Count Non-Decreasing Subarrays After K Operations

Count Non-Decreasing Subarrays After K Operations - Problem

Array Stack Segment Tree Queue Sliding Window Monotonic Stack Monotonic Queue Hard

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: 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.

Input & Output

Example 1 — Basic Case

$ Input: nums = [4,2,5,3], k = 3

› Output: 8

💡 Note: Valid subarrays: [4] (0 ops), [2] (0 ops), [5] (0 ops), [3] (0 ops), [4,2] (2 ops: make 2→4), [2,5] (0 ops), [5,3] (2 ops: make 3→5), [2,5,3] (2 ops: make 3→5). Total: 8 subarrays.

Example 2 — No Operations Needed

$ Input: nums = [1,2,3,4], k = 2

› Output: 10

💡 Note: Array is already non-decreasing, so all 10 possible subarrays ([1], [1,2], [1,2,3], [1,2,3,4], [2], [2,3], [2,3,4], [3], [3,4], [4]) need 0 operations ≤ 2.

Example 3 — Limited Operations

$ Input: nums = [5,1,3], k = 1

› Output: 4

💡 Note: Valid subarrays: [5] (0 ops), [1] (0 ops), [3] (0 ops), [1,3] (0 ops). Subarray [5,1] needs 4 ops (make 1→5), [5,1,3] needs 4 ops, [5,3] needs 2 ops - all exceed k=1.

Constraints

1 ≤ nums.length ≤ 10⁵
1 ≤ nums[i] ≤ 10⁹
0 ≤ k ≤ 10¹²

Visualization

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Understanding the Visualization

Input

Array [4,2,5,3] with k=3 operations allowed per subarray

Process

For each subarray, calculate minimum operations to make non-decreasing

Output

Count of subarrays that need ≤ k operations

Key Takeaway

🎯 Key Insight: Calculate minimum increments needed for each subarray to become non-decreasing, then count those requiring ≤ k operations

Asked in

G Google 45 f Meta 38 a Amazon 32 M Microsoft 28

The key insight is to calculate the minimum operations needed to make each subarray non-decreasing by ensuring each element is at least as large as the previous maximum. The optimal approach uses sliding window with early termination when operations exceed k. Time: O(n²), Space: O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force - Check All Subarrays	O(n³)	O(1)	Generate all subarrays and calculate operations needed for each
Sliding Window Optimization	O(n²)	O(1)	Use sliding window to efficiently calculate operations for overlapping subarrays
Monotonic Stack Optimization	O(n²)	O(n)	Use monotonic stack to efficiently track elements that need to be incremented

Brute Force - Check All Subarrays — Algorithm Steps

Generate all possible subarrays
For each subarray, calculate minimum operations needed
Count subarrays that need ≤ k operations

Visualization

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Step-by-Step Walkthrough

Generate Subarrays

Create all possible subarrays

Calculate Operations

Count increments needed for non-decreasing

Count Valid

Count subarrays needing ≤ k operations

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int* nums, int numsSize, int k) {
    int count = 0;
    
    // Check all possible subarrays
    for (int i = 0; i < numsSize; i++) {
        for (int j = i; j < numsSize; j++) {
            // Calculate operations needed for subarray nums[i:j+1]
            int operations = 0;
            int tempMax = nums[i];
            
            for (int idx = i; idx <= j; idx++) {
                if (nums[idx] < tempMax) {
                    operations += tempMax - nums[idx];
                } else {
                    tempMax = nums[idx];
                }
            }
            
            if (operations <= k) {
                count++;
            }
        }
    }
    
    return count;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array manually
    int nums[1000];
    int numsSize = 0;
    char* token = strtok(line, "[],");
    while (token != NULL) {
        if (strlen(token) > 0 && token[0] != '\n') {
            nums[numsSize++] = atoi(token);
        }
        token = strtok(NULL, "[],");
    }
    
    int k;
    scanf("%d", &k);
    
    int result = solution(nums, numsSize, k);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

Three nested loops: O(n²) subarrays, each taking O(n) to process

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra variables

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Check All Subarrays — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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