Tree Decomposition Path Validator 3 — Problem Statement & Solution Guide

StackHardMonotonic Queue Sliding Horizon

TimeO(n)

SpaceO(1)

Problem Description

You are tasked with optimizing a data pipeline that processes a continuous stream of sensor readings. The system requires identifying the most significant contiguous segment of data within a fixed sliding window of size K. Specifically, given an array of integers representing the sensor values and an integer K, determine the maximum sum of any contiguous subarray of exactly length K.

To achieve optimal performance for large datasets, you must utilize a Monotonic Queue (or Deque) to maintain the indices of elements in a way that allows for O(1) amortized time complexity for both insertion and removal operations. The queue should store indices such that the corresponding values in the array are in decreasing order. This structure ensures that the front of the queue always holds the index of the maximum value within the current valid window, allowing for efficient tracking of the window's sum as it slides across the array.

Your function should return the maximum sum encountered among all valid windows of size K. If the array length is less than K, return 0 or handle it as per standard edge-case conventions, though constraints guarantee N >= K.

DSA Pattern Breakdown

"Tree Decomposition Path Validator 3"

hard

WHY DOES IT MATTER?

The sliding window pattern reduces time complexity from O(n*K) to O(n), which is critical when processing high‑volume sensor data streams or real‑time analytics. It also keeps memory usage minimal, enabling deployment on resource‑constrained edge devices.

OPTIMIZATION CHALLENGE

The key insight is that consecutive subarrays of length K share K-1 elements. By reusing the previous sum and updating it with two constant‑time operations (subtract and add), we avoid recomputing the entire sum each time.

REAL-WORLD CONNECTION

Think of a camera capturing a moving scene: you only need to process the current frame and discard the old one. Similarly, the sliding window processes only the current segment of data, discarding the past, which mirrors how streaming systems like Kafka or Flink handle windowed aggregations.

When explaining this in an interview, emphasize the overlap between windows and show a quick diagram or code snippet that updates the sum in O(1). Also mention edge cases like negative numbers and K equal to array length to demonstrate thoroughness.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)

💾 Space:O(1)

Core Theory — Why This Approach?

The problem asks for the maximum sum of any contiguous subarray of exactly length K in an array of integers. A naive approach would iterate over every possible starting index i (0 ≤ i ≤ n-K) and compute the sum of the K elements starting at i, resulting in an O(n*K) time complexity. This becomes infeasible for large n (e.g., 10^6) because the number of operations grows quadratically.

The optimal paradigm is the sliding‑window technique. By first computing the sum of the first K elements, we can then move the window one step at a time: subtract the element that leaves the window and add the new element that enters. Each step takes O(1) time, so the entire array is processed in O(n) time. The space usage is O(1) because we only store the current window sum and the maximum found so far.

This pattern is a classic example of transforming a nested‑loop problem into a single pass by exploiting the overlap between consecutive subarrays. It demonstrates how careful bookkeeping can reduce time complexity from quadratic to linear, which is essential for real‑time data pipelines and large‑scale analytics.

Interview Questions on This Problem

Q1How would you modify the sliding window solution if the window size K could vary during runtime?

If K changes, you can maintain a prefix sum array so that any subarray sum can be retrieved in O(1). For a dynamic K, you would recompute the initial window sum for the new K and then slide as before. Alternatively, you could use a deque to maintain the last K elements and adjust its size when K changes, but this adds complexity and may not be necessary if K changes infrequently.

Q2A fintech platform needs to detect the highest 5‑minute trading volume in a 24‑hour stream of trades. Which algorithmic pattern would you use and why?

I would use the sliding window pattern because the window size (5 minutes) is fixed and the stream is continuous. By maintaining a running sum of trade volumes within the current 5‑minute window and updating it as new trades arrive and old ones expire, we can compute the maximum volume in O(1) per trade, ensuring real‑time performance.

Q3During a coding interview at a high‑growth startup, you are asked to find the maximum sum subarray of length K in an array that may contain negative numbers. What pitfalls should you watch out for?

You must initialize the maximum sum to the sum of the first K elements, not to zero, because all numbers could be negative. Also, ensure that you handle the case where K equals the array length, and avoid integer overflow by using a 64‑bit integer type if the sums can be large.

Examples

Example 1

Input

nums = [2, 1, 5, 1, 3, 2], K = 3

Output

Explanation: The valid windows are [2,1,5] (sum=8), [1,5,1] (sum=7), [5,1,3] (sum=9), and [1,3,2] (sum=6). The maximum sum is 9.

