Protocol Pipeline Partition 37 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing protocol and pipeline metrics, construct an optimal algorithm to evaluate and compute the target partition value under given operational constraints.
Examples
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]
Output
0
Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20], we need to find the sum of all elements greater than K. In this case, K is not specified, so we assume K = 15. Since there are no elements greater than 15, the correct output is 0.
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21]
Output
21
Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21], we need to find the sum of all elements greater than K. In this case, K = 15. Since there are elements greater than 15 (21), the correct output is 21.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use DFS Traversal 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) {
if (nums.length === 0 || nums.length === 1) return 0;
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] > K) sum += nums[i];
}
return sum;
}class Solution {
public:
int solution(vector<int> nums, int K) {
if (nums.size() == 0 || nums.size() == 1) return 0;
int sum = 0;
for (int i = 0; i < nums.size(); i++) {
if (nums[i] > K) sum += nums[i];
}
return sum;
}
};class Solution {
public int solution(int[] nums, int K) {
if (nums.length == 0 || nums.length == 1) return 0;
int sum = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] > K) sum += nums[i];
}
return sum;
}
}def solution(nums, K):
if len(nums) == 0 or len(nums) == 1:
return 0
sum = 0
for i in range(len(nums)):
if nums[i] > K:
sum += nums[i]
return sumfunction solution(nums, K) {
if (nums.length === 0 || nums.length === 1) return 0;
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] > K) sum += nums[i];
}
return sum;
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