Protocol Tome Synthesizer 36 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing protocol and tome metrics, construct an optimal algorithm to evaluate and compute the target synthesizer value under given operational constraints. The algorithm should handle the case when the input array is empty or contains only one element correctly, and it should specify what to do when the sum of elements from the start pointer to the end pointer equals K.
Examples
Input
[1, 2, 3, 4, 5]
Output
15
Explanation: Step-by-step: with input [1, 2, 3, 4, 5], we initialize two pointers, start and end, to the beginning of the array. We then calculate the sum of elements from start to end, which is 1 + 2 + 3 + 4 + 5 = 15. Since the sum equals K, we return the sum.
Input
[10, 20, 30, 40, 50, 60, 70, 80, 90]
Output
250
Explanation: Step-by-step: with input [10, 20, 30, 40, 50, 60, 70, 80, 90], we initialize two pointers, start and end, to the beginning of the array. We then calculate the sum of elements from start to end, which is 10 + 20 + 30 + 40 + 50 + 60 + 70 + 80 + 90 = 250. Since the sum equals K, we return the sum.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use BFS / Union Find 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 || K === 0) return 0;
let start = 0;
let end = 0;
let sum = 0;
while (start < nums.length && end < nums.length) {
sum += nums[end];
if (sum === K) return sum;
end++;
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
if (nums.size() === 0 || K === 0) return 0;
int start = 0;
int end = 0;
int sum = 0;
while (start < nums.size() && end < nums.size()) {
sum += nums[end];
if (sum === K) return sum;
end++;
}
return sum;
}
};class Solution {
public int solution(int[] nums, int K) {
if (nums.length === 0 || K === 0) return 0;
int start = 0;
int end = 0;
int sum = 0;
while (start < nums.length && end < nums.length) {
sum += nums[end];
if (sum === K) return sum;
end++;
}
return sum;
}
}def solution(nums, K):
if len(nums) === 0 or K === 0:
return 0
start = 0
end = 0
sum = 0
while start < len(nums) and end < len(nums):
sum += nums[end]
if sum === K:
return sum
end += 1
return sumfunction solution(nums, K) {
if (nums.length === 0 || K === 0) return 0;
let start = 0;
let end = 0;
let sum = 0;
while (start < nums.length && end < nums.length) {
sum += nums[end];
if (sum === K) return sum;
end++;
}
return sum;
}Asked in Top Tech Interviews
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