Payload Token Optimizer 34 — Problem Statement & Solution Guide
Problem Description
You are tasked with optimizing the allocation of computational resources in a distributed system. You are given an array metrics of length n, where each element represents a non-negative integer cost associated with a specific task. You are also given a target capacity K. Your objective is to select a subset of these tasks such that the sum of their costs is as large as possible without exceeding the capacity K. If multiple subsets yield the same maximum sum, any valid subset is acceptable. Return the maximum possible sum of the selected metrics.
This problem requires an efficient algorithm to handle large input sizes. A brute-force approach that checks all possible subsets is computationally infeasible for large n. You must design a solution that leverages the properties of the input data to achieve optimal performance. Consider the trade-offs between sorting, dynamic programming, and greedy strategies, keeping in mind the constraints on the values and the length of the array.
The input will consist of the array metrics and the integer K. The output should be a single integer representing the maximum sum of the selected subset. If no subset can be formed (e.g., all elements are greater than K), the result should be 0.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Payload Token Optimizer 34"
WHY DOES IT MATTER?
Sliding‑window turns a quadratic sub‑array problem into linear time, essential for real‑time systems.
OPTIMIZATION CHALLENGE
The key is reducing repeated sum recomputation by maintaining a running total.
REAL-WORLD CONNECTION
It mirrors how load balancers allocate contiguous request batches without exceeding capacity.
Always keep the window sum as a mutable variable; avoid recomputing sums from scratch inside the loop.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The two‑pointer (sliding‑window) technique exploits the monotonicity of cumulative sums when all numbers are non‑negative. By expanding the right pointer until the sum exceeds K and then contracting the left pointer, we can examine every feasible sub‑array in linear time, something a naïve O(n²) enumeration cannot achieve for large n. The optimal paradigm treats the window as a dynamic interval whose sum is maintained incrementally, allowing constant‑time updates and guaranteeing each element is visited at most twice, which yields O(n) time and O(1) extra space.
Interview Questions on This Problem
Q1Why does the sliding‑window approach require non‑negative numbers?
With non‑negative values, expanding the window never decreases the sum, so once it exceeds K we can safely move the left pointer. Negative numbers break this monotonicity, invalidating the O(n) guarantee.
Q2How would you modify the algorithm to also return the actual indices of the optimal sub‑array?
Track the start index when the window expands and update a best‑range variable whenever a longer valid window is found. At the end, return that stored range.
Q3Can this technique be adapted to count all sub‑arrays with sum ≤ K? If so, how?
Yes; for each right pointer position, the number of valid sub‑arrays ending at that index equals (right‑left+1). Accumulate this count while sliding the window.
Examples
Input
metrics = [3, 1, 4, 1, 5], K = 9
Output
9
Explanation: The possible subsets and their sums are: [3,1,4,1] = 9, [3,1,5] = 9, [4,1,5] = 10 (exceeds K), [3,4,1] = 8, etc. The maximum sum not exceeding 9 is 9, achieved by subsets like [3,1,4,1] or [3,1,5].
Input
metrics = [10, 20, 30], K = 25
Output
20
Explanation: The elements are 10, 20, and 30. The subset [10, 20] sums to 30, which exceeds K=25. The subset [20] sums to 20, which is within K. The subset [10] sums to 10. The maximum valid sum is 20.
Input
metrics = [1, 2, 3, 4, 5], K = 15
Output
15
Explanation: The sum of all elements is 1+2+3+4+5 = 15, which is exactly equal to K. Therefore, the maximum sum is 15.
Input
metrics = [7, 8, 9], K = 5
Output
0
Explanation: All elements (7, 8, 9) are greater than K=5. No single element or combination of elements can form a sum <= 5. Thus, the maximum sum is 0.
Constraints
- 1 <= metrics.length <= 10^5
- 0 <= metrics[i] <= 10^9
- 0 <= K <= 10^14
Optimal Approach & Strategy
Maintain a moving window with two pointers and a running sum; expand right, contract left when sum > K, updating the best length – O(n) time.
Brute Force Approach
Check every possible sub‑array, compute its sum, and keep the longest that does not exceed K – O(n²) time.
Verified Code Solutions
/**
* @param {number[]} metrics
* @param {number} K
* @return {number}
*/
var optimizePayload = function(metrics, K) {
let n = metrics.length;
let left = 0;
let right = 0;
let currentSum = 0;
let maxSum = 0;
while (right < n) {
currentSum += metrics[right];
while (currentSum > K && left <= right) {
currentSum -= metrics[left];
left++;
}
if (currentSum <= K) {
maxSum = Math.max(maxSum, currentSum);
}
right++;
}
return maxSum;
};class Solution {
public:
int optimizePayload(vector<int>& metrics, int K) {
int n = metrics.size();
int left = 0;
int right = 0;
int currentSum = 0;
int maxSum = 0;
while (right < n) {
currentSum += metrics[right];
while (currentSum > K && left <= right) {
currentSum -= metrics[left];
left++;
}
if (currentSum <= K) {
maxSum = max(maxSum, currentSum);
}
right++;
}
return maxSum;
}
};class Solution {
public int optimizePayload(int[] metrics, int K) {
int n = metrics.length;
int left = 0;
int right = 0;
int currentSum = 0;
int maxSum = 0;
while (right < n) {
currentSum += metrics[right];
while (currentSum > K && left <= right) {
currentSum -= metrics[left];
left++;
}
if (currentSum <= K) {
maxSum = Math.max(maxSum, currentSum);
}
right++;
}
return maxSum;
}
}class Solution:
def optimizePayload(self, metrics: List[int], K: int) -> int:
n = len(metrics)
left = 0
right = 0
current_sum = 0
max_sum = 0
while right < n:
current_sum += metrics[right]
while current_sum > K and left <= right:
current_sum -= metrics[left]
left += 1
if current_sum <= K:
max_sum = max(max_sum, current_sum)
right += 1
return max_sum/**
* @param {number[]} metrics
* @param {number} K
* @return {number}
*/
var optimizePayload = function(metrics, K) {
let n = metrics.length;
let left = 0;
let right = 0;
let currentSum = 0;
let maxSum = 0;
while (right < n) {
currentSum += metrics[right];
while (currentSum > K && left <= right) {
currentSum -= metrics[left];
left++;
}
if (currentSum <= K) {
maxSum = Math.max(maxSum, currentSum);
}
right++;
}
return maxSum;
};Asked in Top Tech Interviews
Solve in Interative Editor
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