Matrix Stream Consolidator 38 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing matrix and stream metrics, construct an optimal algorithm to evaluate and compute the target consolidator value under given operational constraints. The algorithm should maintain a sliding window of size 3 and return the maximum sum of the window.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Matrix Stream Consolidator 38"
WHY DOES IT MATTER?
Sliding windows turn overlapping sub‑problem recomputation into incremental updates.
OPTIMIZATION CHALLENGE
The key is reducing the per‑step work from O(k) to O(1) for a fixed k.
REAL-WORLD CONNECTION
Network routers compute moving averages of packet latency using the same principle.
Maintain a running sum and update it in‑place; avoid creating slices or copying arrays.
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 window of size k is recomputed from scratch—into a linear pass by reusing information from the previous window. For a fixed window size of 3, the sum of the next window can be obtained by subtracting the element exiting the window and adding the new entrant, eliminating redundant additions. Naïve approaches fail on large streams because they repeatedly traverse overlapping sub‑arrays, leading to unnecessary arithmetic and cache pressure. The optimal paradigm leverages constant‑time updates and a single traversal, guaranteeing O(n) time while using O(1) auxiliary space, which is essential for high‑throughput streaming data.
Interview Questions on This Problem
Q1How does the sliding window technique achieve O(n) time for fixed‑size window problems?
It updates the window sum by removing the leftmost element and adding the new rightmost element. This constant‑time update avoids recomputing the entire sum for each position.
Q2What edge cases must be handled when the input length is less than the window size?
If the array length is smaller than 3, no full window exists, so the function should return a sentinel (e.g., 0 or negative infinity). This prevents out‑of‑bounds access.
Q3Why is O(1) extra space possible even though we track a window?
Only the current sum and a few indices are needed; the window’s elements remain in the original array. No additional data structures are allocated.
Examples
Input
[10, 20, 30, 40, 50, 60, 70, 80, 90, 100]
Output
90
Explanation: Step-by-step: with input [10, 20, 30, 40, 50, 60, 70, 80, 90, 100], we maintain a window of size 3. The maximum sum of the window is 90, which is the target consolidator value.
Input
[5, 10, 15, 20, 25, 30, 35, 40, 45, 50]
Output
40
Explanation: Step-by-step: with input [5, 10, 15, 20, 25, 30, 35, 40, 45, 50], we maintain a window of size 3. The maximum sum of the window is 40, which is the target consolidator value.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Compute the sum of the first three elements, then slide the window, updating the sum by subtracting the leftmost element and adding the new rightmost element, updating the max each step.
Brute Force Approach
Iterate over every possible start index, sum the three elements each time, and track the maximum.
Verified Code Solutions
function solution(nums) {
let windowSum = 0;
let maxSum = -Infinity;
let windowStart = 0;
for (let windowEnd = 0; windowEnd < nums.length; windowEnd++) {
windowSum += nums[windowEnd];
if (windowEnd >= 2) {
windowSum -= nums[windowStart];
windowStart++;
}
maxSum = Math.max(maxSum, windowSum);
}
return maxSum;
}class Solution {
public:
int solution(vector<int>& nums) {
int windowSum = 0;
int maxSum = INT_MIN;
int windowStart = 0;
for (int windowEnd = 0; windowEnd < nums.size(); windowEnd++) {
windowSum += nums[windowEnd];
if (windowEnd >= 2) {
windowSum -= nums[windowStart];
windowStart++;
}
maxSum = max(maxSum, windowSum);
}
return maxSum;
}
};class Solution {
public int solution(int[] nums) {
int windowSum = 0;
int maxSum = Integer.MIN_VALUE;
int windowStart = 0;
for (int windowEnd = 0; windowEnd < nums.length; windowEnd++) {
windowSum += nums[windowEnd];
if (windowEnd >= 2) {
windowSum -= nums[windowStart];
windowStart++;
}
maxSum = Math.max(maxSum, windowSum);
}
return maxSum;
}
}def solution(nums):
window_sum = 0
max_sum = float('-inf')
window_start = 0
for window_end in range(len(nums)):
window_sum += nums[window_end]
if window_end >= 2:
window_sum -= nums[window_start]
window_start += 1
max_sum = max(max_sum, window_sum)
return max_sumfunction solution(nums) {
let windowSum = 0;
let maxSum = -Infinity;
let windowStart = 0;
for (let windowEnd = 0; windowEnd < nums.length; windowEnd++) {
windowSum += nums[windowEnd];
if (windowEnd >= 2) {
windowSum -= nums[windowStart];
windowStart++;
}
maxSum = Math.max(maxSum, windowSum);
}
return maxSum;
}Asked in Top Tech Interviews
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