we increment the count of non-decreasing subarrays. We can optimize this process by maintaining a window that we slide from left to right. We keep track of the sum of elements in the window and the maximum element in the window. If the difference between the maximum element and the sum of elements is less than or equal to k, we count the subarrays that can be made non-decreasing. The number of subarrays that can be made non-decreasing for a window of size n is n*(n+1)/2.

Time Complexity: O(n) Space Complexity: O(1)

Solution Code

class Solution {
    public int countNonDecreasing(int[] nums, int k) {
        int n = nums.length;
        int count = 0;
        int sum = 0;
        int maxNum = Integer.MIN_VALUE;
        int left = 0;
        
        for (int right = 0; right < n; right++) {
            sum += nums[right];
            maxNum = Math.max(maxNum, nums[right]);
            
            while (maxNum * (right - left + 1) - sum > k) {
                sum -= nums[left];
                maxNum = Integer.MIN_VALUE;
                left++;
                for (int i = left; i <= right; i++) {
                    maxNum = Math.max(maxNum, nums[i]);
                }
            }
            
            count += right - left + 1;
        }
        
        return count;
    }
}

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