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239. Sliding Window Maximum

Description

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

Solutions

Solution 1

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class Solution:
    def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:
        q = [(-v, i) for i, v in enumerate(nums[: k - 1])]
        heapify(q)
        ans = []
        for i in range(k - 1, len(nums)):
            heappush(q, (-nums[i], i))
            while q[0][1] <= i - k:
                heappop(q)
            ans.append(-q[0][0])
        return ans

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class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        PriorityQueue<int[]> q
            = new PriorityQueue<>((a, b) -> a[0] == b[0] ? a[1] - b[1] : b[0] - a[0]);
        int n = nums.length;
        for (int i = 0; i < k - 1; ++i) {
            q.offer(new int[] {nums[i], i});
        }
        int[] ans = new int[n - k + 1];
        for (int i = k - 1, j = 0; i < n; ++i) {
            q.offer(new int[] {nums[i], i});
            while (q.peek()[1] <= i - k) {
                q.poll();
            }
            ans[j++] = q.peek()[0];
        }
        return ans;
    }
}

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class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        priority_queue<pair<int, int>> q;
        int n = nums.size();
        for (int i = 0; i < k - 1; ++i) {
            q.push({nums[i], -i});
        }
        vector<int> ans;
        for (int i = k - 1; i < n; ++i) {
            q.push({nums[i], -i});
            while (-q.top().second <= i - k) {
                q.pop();
            }
            ans.emplace_back(q.top().first);
        }
        return ans;
    }
};

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func maxSlidingWindow(nums []int, k int) (ans []int) {
    q := hp{}
    for i, v := range nums[:k-1] {
        heap.Push(&q, pair{v, i})
    }
    for i := k - 1; i < len(nums); i++ {
        heap.Push(&q, pair{nums[i], i})
        for q[0].i <= i-k {
            heap.Pop(&q)
        }
        ans = append(ans, q[0].v)
    }
    return
}

type pair struct{ v, i int }

type hp []pair

func (h hp) Len() int { return len(h) }
func (h hp) Less(i,