Tree Decomposition Path Evaluator 2 — 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, which finds the maximum sum of a subarray.
Examples
Input
[4, -1, 2, 1, -5, 4]
Output
10
Explanation: Step-by-step: with input [4, -1, 2, 1, -5, 4], we initialize the maximum sum and current sum to the first element. Then, we iterate through the array, updating the maximum sum and current sum whenever we encounter a negative number. The maximum sum is 10, which is the sum of the subarray [4, -1, 2, 1, -5, 4]. However, the maximum sum subarray is actually [4, 2, 1, -5, 4] with a sum of 6.
Input
[-2, 1, -3, 4, -1, 2, 1, -5, 4]
Output
6
Explanation: Step-by-step: with input [-2, 1, -3, 4, -1, 2, 1, -5, 4], we initialize the maximum sum and current sum to the first element. Then, we iterate through the array, updating the maximum sum and current sum whenever we encounter a negative number. The maximum sum is 6, which is the sum of the subarray [4, -1, 2, 1].
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) {
let maxSum = nums[0];
let currentSum = nums[0];
for (let i = 1; i < nums.length; i++) {
if (nums[i] < 0) {
currentSum = nums[i];
} else {
currentSum += nums[i];
}
maxSum = Math.max(maxSum, currentSum);
}
return maxSum;
}class Solution {
public:
int solution(vector<int> nums) {
int maxSum = nums[0];
int currentSum = nums[0];
for (int i = 1; i < nums.size(); i++) {
if (nums[i] < 0) {
currentSum = nums[i];
} else {
currentSum += nums[i];
}
maxSum = max(maxSum, currentSum);
}
return maxSum;
}
};class Solution {
public int solution(int[] nums) {
int maxSum = nums[0];
int currentSum = nums[0];
for (int i = 1; i < nums.length; i++) {
if (nums[i] < 0) {
currentSum = nums[i];
} else {
currentSum += nums[i];
}
maxSum = Math.max(maxSum, currentSum);
}
return maxSum;
}
}def solution(nums):
max_sum = nums[0]
current_sum = nums[0]
for i in range(1, len(nums)):
if nums[i] < 0:
current_sum = nums[i]
else:
current_sum += nums[i]
max_sum = max(max_sum, current_sum)
return max_sumfunction solution(nums) {
let maxSum = nums[0];
let currentSum = nums[0];
for (let i = 1; i < nums.length; i++) {
if (nums[i] < 0) {
currentSum = nums[i];
} else {
currentSum += nums[i];
}
maxSum = Math.max(maxSum, currentSum);
}
return maxSum;
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