239. Sliding Window Maximum¶

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239. Sliding Window Maximum

Hard

You are given an array of integers 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.

Example 1:

Input: nums = [1,3,-1,-3,5,3,6,7], 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

Example 2:

Input: nums = [1], k = 1
Output: [1]

Constraints:

• 1 <= nums.length <= 105
• -104 <= nums[i] <= 104
• 1 <= k <= nums.length

Solution¶

class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        int n = nums.length;
        int res[] = new int[n - k + 1];
        int p = 0, left = 0;;
        TreeSet<Integer> set = new TreeSet<>();
        HashMap<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < k; i++) {
            map.put(nums[i] , map.getOrDefault(nums[i] , 0) + 1);
            set.add(nums[i]);
        }
        res[p++] = set.last();
        for (int i = k; i < n; i++) {
            map.put(nums[left] , map.getOrDefault(nums[left], 0) - 1);
            if (map.getOrDefault(nums[left], 0) == 0) set.remove(nums[left]);
            map.put(nums[i] , map.getOrDefault(nums[i] , 0) + 1);
            set.add(nums[i]);
            res[p++] = set.last();
            left++;
        }
        return res;
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here