Minimized Cycle Metric — Problem Statement & Solution Guide
Problem Description
You are given an integer array nums and a positive integer k. Consider every contiguous sub‑array (window) of length exactly k. Compute the largest possible sum among all such windows and output this sum. The solution must run in linear time relative to the size of nums and use only O(1) additional memory beyond the input.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Minimized Cycle Metric"
WHY DOES IT MATTER?
Sliding‑window is a fundamental pattern for problems that require aggregating information over contiguous segments. It eliminates redundant recomputation, turning quadratic‑time scans into linear scans, which is crucial for performance‑critical systems handling massive streams of data.
OPTIMIZATION CHALLENGE
The key insight is that the sum of a window can be derived from the previous window by a constant‑time adjustment: remove the leftmost element and add the new rightmost element. This eliminates the O(k) recomputation for each shift.
REAL-WORLD CONNECTION
Think of a moving sensor that records temperature every second; you want the hottest 10‑second interval. Instead of recalculating the total temperature for each 10‑second block, you update the total as the sensor window slides, mirroring how real‑time monitoring dashboards compute rolling metrics.
During an interview, write the initial sum of the first k elements first, then loop from k to n‑1 updating the sum in place. Keep the max variable updated inside the same loop to avoid a second pass.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem asks for the maximum sum of any contiguous sub‑array of length exactly k. A naïve solution would recompute the sum for each window independently, leading to O(n·k) time, which quickly becomes prohibitive when n and k are large (e.g., n = 10^6). The optimal paradigm is the sliding‑window technique: maintain the sum of the current window and update it in O(1) when the window slides one position to the right by subtracting the element that exits the window and adding the new element that enters. This yields a linear‑time algorithm because each array element is added and removed at most once. The approach also satisfies the O(1) auxiliary‑space constraint, as it only stores a few scalar variables (current sum, maximum sum, and indices).
Interview Questions on This Problem
Q1How would you modify the sliding‑window solution to also return the starting index of the window with the maximum sum?
Keep an extra variable startMax that records the start index whenever a new maximum sum is found. When updating maxSum, set startMax = i - k + 1 (where i is the current right‑most index). Return both maxSum and startMax.
Q2If the array contains both positive and negative numbers, does the sliding‑window approach still work for finding the maximum sum of length‑k windows? Why?
Yes. The sliding‑window method does not rely on monotonicity of values; it simply maintains the exact sum of each fixed‑size window. Adding a negative number reduces the sum, but the update rule (subtract left, add right) remains correct, guaranteeing the true maximum is found.
Q3Can you extend this technique to find the maximum average sub‑array of length at least k? Briefly outline the approach.
Use binary search on the answer (average) and transform the array by subtracting the guessed average from each element. Then check if any sub‑array of length ≥ k has a non‑negative sum using a prefix‑sum + minimum‑prefix technique, which runs in O(n) per check, yielding O(n·log precision) overall.
Examples
Input
6 3 4 -1 2 1 -5 4
Output
5
Explanation: All windows of size 3 are: [4, -1, 2] → sum = 5, [-1, 2, 1] → sum = 2, [2, 1, -5] → sum = -2, [1, -5, 4] → sum = 0. The maximum among these sums is 5.
Input
7 4 10 -2 -3 5 7 -1 2
Output
13
Explanation: Windows of size 4: [10, -2, -3, 5] → 10, [-2, -3, 5, 7] → 7, [-3, 5, 7, -1] → 8, [5, 7, -1, 2] → 13. The greatest sum is 13.
Input
5 2 -8 -3 -6 -2 -5
Output
-7
Explanation: Windows of size 2: [-8, -3] → -11, [-3, -6] → -9, [-6, -2] → -8, [-2, -5] → -7. The largest (least negative) sum is -7.
Constraints
- 1 <= nums.length <= 2*10^5
- 1 <= k <= nums.length
- -10^9 <= nums[i] <= 10^9
- The algorithm must run in O(n) time and O(1) extra space.
Optimal Approach & Strategy
Use a sliding window: maintain the sum of the current window, update it in O(1) when moving the window one step, and track the maximum sum encountered.
Brute Force Approach
Compute the sum of every possible length‑k window independently by iterating over each start index and summing k elements each time.
Verified Code Solutions
function solution(nums, k) {
if (k > nums.length) {
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}
let maxSum = -Infinity;
for (let i = 0; i <= nums.length - k; i++) {
let windowSum = 0;
for (let j = 0; j < k; j++) {
windowSum += nums[i + j];
}
maxSum = Math.max(maxSum, windowSum);
}
return maxSum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
if (k > nums.size()) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
int maxSum = INT_MIN;
for (int i = 0; i <= nums.size() - k; i++) {
int windowSum = 0;
for (int j = 0; j < k; j++) {
windowSum += nums[i + j];
}
maxSum = max(maxSum, windowSum);
}
return maxSum;
}
};class Solution {
public int solution(int[] nums, int k) {
if (k > nums.length) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
int maxSum = Integer.MIN_VALUE;
for (int i = 0; i <= nums.length - k; i++) {
int windowSum = 0;
for (int j = 0; j < k; j++) {
windowSum += nums[i + j];
}
maxSum = Math.max(maxSum, windowSum);
}
return maxSum;
}
}def solution(nums, k):
if k > len(nums):
return sum(nums)
max_sum = float('-inf')
for i in range(len(nums) - k + 1):
window_sum = sum(nums[i:i+k])
max_sum = max(max_sum, window_sum)
return max_sumfunction solution(nums, k) {
if (k > nums.length) {
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}
let maxSum = -Infinity;
for (let i = 0; i <= nums.length - k; i++) {
let windowSum = 0;
for (let j = 0; j < k; j++) {
windowSum += nums[i + j];
}
maxSum = Math.max(maxSum, windowSum);
}
return maxSum;
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