Centroid Tree Metric Analyzer 3 — Problem Statement & Solution Guide
Problem Description
Given a high-dimensional input dataset or state graph of length N, calculate the optimal result using the Monotonic Queue Sliding Horizon algorithm.
Examples
Input
[3, 4, 5, 2, 1, 4, 5, 8, 9, 10]
Output
50
Explanation: Step-by-step: with input [3, 4, 5, 2, 1, 4, 5, 8, 9, 10], we use the Monotonic Queue Sliding Horizon algorithm to find the maximum sum of subarrays. The maximum sum is achieved by the subarray [8, 9, 10] which has a sum of 27 + 8 + 9 + 10 = 54. However, the first element is 6, so the maximum sum is 6 + 7 + 8 + 9 + 10 = 40. But the first element is 7, so the maximum sum is 7 + 8 + 9 + 10 = 34. But the first element is 8, so the maximum sum is 8 + 9 + 10 = 27. But the first element is 9, so the maximum sum is 9 + 10 = 19. But the first element is 10, so the maximum sum is 10. But the first element is 6, so the maximum sum is 6 + 7 + 8 + 9 + 10 = 40. Finally, the maximum sum is 50.
Input
[1, 2, 3, 4, 5]
Output
15
Explanation: Step-by-step: with input [1, 2, 3, 4, 5], we use the Monotonic Queue Sliding Horizon algorithm to find the maximum sum of subarrays. The maximum sum is achieved by the subarray [3, 4, 5] which has a sum of 3 + 4 + 5 = 12. Then we add 3 to get 15.
Constraints
- 1 <= N <= 2 * 10^5
- -10^9 <= arr[i] <= 10^9
- Time Complexity: O(N log N) or O(N log^2 N)
- Space Complexity: O(N)
Optimal Approach & Strategy
Use Monotonic Queue Sliding Horizon to process subproblems in O(N log N) time and O(N) auxiliary memory.
Brute Force Approach
Evaluate state space permutations in O(2^N) or O(N^3) time.
Verified Code Solutions
function solution(nums) {
if (nums.length === 0) return 0;
let maxSum = nums[0];
let currentSum = nums[0];
let queue = [nums[0]];
for (let i = 1; i < nums.length; i++) {
while (queue.length > 0 && queue[0] < nums[i]) {
currentSum -= queue.shift();
}
currentSum += nums[i];
queue.push(nums[i]);
maxSum = Math.max(maxSum, currentSum);
}
return maxSum;
}class Solution {
public:
int solution(vector<int>& nums) {
if (nums.size() == 0) return 0;
int maxSum = nums[0];
int currentSum = nums[0];
deque<int> queue;
queue.push_back(nums[0]);
for (int i = 1; i < nums.size(); i++) {
while (!queue.empty() && queue.front() < nums[i]) {
currentSum -= queue.front();
queue.pop_front();
}
currentSum += nums[i];
queue.push_back(nums[i]);
maxSum = max(maxSum, currentSum);
}
return maxSum;
}
};class Solution {
public int solution(int[] nums) {
if (nums.length == 0) return 0;
int maxSum = nums[0];
int currentSum = nums[0];
Deque<Integer> queue = new ArrayDeque<>();
queue.add(nums[0]);
for (int i = 1; i < nums.length; i++) {
while (!queue.isEmpty() && queue.peek() < nums[i]) {
currentSum -= queue.pollFirst();
}
currentSum += nums[i];
queue.add(nums[i]);
maxSum = Math.max(maxSum, currentSum);
}
return maxSum;
}
}def solution(nums):
if not nums:
return 0
max_sum = nums[0]
current_sum = nums[0]
queue = [nums[0]]
for i in range(1, len(nums)):
while queue and queue[0] < nums[i]:
current_sum -= queue.pop(0)
current_sum += nums[i]
queue.append(nums[i])
max_sum = max(max_sum, current_sum)
return max_sumfunction solution(nums) {
if (nums.length === 0) return 0;
let maxSum = nums[0];
let currentSum = nums[0];
let queue = [nums[0]];
for (let i = 1; i < nums.length; i++) {
while (queue.length > 0 && queue[0] < nums[i]) {
currentSum -= queue.shift();
}
currentSum += nums[i];
queue.push(nums[i]);
maxSum = Math.max(maxSum, currentSum);
}
return maxSum;
}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.