Pipeline Beacon Partition 19 — Problem Statement & Solution Guide

Binary TreesMediumFixed/Dynamic Window

TimeO(n)

SpaceO(1)

Problem Description

You are tasked with analyzing a sequence of integer metrics collected from a distributed pipeline system. The system partitions the data stream into fixed-size windows to compute local aggregates for real-time monitoring. Given an array of integers representing the raw metrics and a window size k, your objective is to compute the sum of elements in each contiguous subarray of length k as the window slides from left to right across the array.

The input consists of an integer array nums and an integer k. You must return an array of integers where each element at index i represents the sum of the subarray nums[i ... i+k-1]. The window moves one position to the right at each step until it reaches the end of the array. If the array length is less than k, return an empty array.

This problem requires efficient computation of sliding window sums. A naive approach would recalculate the sum for every window in O(n*k) time, but an optimized approach can achieve O(n) time complexity by reusing the previous window's sum, subtracting the element that exits the window and adding the element that enters it.

DSA Pattern Breakdown

"Pipeline Beacon Partition 19"

medium

WHY DOES IT MATTER?

Sliding‑window patterns turn quadratic work into linear work by reusing previous computation.

OPTIMIZATION CHALLENGE

The key is to eliminate the inner loop by maintaining a running aggregate as the window slides.

REAL-WORLD CONNECTION

Network routers compute moving averages of packet latency using the same principle.

Initialize the first window sum once, then loop from k to n‑1 updating the sum in place.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)

💾 Space:O(1)

Core Theory — Why This Approach?

The sliding‑window sum problem asks for the sum of every contiguous subarray of length k. A naive solution recomputes each sum from scratch, leading to O(n·k) time, which becomes prohibitive when n and k are large (e.g., streaming telemetry). The optimal paradigm leverages the overlapping nature of windows: when the window slides one position, the leftmost element exits and the next element enters, allowing the new sum to be derived from the previous sum in O(1) time. This yields a linear‑time algorithm that processes the stream in a single pass while using constant auxiliary space, which is essential for real‑time monitoring systems.

Interview Questions on This Problem

Q1How does the sliding‑window technique reduce time complexity compared to recomputing each window sum?

It updates the sum by subtracting the element leaving the window and adding the new element entering. This constant‑time update avoids the O(k) work per window.

Q2What edge cases must you handle when implementing a sliding‑window sum?

When k is larger than the array length, no window exists and the result should be empty. Also, k equal to 0 should be treated as an invalid input or return zeros per specification.

Q3Can the sliding‑window approach be extended to compute other aggregates like max or min?

Yes, but max/min require a deque or monotonic queue to maintain candidates in O(1) amortized time. Simple addition works only for associative and invertible operations like sum.

Examples

Example 1

Input

nums = [1, 3, 2, 5, 4], k = 3

Output

[6, 10, 11]

Explanation: Window 1: [1, 3, 2] -> sum = 1 + 3 + 2 = 6. Window 2: [3, 2, 5] -> sum = 3 + 2 + 5 = 10. Window 3: [2, 5, 4] -> sum = 2 + 5 + 4 = 11. Total windows = 5 - 3 + 1 = 3.

Example 2

Input

nums = [10, -2, 7, 0, 5], k = 2

Output

[8, 5, 7, 5]

Explanation: Window 1: [10, -2] -> sum = 10 + (-2) = 8. Window 2: [-2, 7] -> sum = -2 + 7 = 5. Window 3: [7, 0] -> sum = 7 + 0 = 7. Window 4: [0, 5] -> sum = 0 + 5 = 5. Total windows = 5 - 2 + 1 = 4.

Example 3

Input

nums = [4, 4, 4], k = 3

Output

[12]

Explanation: Window 1: [4, 4, 4] -> sum = 4 + 4 + 4 = 12. Total windows = 3 - 3 + 1 = 1.

Example 4

Input

nums = [1, 2], k = 3

Output

[]

Explanation: The array length (2) is less than the window size (3), so no valid windows exist. Return an empty array.

Constraints

1 <= nums.length <= 10^5
-10^9 <= nums[i] <= 10^9
1 <= k <= nums.length
The answer is guaranteed to fit in a 64-bit integer.

Optimal Approach & Strategy

Calculate the first window sum, then update it in O(1) as the window slides, achieving O(n) total time.

Brute Force Approach

Compute each window sum independently by iterating k elements for every start index, resulting in O(n·k) time.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n)

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {number[]}
 */
var slidingWindowMax = function(nums, k) {
    const result = [];
    const dq = [];
    for (let i = 0; i < nums.length; i++) {
        while (dq.length > 0 && nums[dq[dq.length - 1]] <= nums[i]) {
            dq.pop();
        }
        dq.push(i);
        if (dq[0] <= i - k) {
            dq.shift();
        }
        if (i >= k - 1) {
            result.push(nums[dq[0]]);
        }
    }
    return result;
};

class Solution {
public:
    vector<int> slidingWindowMax(vector<int>& nums, int k) {
        vector<int> result;
        deque<int> dq;
        for (int i = 0; i < nums.size(); ++i) {
            while (!dq.empty() && nums[dq.back()] <= nums[i]) {
                dq.pop_back();
            }
            dq.push_back(i);
            if (dq.front() <= i - k) {
                dq.pop_front();
            }
            if (i >= k - 1) {
                result.push_back(nums[dq.front()]);
            }
        }
        return result;
    }
};

class Solution {
    public int[] slidingWindowMax(int[] nums, int k) {
        int n = nums.length;
        int[] result = new int[n - k + 1];
        int[] deque = new int[n];
        int head = 0, tail = 0;
        for (int i = 0; i < n; i++) {
            while (head < tail && nums[deque[tail - 1]] <= nums[i]) {
                tail--;
            }
            deque[tail++] = i;
            if (deque[head] <= i - k) {
                head++;
            }
            if (i >= k - 1) {
                result[i - k + 1] = nums[deque[head]];
            }
        }
        return result;
    }
}

class Solution:
    def slidingWindowMax(self, nums: List[int], k: int) -> List[int]:
        from collections import deque
        result = []
        dq = deque()
        for i, num in enumerate(nums):
            while dq and nums[dq[-1]] <= num:
                dq.pop()
            dq.append(i)
            if dq[0] <= i - k:
                dq.popleft()
            if i >= k - 1:
                result.append(nums[dq[0]])
        return result

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {number[]}
 */
var slidingWindowMax = function(nums, k) {
    const result = [];
    const dq = [];
    for (let i = 0; i < nums.length; i++) {
        while (dq.length > 0 && nums[dq[dq.length - 1]] <= nums[i]) {
            dq.pop();
        }
        dq.push(i);
        if (dq[0] <= i - k) {
            dq.shift();
        }
        if (i >= k - 1) {
            result.push(nums[dq[0]]);
        }
    }
    return result;
};

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Pipeline Beacon Partition 19 Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Input

Output

Constraints

More Binary Trees

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Pipeline Beacon Partition 19 — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor