Node Vault Synthesizer 40 — Problem Statement & Solution Guide

StackEasyDFS Traversal

TimeO(n)

SpaceO(k)

Problem Description

Given a sequence of data elements representing node and vault metrics, construct an optimal algorithm to evaluate and compute the target synthesizer value under given operational constraints. The algorithm should handle cases where the input array has less than k elements by returning the sum of all elements in the array.

DSA Pattern Breakdown

"Node Vault Synthesizer 40"

easy

WHY DOES IT MATTER?

Maintaining a fixed‑size window efficiently is a core pattern for real‑time analytics and streaming data.

OPTIMIZATION CHALLENGE

The key is reducing repeated summations from O(k) per step to O(1) by using a stack to track and discard the oldest value.

REAL-WORLD CONNECTION

Think of a sensor that reports the last k readings to compute a moving average for control systems.

Initialize a running sum variable and update it alongside stack operations to avoid extra traversals.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)

💾 Space:O(k)

Core Theory — Why This Approach?

The problem reduces to maintaining a running total of the first k elements in a stream of numbers. A naïve approach would repeatedly sum the first k items for each query, leading to O(n·k) time, which explodes for large n or k. By leveraging a stack, we can push each incoming element while keeping a parallel cumulative sum; when the stack size exceeds k, we pop the oldest element and subtract its value, thus preserving the exact sum of the most recent k elements in O(1) amortized time per operation. This technique embodies the optimal sliding‑window paradigm: constant‑time updates using a data structure that supports both push and pop, which a stack (or deque) provides efficiently.

Interview Questions on This Problem

Q1Why is a simple loop that recomputes the sum for every new element inefficient for large inputs?

It repeats work already done, resulting in O(n·k) time instead of linear. Each iteration re‑adds up to k values that were previously summed.

Q2How does a stack enable O(1) updates when maintaining a window of size k?

The stack stores elements in order, allowing the newest element to be pushed and the oldest to be popped in constant time. Adjusting the running sum with the pushed and popped values yields an O(1) update.

Q3What edge case must be handled when the array length is smaller than k?

The algorithm should return the sum of all available elements rather than attempting to pop nonexistent items. This is typically handled by checking the stack size before popping.

Examples

Example 1

Input

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10] and k = 3

Output

Explanation: Step-by-step: First, sort the array in descending order. Then, sum the first k elements. In this case, the sorted array is [10, 9, 8, 7, 6, 5, 4, 3, 2, 1] and the sum of the first 3 elements is 10 + 9 + 8 = 27. However, we need to consider the next 2 elements as well, which are 7 and 6. Therefore, the final sum is 27 + 7 + 6 = 40. But since the problem asks for the sum of the 3 largest values, we should only consider the first 3 elements, which are 10, 9, and 8. So the correct sum is 10 + 9 + 8 = 27. But since the problem asks for the sum of the 3 largest values, we should only consider the first 3 elements, which are 10, 9, and 8. So the correct sum is 10 + 9 + 8 = 27. But since the problem asks for the sum of the 3 largest values, we should only consider the first 3 elements, which are 10, 9, and 8. So the correct sum is 10 + 9 + 8 = 27.

Example 2

Input

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10] and k = 5

Output

Constraints

1 <= N <= 10^5
-10^4 <= metrics[i] <= 10^4
1 <= K <= N

Optimal Approach & Strategy

Use a stack with a running sum, pushing new elements and popping when size > k, achieving O(n) time and O(k) space.

Brute Force Approach

Iterate over the first k elements for each query and recompute the sum from scratch, leading to O(n·k) time.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n)

function solution(nums, k) {
    nums.sort((a, b) => b - a);
    let sum = 0;
    for (let i = 0; i < k; i++) {
      sum += nums[i];
    }
    return sum;
}

class Solution {
public:
    int solution(vector<int>& nums, int k) {
        sort(nums.begin(), nums.end(), greater<int>());
        int sum = 0;
        for (int i = 0; i < k; i++) {
            sum += nums[i];
        }
        return sum;
    }
};

class Solution {
    public int solution(int[] nums, int k) {
        Arrays.sort(nums);
        int sum = 0;
        for (int i = 0; i < k; i++) {
            sum += nums[i];
        }
        return sum;
    }
}

def solution(nums, k):
    nums.sort(reverse=True)
    return sum(nums[:k])

function solution(nums, k) {
    nums.sort((a, b) => b - a);
    let sum = 0;
    for (let i = 0; i < k; i++) {
      sum += nums[i];
    }
    return sum;
}

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Node Vault Synthesizer 40 Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Stack

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Node Vault Synthesizer 40 — 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