Pipeline Grid Resolver 31 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing pipeline and grid metrics, construct an optimal algorithm to evaluate and compute the target resolver value under given operational constraints. Handle edge cases where the input array is empty or null, or where the value of K is greater than the length of the input array. If the sum of the first K elements is 0, handle it according to the problem statement.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Pipeline Grid Resolver 31"
WHY DOES IT MATTER?
Sliding windows turn quadratic scans into linear ones, crucial for real‑time analytics.
OPTIMIZATION CHALLENGE
The key is to eliminate repeated work by reusing the previous window's state.
REAL-WORLD CONNECTION
Network routers maintain a moving average of packet latency using the same principle.
Always validate K and array length first; then keep a single running sum variable.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The sliding window technique transforms a naïve O(N·K) scan—where each possible subarray of length K is summed independently—into a linear pass by reusing the previous window's sum. By subtracting the element exiting the window and adding the new entrant, we maintain the current sum in O(1) time, yielding an overall O(N) algorithm. Naïve approaches fail on large inputs because they recompute overlapping sums, causing redundant work and exceeding time limits. The optimal paradigm leverages constant‑time incremental updates and minimal auxiliary storage, embodying the essence of amortized analysis in array processing.
Interview Questions on This Problem
Q1How does the sliding window reduce time complexity compared to the brute‑force method?
It updates the sum by removing the leftmost element and adding the new rightmost element, avoiding recomputation of the entire window. This changes the per‑window cost from O(K) to O(1).
Q2What edge cases must be handled when K is larger than the array length?
If K > N, the problem is undefined; most solutions return 0 or an error. You should explicitly check and handle this before processing.
Q3Can the sliding window be adapted to find the maximum average instead of sum?
Yes, compute the window sum as usual and divide by K to get the average; the max average follows the same window updates. The complexity remains O(N).
Examples
Input
[10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130, 140, 150, 160, 170, 180, 190, 200]
Output
150
Explanation: Step-by-step: Given the input array [10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130, 140, 150, 160, 170, 180, 190, 200], we need to find the target resolver value. The sum of the first K elements (K = 15) is 1500. However, the problem statement requires handling the case where the sum of the first K elements is 0. In this case, we should return 0. Therefore, the output is 0.
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Output
0
Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], we need to find the target resolver value. The sum of the first K elements (K = 10) is 55. However, the problem statement requires handling the case where the sum of the first K elements is 0. Therefore, the output is 0.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Initialize the sum of the first K elements, then slide the window, updating the sum by subtracting the leftmost element and adding the new rightmost one, tracking the max; this is O(N) time.
Brute Force Approach
Iterate over every possible start index, sum K elements each time, and keep the maximum; this is O(N·K) time.
Verified Code Solutions
function solution(nums, k) {
if (k > nums.length || nums.length === 0) {
return 0;
}
let sum = 0;
for (let i = 0; i < k; i++) {
sum += nums[i];
}
if (sum === 0) {
return 0;
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
if (k > nums.size() || nums.size() == 0) {
return 0;
}
int sum = 0;
for (int i = 0; i < k; i++) {
sum += nums[i];
}
if (sum == 0) {
return 0;
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
if (k > nums.length || nums.length == 0) {
return 0;
}
int sum = 0;
for (int i = 0; i < k; i++) {
sum += nums[i];
}
if (sum == 0) {
return 0;
}
return sum;
}
}def solution(nums, k):
if k > len(nums) or len(nums) == 0:
return 0
total = 0
for i in range(k):
total += nums[i]
if total == 0:
return 0
return totalfunction solution(nums, k) {
if (k > nums.length || nums.length === 0) {
return 0;
}
let sum = 0;
for (let i = 0; i < k; i++) {
sum += nums[i];
}
if (sum === 0) {
return 0;
}
return sum;
}Asked in Top Tech Interviews
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