Network Network Optimizer 36 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing network and network metrics, construct an optimal algorithm to evaluate and compute the target optimizer value under given operational constraints. The target optimizer value is the maximum value in the array that is less than or equal to the target value K.
Examples
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 30, 31, 32, 33, 34, 35, 36]
Output
30
Explanation: Step-by-step: Given the array [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 30, 31, 32, 33, 34, 35, 36] and target value K = 30, we perform a binary search to find the maximum value less than or equal to 30. The binary search process is as follows: 1. We start by comparing the middle element (15) with the target value (30). Since 15 is less than 30, we move to the right half of the array. 2. We repeat the process by comparing the middle element (30) with the target value (30). Since 30 is equal to 30, we return the previous element (29) as the maximum value less than or equal to 30.
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 30, 31, 32, 33, 34, 35, 36]
Output
29
Explanation: Step-by-step: Given the array [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 30, 31, 32, 33, 34, 35, 36] and target value K = 29, we perform a binary search to find the maximum value less than or equal to 29. The binary search process is as follows: 1. We start by comparing the middle element (15) with the target value (29). Since 15 is less than 29, we move to the right half of the array. 2. We repeat the process by comparing the middle element (29) with the target value (29). Since 29 is equal to 29, we return the previous element (28) as the maximum value less than or equal to 29.
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) {
let left = 0;
let right = nums.length - 1;
while (left <= right) {
let mid = Math.floor((left + right) / 2);
if (nums[mid] <= K) {
if (mid === nums.length - 1 || nums[mid + 1] > K) {
return nums[mid];
}
left = mid + 1;
} else {
right = mid - 1;
}
}
return -1;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
int left = 0;
int right = nums.size() - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] <= K) {
if (mid == nums.size() - 1 || nums[mid + 1] > K) {
return nums[mid];
}
left = mid + 1;
} else {
right = mid - 1;
}
}
return -1;
}
};class Solution {
public int solution(int[] nums, int K) {
int left = 0;
int right = nums.length - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] <= K) {
if (mid == nums.length - 1 || nums[mid + 1] > K) {
return nums[mid];
}
left = mid + 1;
} else {
right = mid - 1;
}
}
return -1;
}
}def solution(nums, K):
left = 0
right = len(nums) - 1
while left <= right:
mid = (left + right) // 2
if nums[mid] <= K:
if mid == len(nums) - 1 or nums[mid + 1] > K:
return nums[mid]
left = mid + 1
else:
right = mid - 1
return -1function solution(nums, K) {
let left = 0;
let right = nums.length - 1;
while (left <= right) {
let mid = Math.floor((left + right) / 2);
if (nums[mid] <= K) {
if (mid === nums.length - 1 || nums[mid + 1] > K) {
return nums[mid];
}
left = mid + 1;
} else {
right = mid - 1;
}
}
return -1;
}Asked in Top Tech Interviews
Solve in Interative Editor
Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.