Problem
Slide a window of size k across the list one step at a time and report the maximum inside each window. A deque keeps this O(n).
Implement window_maximums(numbers, k).
Input
numbers: The list of integers. k: The window size.
Output
Return the maximum of each window, left to right.
Constraints
1 ≤ k ≤ length ≤ 200,000
Examples
Input:
k = 3
numbers = [1, 3, -1, -3, 5, 3, 6, 7]
Expected output:
expected = [3, 3, 5, 5, 6, 7]Input:
long long k = 3;
std::vector<long long> numbers = {1, 3, -1, -3, 5, 3, 6, 7};
Expected output:
std::vector<long long> expected = {3, 3, 5, 5, 6, 7};Input:
long k = 3;
long[] numbers = new long[] {1, 3, -1, -3, 5, 3, 6, 7};
Expected output:
long[] expected = new long[] {3, 3, 5, 5, 6, 7};Explanation
Windows [1,3,-1], [3,-1,-3], [-1,-3,5], … have maxima 3, 3, 5, 5, 6, 7.
Input:
k = 1
numbers = [4, 2]
Expected output:
expected = [4, 2]Input:
long long k = 1;
std::vector<long long> numbers = {4, 2};
Expected output:
std::vector<long long> expected = {4, 2};Input:
long k = 1;
long[] numbers = new long[] {4, 2};
Expected output:
long[] expected = new long[] {4, 2};Explanation
Windows of one element are the elements themselves: [4, 2].
How to approach it
Try on your own first, then open one hint at a time.
Hint 1
The maximum can disappear when it leaves the window, so you need a structure that updates by both value and index.
Hint 2
Keep candidate indices in decreasing order of their values, dropping anything outside the current window.
Hint 3
Watch the first full window and windows of size one; every answer must correspond to a fully formed window.
Expected complexity
O(n) time, O(k) extra space
References
This challenge is adapted for CodeUtility. Follow the statement and constraints above.