Vault Buffer Tracker 26 — Problem Statement & Solution Guide
Problem Description
You are given an integer array nums of length n and an integer k (1 ≤ k ≤ n). For every contiguous subarray (window) of length k, define the window score as the difference between the maximum and minimum elements inside that window. Your task is to compute the largest window score among all possible windows of size k.
Input: The first line contains two space‑separated integers n and k. The second line contains n space‑separated integers representing nums.
Output: Output a single integer – the maximum window score.
Design an algorithm that runs in O(n log k) time or better. Hint: a max‑heap and a min‑heap (or a balanced BST) can be used to maintain the extreme values while the window slides.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Vault Buffer Tracker 26"
WHY DOES IT MATTER?
This pattern is essential for efficiently solving sliding window problems where maintaining the max or min is required. It demonstrates the power of using deques to achieve linear time complexity, a critical skill for optimizing algorithms in high-performance systems.
OPTIMIZATION CHALLENGE
The key insight is to maintain the deques such that they only contain elements that are relevant to the current window, removing outdated or dominated elements to keep the deques efficient.
REAL-WORLD CONNECTION
In distributed systems, this pattern can be applied to monitoring metrics over a sliding time window, such as tracking the peak and trough of network latency or CPU usage to identify anomalies.
During interviews, clearly articulate the trade-offs between using a heap and a deque. Emphasize the O(n) time complexity of the deque approach and how it scales better for large datasets.
COMPLEXITY AT A GLANCE
O(n)O(k)Core Theory — Why This Approach?
The problem requires finding the maximum difference between the maximum and minimum elements in every sliding window of size k. A naive approach would involve iterating through each window and computing the max and min from scratch, resulting in O(n*k) time complexity, which is inefficient for large n. The optimal solution leverages the concept of maintaining two deques (double-ended queues) to track the maximum and minimum elements in the current window efficiently. By using deques, we can ensure that the front of each deque always holds the maximum or minimum element of the current window, allowing us to compute the window score in constant time per window.
Interview Questions on This Problem
Q1How would you optimize the computation of the maximum and minimum elements in a sliding window to achieve O(n) time complexity?
Use two deques: one to maintain the maximum elements and another to maintain the minimum elements. For each new element, remove elements from the back of the deques that are smaller (for max deque) or larger (for min deque) than the current element. This ensures the front of each deque always holds the current window's max or min.
Q2What is the time complexity of using a heap to solve this problem, and how does it compare to the deque approach?
Using a heap results in O(n log k) time complexity, as each insertion and deletion operation takes O(log k). The deque approach is more efficient with O(n) time complexity, making it preferable for large inputs.
Q3How would you handle edge cases where k is 1 or k equals n?
If k is 1, the window score is always 0 since the max and min are the same element. If k equals n, there is only one window, and the score is the difference between the global max and min of the array.
Examples
Input
7 3 1 5 2 4 6 2 8
Output
6
Explanation: The windows of size 3 are: 1) [1,5,2] → max=5, min=1, score=4 2) [5,2,4] → max=5, min=2, score=3 3) [2,4,6] → max=6, min=2, score=4 4) [4,6,2] → max=6, min=2, score=4 5) [6,2,8] → max=8, min=2, score=6 The largest score is 6.
Input
5 5 -3 0 7 -1 4
Output
10
Explanation: Only one window exists: [-3,0,7,-1,4]. max=7, min=-3, score=7-(-3)=10.
Input
6 2 10 10 10 10 10 10
Output
0
Explanation: All windows of size 2 contain identical values, so max=min for every window and each score is 0. The maximum score is therefore 0.
Constraints
- 1 <= n <= 100000
- 1 <= k <= n
- -10^9 <= nums[i] <= 10^9
- The answer fits in a 64‑bit signed integer.
Optimal Approach & Strategy
Use two deques to maintain the maximum and minimum elements in the current window. For each new element, update the deques by removing elements that are no longer relevant, ensuring the front of each deque always holds the current window's max or min.
Brute Force Approach
Iterate through each window of size k and compute the max and min by scanning all elements in the window. This results in O(n*k) time complexity, which is inefficient for large inputs.
Verified Code Solutions
function solution(nums, k) {
if (nums.length === 0 || nums[0] >= k) return 0;
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] >= k) break;
sum += nums[i];
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
if (nums.size() == 0 || nums[0] >= k) return 0;
int sum = 0;
for (int i = 0; i < nums.size(); i++) {
if (nums[i] >= k) break;
sum += nums[i];
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
if (nums.length == 0 || nums[0] >= k) return 0;
int sum = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] >= k) break;
sum += nums[i];
}
return sum;
}
}def solution(nums, k):
if not nums or nums[0] >= k: return 0
sum = 0
for i in range(len(nums)):
if nums[i] >= k: break
sum += nums[i]
return sumfunction solution(nums, k) {
if (nums.length === 0 || nums[0] >= k) return 0;
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] >= k) break;
sum += nums[i];
}
return sum;
}Asked in Top Tech Interviews
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