Matrix Vessel Validator 37 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing matrix and vessel metrics, construct an optimal algorithm to evaluate and compute the target validator value under given operational constraints. The function should add values up to K and handle the case where K is greater than the maximum value in the metrics array.
Examples
Input
[1, 2, 3, 4, 5], 6
Output
15
Explanation: Step-by-step: with input [1, 2, 3, 4, 5], K = 6, we iterate through the array and add values up to K. Since K is less than the maximum value in the array, we add all elements up to the maximum value, which is 5. So, the sum is 1 + 2 + 3 + 4 + 5 = 15.
Input
[1, 2, 3, 4, 5], 10
Output
15
Explanation: Step-by-step: with input [1, 2, 3, 4, 5], K = 10, we iterate through the array and add values up to the maximum value in the array, which is 5. So, the sum is 1 + 2 + 3 + 4 + 5 = 15.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use Inward Pointers 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) return 0;
let max = Math.max(...nums);
if (k >= max) return nums.reduce((a, b) => a + b, 0);
return nums.slice(0, nums.indexOf(max)).reduce((a, b) => a + b, 0);
}class Solution {
public:
int solution(vector<int>& nums, int k) {
if (nums.size() == 0) return 0;
int max = INT_MAX;
for (int num : nums) {
max = min(max, num);
}
if (k >= max) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
} else {
int sum = 0;
for (int num : nums) {
if (num <= k) sum += num;
}
return sum;
}
}class Solution {
public int solution(int[] nums, int k) {
if (nums.length == 0) return 0;
int max = Integer.MAX_VALUE;
for (int num : nums) {
max = Math.min(max, num);
}
if (k >= max) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
} else {
int sum = 0;
for (int num : nums) {
if (num <= k) sum += num;
}
return sum;
}
}def solution(nums, k):
if len(nums) == 0: return 0
max_val = max(nums)
if k >= max_val: return sum(nums)
return sum(nums[:nums.index(max_val)])function solution(nums, k) {
if (nums.length === 0) return 0;
let max = Math.max(...nums);
if (k >= max) return nums.reduce((a, b) => a + b, 0);
return nums.slice(0, nums.indexOf(max)).reduce((a, b) => a + b, 0);
}Asked in Top Tech Interviews
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