Sensor Cluster Tracker 36 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing sensor and cluster metrics, construct an optimal algorithm to evaluate and compute the target tracker value under given operational constraints.
Examples
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 15
Output
40
Explanation: Step-by-step: Given the input [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] and K = 15, we need to find the maximum sum of a subarray that does not exceed K. The subarray [6, 7, 8, 9, 10] has a sum of 40, which is greater than K. However, the subarray [1, 2, 3, 4, 5] has a sum of 15, which is equal to K. Therefore, the output is 40.
Input
[1, 2, 3, 4, 5], 15
Output
15
Explanation: Step-by-step: Given the input [1, 2, 3, 4, 5] and K = 15, we need to find the maximum sum of a subarray that does not exceed K. The subarray [1, 2, 3, 4, 5] has a sum of 15, which is equal to K. Therefore, the output is 15.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use 2D Grid DP technique to process inputs in O(N) linear time.
Brute Force Approach
Check all possible combinations in O(N^2) time.
Verified Code Solutions
function solution(nums, K) {
let maxSum = 0;
let currentSum = 0;
let windowStart = 0;
for (let windowEnd = 0; windowEnd < nums.length; windowEnd++) {
currentSum += nums[windowEnd];
while (currentSum > K && windowStart <= windowEnd) {
currentSum -= nums[windowStart];
windowStart++;
}
maxSum = Math.max(maxSum, currentSum);
}
return maxSum;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
int maxSum = 0;
int currentSum = 0;
int windowStart = 0;
for (int windowEnd = 0; windowEnd < nums.size(); windowEnd++) {
currentSum += nums[windowEnd];
while (currentSum > K && windowStart <= windowEnd) {
currentSum -= nums[windowStart];
windowStart++;
}
maxSum = max(maxSum, currentSum);
}
return maxSum;
}
};class Solution {
public int solution(int[] nums, int K) {
int maxSum = 0;
int currentSum = 0;
int windowStart = 0;
for (int windowEnd = 0; windowEnd < nums.length; windowEnd++) {
currentSum += nums[windowEnd];
while (currentSum > K && windowStart <= windowEnd) {
currentSum -= nums[windowStart];
windowStart++;
}
maxSum = Math.max(maxSum, currentSum);
}
return maxSum;
}
}def solution(nums, K):
max_sum = 0
current_sum = 0
window_start = 0
for window_end in range(len(nums)):
current_sum += nums[window_end]
while current_sum > K and window_start <= window_end:
current_sum -= nums[window_start]
window_start += 1
max_sum = max(max_sum, current_sum)
return max_sumfunction solution(nums, K) {
let maxSum = 0;
let currentSum = 0;
let windowStart = 0;
for (let windowEnd = 0; windowEnd < nums.length; windowEnd++) {
currentSum += nums[windowEnd];
while (currentSum > K && windowStart <= windowEnd) {
currentSum -= nums[windowStart];
windowStart++;
}
maxSum = Math.max(maxSum, currentSum);
}
return maxSum;
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