Example 2

Input

nums = [1, 2, 3, 4, 5], K = 2

Output

Explanation: The valid windows are [1,2] (sum=3), [2,3] (sum=5), [3,4] (sum=7), and [4,5] (sum=9). The maximum sum is 9.

Example 3

Input

nums = [5, 4, 3, 2, 1], K = 3

Output

Explanation: The valid windows are [5,4,3] (sum=12), [4,3,2] (sum=9), and [3,2,1] (sum=6). The maximum sum is 12.

Example 4

Input

nums = [10, -5, 10, -5, 10], K = 3

Output

Explanation: The valid windows are [10,-5,10] (sum=15), [-5,10,-5] (sum=0), and [10,-5,10] (sum=15). The maximum sum is 15.

Constraints

1 <= K <= nums.length <= 10^5
-10^9 <= nums[i] <= 10^9
The sum of all elements in any window may exceed 32-bit integer limits, so use 64-bit integers for accumulation.

Optimal Approach & Strategy

Maintain a running sum of the current window. Initialize it with the first K elements, then slide the window by subtracting the element that exits and adding the one that enters, updating the maximum as you go. This runs in O(n) time and O(1) space.

Brute Force Approach

Compute the sum of every subarray of length K by nested loops: for each starting index, sum the next K elements. This takes O(n*K) time and O(1) space.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n)

/**
 * @param {number[]} nums
 * @param {number} K
 * @return {number}
 */
var maxSumSubarray = function(nums, K) {
    const n = nums.length;
    if (n < K) return 0;
    
    let currentSum = 0;
    let maxSum = -Infinity;
    
    // Calculate sum of first window
    for (let i = 0; i < K; i++) {
        currentSum += nums[i];
    }
    maxSum = currentSum;
    
    // Slide the window
    for (let i = K; i < n; i++) {
        currentSum += nums[i] - nums[i - K];
        if (currentSum > maxSum) {
            maxSum = currentSum;
        }
    }
    
    return maxSum;
};

class Solution {
public:
    int maxSumSubarray(vector<int>& nums, int K) {
        int n = nums.size();
        if (n < K) return 0;
        
        long long currentSum = 0;
        long long maxSum = LLONG_MIN;
        
        // Calculate sum of first window
        for (int i = 0; i < K; ++i) {
            currentSum += nums[i];
        }
        maxSum = currentSum;
        
        // Slide the window
        for (int i = K; i < n; ++i) {
            currentSum += nums[i] - nums[i - K];
            if (currentSum > maxSum) {
                maxSum = currentSum;
            }
        }
        
        return (int)maxSum;
    }
};

class Solution {
    public int maxSumSubarray(int[] nums, int K) {
        int n = nums.length;
        if (n < K) return 0;
        
        long currentSum = 0;
        long maxSum = Long.MIN_VALUE;
        
        // Calculate sum of first window
        for (int i = 0; i < K; i++) {
            currentSum += nums[i];
        }
        maxSum = currentSum;
        
        // Slide the window
        for (int i = K; i < n; i++) {
            currentSum += nums[i] - nums[i - K];
            if (currentSum > maxSum) {
                maxSum = currentSum;
            }
        }
        
        return (int) maxSum;
    }
}

class Solution:
    def maxSumSubarray(self, nums: List[int], K: int) -> int:
        n = len(nums)
        if n < K:
            return 0
        
        current_sum = 0
        max_sum = float('-inf')
        
        # Calculate sum of first window
        for i in range(K):
            current_sum += nums[i]
        max_sum = current_sum
        
        # Slide the window
        for i in range(K, n):
            current_sum += nums[i] - nums[i - K]
            if current_sum > max_sum:
                max_sum = current_sum
        
        return max_sum

/**
 * @param {number[]} nums
 * @param {number} K
 * @return {number}
 */
var maxSumSubarray = function(nums, K) {
    const n = nums.length;
    if (n < K) return 0;
    
    let currentSum = 0;
    let maxSum = -Infinity;
    
    // Calculate sum of first window
    for (let i = 0; i < K; i++) {
        currentSum += nums[i];
    }
    maxSum = currentSum;
    
    // Slide the window
    for (let i = K; i < n; i++) {
        currentSum += nums[i] - nums[i - K];
        if (currentSum > maxSum) {
            maxSum = currentSum;
        }
    }
    
    return maxSum;
};

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Tree Decomposition Path Validator 3 Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Input

Output

Constraints

More Stack

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Tree Decomposition Path Validator 3 — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor