Tree Decomposition Path Evaluator — Problem Statement & Solution Guide

StackHardMonotonic Queue Sliding Horizon

TimeO(N)

SpaceO(1)

Quick Answer & Algorithm Key Takeaway

Master coding challenges related to Stack and solve the Tree Decomposition Path Evaluator 2 problem optimally.

TopicStack

PatternMonotonic Queue Sliding Horizon

TimeO(N)

SpaceO(1)

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.

DSA Pattern Breakdown

"Tree Decomposition Path Evaluator"

hard

WHY DOES IT MATTER?

Kadane's Algorithm is a foundational dynamic programming pattern that demonstrates how to solve optimization problems by breaking them into smaller subproblems and reusing previously computed results. It is essential for understanding how to reduce time complexity from quadratic to linear in many array-based problems.

OPTIMIZATION CHALLENGE

The key insight is that if the sum of the subarray ending at the previous index is negative, it will only decrease the sum of any subsequent subarray. Therefore, it is optimal to start a new subarray at the current index rather than extending the previous one.

REAL-WORLD CONNECTION

This pattern is analogous to financial risk management, where you track the maximum profit from a series of transactions. You decide at each step whether to continue holding an asset or to reset your position, ensuring you capture the best possible return without looking back at all previous states.

In interviews, clearly articulate the state transition: 'current_sum = max(arr[i], current_sum + arr[i])'. This shows you understand the dynamic programming recurrence relation and can justify why it works.

COMPLEXITY AT A GLANCE

⏱ Time:O(N)

💾 Space:O(1)

Core Theory — Why This Approach?

The problem statement contains a significant conceptual mismatch: it references a 'Monotonic Queue Sliding Horizon' for finding the maximum sum of a subarray, but the topic is 'Stack' and the title suggests tree decomposition. However, the core algorithmic task described—finding the maximum sum of a contiguous subarray—is classically solved using Kadane's Algorithm, which is a dynamic programming approach, not a stack or monotonic queue problem. Monotonic queues are typically used for sliding window maximum/minimum problems, not maximum subarray sum. If we strictly adhere to the 'Stack' topic, a stack-based approach for maximum subarray sum is non-standard and inefficient compared to Kadane's. Therefore, the optimal paradigm here is Kadane's Algorithm, which leverages the principle of optimal substructure: the maximum subarray ending at index i is either the element itself or the sum of the element and the maximum subarray ending at i-1. This reduces the problem to a linear scan with constant space.

Interview Questions on This Problem

Q1How would you adapt Kadane's Algorithm to return the indices of the maximum subarray, not just the sum?

Maintain two additional variables: start and end to track the current subarray boundaries, and globalStart and globalEnd to track the best subarray found so far. When the current sum drops below zero, reset the current sum to zero and update start to i+1. When the current sum exceeds the global maximum, update the global maximum and copy start/end to globalStart/globalEnd.

Q2What is the time and space complexity of Kadane's Algorithm, and why is it superior to a brute-force approach?

Kadane's Algorithm runs in O(N) time and O(1) space. It is superior to the brute-force O(N^2) or O(N^3) approaches because it exploits the optimal substructure of the problem, avoiding redundant calculations by reusing the result of the previous subproblem.

Q3How would you handle the case where all elements in the array are negative?

Kadane's Algorithm naturally handles this case by initializing the maximum sum to the first element and updating it if a larger (less negative) element is found. The key is to ensure that the current sum is reset only when it becomes negative, but the global maximum is updated correctly even if all values are negative.

Examples

Example 1

Input

[4, -1, 2, 1, -5, 4]

Output

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.

Example 2

Input

[-2, 1, -3, 4, -1, 2, 1, -5, 4]

Output

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

The optimized approach uses Kadane's Algorithm, which iterates through the array once, maintaining a running sum that resets to zero if it becomes negative. This ensures that only the most promising subarrays are considered, reducing the time complexity to O(N).

Brute Force Approach

The brute-force approach involves checking all possible subarrays by using two nested loops to calculate the sum of each subarray and keeping track of the maximum sum found. This results in a time complexity of O(N^2) or O(N^3) depending on implementation.

Code Solutions

Language:

JavaScript Solution

Time: O(N)

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_sum

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;
}

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