What is the time complexity of Count Non-Decreasing Subarrays After K Operations?

The optimal solution for Count Non-Decreasing Subarrays After K Operations runs in O(n) time and uses O(1) space. The time complexity is O(n) because each element is processed at most twice (once by the right pointer and once by the left pointer).

What is the brute force solution for Count Non-Decreasing Subarrays After K Operations?

The brute force approach for Count Non-Decreasing Subarrays After K Operations is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible subarray and calculates how many operations are needed to make it non-decreasing. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Count Non-Decreasing Subarrays After K Operations?

Explain your thought process clearly while coding. Consider edge cases, such as very small or very large arrays. Practice similar problems to become familiar with the two-pointer technique.

What are common mistakes when solving Count Non-Decreasing Subarrays After K Operations?

Not resetting the left pointer correctly when operations exceed k. Failing to account for all valid subarrays when counting.

#3420

Count Non-Decreasing Subarrays After K Operations

Hard

ArrayStackSegment TreeQueueSliding WindowMonotonic StackMonotonic QueueTwo PointersSliding Window

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution uses a two-pointer technique to efficiently count valid subarrays. This approach reduces the time complexity significantly by avoiding redundant calculations.

⚙️

Algorithm

4 steps

1Step 1: Initialize two pointers, left and right, both at the start of the array.
2Step 2: Expand the right pointer to include elements in the current subarray while maintaining the count of operations needed to make it non-decreasing.
3Step 3: If the operations exceed k, increment the left pointer to reduce the size of the subarray until operations are within k.
4Step 4: For each valid subarray, count the number of subarrays that can be formed with the current right pointer.

solution.py13 lines

1def countNonDecreasingSubarrays(nums, k):
2    count = 0
3    left = 0
4    operations = 0
5    for right in range(len(nums)):
6        if right > 0 and nums[right] < nums[right - 1]:
7            operations += (nums[right - 1] - nums[right])
8        while operations > k:
9            if left < right and nums[left] < nums[left + 1]:
10                operations -= (nums[left + 1] - nums[left])
11            left += 1
12        count += (right - left + 1)
13    return count

ℹ

Complexity note: The time complexity is O(n) because each element is processed at most twice (once by the right pointer and once by the left pointer).

1Using two pointers can significantly reduce the time complexity.
2Understanding how to maintain a running count of operations is crucial.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.