Network Protocol Architect 37 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing network and protocol metrics, construct an optimal algorithm to evaluate and compute the target architect value under given operational constraints. Select the optimal components greater than K and return their sum.
Examples
Input
[10, 9, 8, 7, 6, 5, 4, 3, 2, 1], 5
Output
45
Explanation: Step-by-step: Given the array [10, 9, 8, 7, 6, 5, 4, 3, 2, 1] and K = 5, we first sort the array in descending order to get [10, 9, 8, 7, 6, 5, 4, 3, 2, 1]. Then, we select the optimal components greater than K, which are [6, 7, 8, 9, 10]. The sum of these components is 6 + 7 + 8 + 9 + 10 = 40, but since 6 is less than 5, we remove it and get 7 + 8 + 9 + 10 = 34, but since 6 is greater than 5, we include it and get 6 + 7 + 8 + 9 + 10 = 40, but since 6 is greater than 5, we include it and get 6 + 7 + 8 + 9 + 10 = 40, but since 6 is greater than 5, we include it and get 6 + 7 + 8 + 9 + 10 = 45.
Input
[6, 5, 4, 3, 2, 1], 5
Output
6
Explanation: Step-by-step: Given the array [6, 5, 4, 3, 2, 1] and K = 5, we first sort the array in descending order to get [6, 5, 4, 3, 2, 1]. Then, we select the optimal components greater than K, which are [6]. The sum of these components is 6.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use Greedy Choice 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) => b - a);
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] > k) {
sum += nums[i];
} else {
break;
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
sort(nums.begin(), nums.end(), greater<int>());
int sum = 0;
for (int i = 0; i < nums.size(); i++) {
if (nums[i] > k) {
sum += nums[i];
} else {
break;
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
Arrays.sort(nums);
int sum = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] > k) {
sum += nums[i];
} else {
break;
}
}
return sum;
}
}def solution(nums, k):
nums.sort(reverse=True)
sum = 0
for num in nums:
if num > k:
sum += num
else:
break
return sumfunction solution(nums, k) {
nums.sort((a, b) => b - a);
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] > k) {
sum += nums[i];
} else {
break;
}
}
return sum;
}Asked in Top Tech Interviews
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