Sliding Window Maximum
Given an array nums, there is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves right by one position. Return the max sliding window.
Follow up: Could you solve it in linear time?
Example:
Input: nums = [1,3,-1,-3,5,3,6,7], and k = 3
Output: [3,3,5,5,6,7]
Explanation:
Window position Max
--------------- -----
[1 3 -1] -3 5 3 6 7 3
1 [3 -1 -3] 5 3 6 7 3
1 3 [-1 -3 5] 3 6 7 5
1 3 -1 [-3 5 3] 6 7 5
1 3 -1 -3 [5 3 6] 7 6
1 3 -1 -3 5 [3 6 7] 7
Constraints:
1 <= nums.length <= 10^5
-10^4 <= nums[i] <= 10^4
1 <= k <= nums.length
// Other way to do this is using leftmax and rightmax arrays with boxed partitions
// https://drive.google.com/file/d/14k1lhaQe9He5EouLXuSlXrMR61HUBMy1/view?usp=sharing
class Solution {
public int[] maxSlidingWindow(int[] nums, int k) {
Deque<Integer> dq = new LinkedList<>();
int[] ans = new int[nums.length - k + 1];
int index = 0;
for (int i = 0; i < nums.length; i++) {
// Removing the elements our of range 'k'
while (dq.size() != 0 && dq.peekFirst() < i - (k - 1))
dq.pollFirst();
// Maintaining increasing deque from right -> left
while (dq.size() != 0 && nums[dq.peekLast()] < nums[i])
dq.pollLast();
dq.addLast(i);
// If our window of size 'k' is ready
if (i >= k - 1) {
ans[index++] = nums[dq.peekFirst()];
}
}
return ans;
}
}
Last updated