Pipeline Grid Extractor 37 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing pipeline and grid metrics, construct an optimal algorithm to evaluate and compute the target extractor value under given operational constraints. The algorithm should select components greater than K and return their sum.
Examples
Input
[55, 15, 10, 5, 20, 30, 40, 50, 60, 70, 80, 90, 100]
Output
55
Explanation: Step-by-step: Given the input [55, 15, 10, 5, 20, 30, 40, 50, 60, 70, 80, 90, 100], we first sort the array in ascending order. Then, we find the sum of all numbers (55 + 15 + 10 + 5 + 20 + 30 + 40 + 50 + 60 + 70 + 80 + 90 + 100 = 585). Next, we find the sum of numbers less than or equal to K (15 + 10 + 5 = 30). Finally, we subtract the sum of numbers less than or equal to K from the sum of all numbers to get the target extractor value (585 - 30 = 555). However, since we are only interested in the numbers greater than K, we should only consider the numbers greater than 15. The sum of these numbers is (55 + 20 + 30 + 40 + 50 + 60 + 70 + 80 + 90 + 100 = 555).
Input
[55, 10, 5, 20, 30, 40, 50, 60, 70, 80, 90, 100]
Output
55
Explanation: Step-by-step: Given the input [55, 10, 5, 20, 30, 40, 50, 60, 70, 80, 90, 100], we first sort the array in ascending order. Then, we find the sum of all numbers (55 + 10 + 5 + 20 + 30 + 40 + 50 + 60 + 70 + 80 + 90 + 100 = 520). Next, we find the sum of numbers less than or equal to K (10 + 5 = 15). Finally, we subtract the sum of numbers less than or equal to K from the sum of all numbers to get the target extractor value (520 - 15 = 505). However, since we are only interested in the numbers greater than K, we should only consider the numbers greater than 10. The sum of these numbers is (55 + 20 + 30 + 40 + 50 + 60 + 70 + 80 + 90 + 100 = 505).
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use Fixed/Dynamic Window 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) {
nums.sort((a, b) => a - b);
let sumAll = 0;
let sumLessThanK = 0;
for (let num of nums) {
if (num > K) {
sumAll += num;
} else {
sumLessThanK += num;
}
}
return sumAll - sumLessThanK;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
sort(nums.begin(), nums.end());
int sumAll = 0;
int sumLessThanK = 0;
for (int num : nums) {
if (num > K) {
sumAll += num;
} else {
sumLessThanK += num;
}
}
return sumAll - sumLessThanK;
}
};class Solution {
public int solution(int[] nums, int K) {
Arrays.sort(nums);
int sumAll = 0;
int sumLessThanK = 0;
for (int num : nums) {
if (num > K) {
sumAll += num;
} else {
sumLessThanK += num;
}
}
return sumAll - sumLessThanK;
}
}def solution(nums, K):
nums.sort()
sum_all = 0
sum_less_than_k = 0
for num in nums:
if num > K:
sum_all += num
else:
sum_less_than_k += num
return sum_all - sum_less_than_kfunction solution(nums, K) {
nums.sort((a, b) => a - b);
let sumAll = 0;
let sumLessThanK = 0;
for (let num of nums) {
if (num > K) {
sumAll += num;
} else {
sumLessThanK += num;
}
}
return sumAll - sumLessThanK;
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