239.md

239. 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.

For example,
Given nums = [1,3,-1,-3,5,3,6,7], and k = 3.

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
Therefore, return the max sliding window as [3,3,5,5,6,7].

Note: 
You may assume k is always valid, ie: 1 ≤ k ≤ input array's size for non-empty array.

Follow up:
Could you solve it in linear time?

my thoughts:

1. min heap.

my solution:

********** 148ms
class Solution(object):
    def maxSlidingWindow(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: List[int]
        """
        d = collections.deque()
        res = []
        for i, n in enumerate(nums):
            while d and nums[d[-1]] < n:
                d.pop()
            d.append(i)
            if i >= k-1:
                res.append(nums[d[0]])
            if d[0] == i - k + 1:
                d.popleft()
        return res

my comments:

from other ppl's solution:

1. I wrote 3 methods I can come up with.

Deque Method: use deque to record the index, remove elements out of window in the head, and remove the smaller elements in the tail which have no change to become max.
    // O(n)
    public int[] maxSlidingWindow_deque(int[] nums, int k) {
        if(nums==null || k>nums.length || k<0) return null;
        if(k==0 || nums.length==0) return new int[0];
        Deque<Integer> dq = new LinkedList<>();
        int[] res = new int[nums.length-k+1];
        int index = 0; // index in r is equal to the left side of window
        for(int i=0; i<nums.length; i++){
            // remove elements out of window
            index = i-k+1;
            while(!dq.isEmpty() && dq.peek()<index){
                dq.poll();
            }
            // remove smaller elements, which have no change to become max
            while(!dq.isEmpty() && nums[dq.peek()]<nums[i]){
                dq.pollLast();
            }
            dq.offer(i);
            if(index>=0){
                res[index] = nums[dq.peek()];
            }
        }
        return res;
    }
Max Heap Method: a naive method. Build a max heap for a window, and remove the element out of window.
    //O((n-k)(lgk+k)) = O((n-k)k)
    public int[] maxSlidingWindow_heap1(int[] nums, int k) {
        if(nums==null || k>nums.length || k<0) return null;
        if(k==0 || nums.length==0) return new int[0];
        PriorityQueue<Integer> heap = new PriorityQueue<>( k, Collections.reverseOrder() );
        int[] res = new int[nums.length - k + 1];
        for(int i=0; i<k; i++)
            heap.offer(nums[i]);
        for(int p=0; p<res.length; p++){
            res[p] = heap.peek();
            heap.remove(nums[p]);
            if(p+k < nums.length)
                heap.offer(nums[p+k]);
        }
        return res;
    }
Max Heap Method 2: Using a class Tuple to record index info.
    // worst case O((n-k)logn) increasing order, best case O((n-k)logk) decreasing order
    public int[] maxSlidingWindow_heap2(int[] nums, int k) {
        if(nums==null || k>nums.length || k<0) return null;
        if(k==0 || nums.length==0) return new int[0];
        PriorityQueue<Tuple> heap = new PriorityQueue<>( k, Collections.reverseOrder() );
        int[] res = new int[nums.length - k + 1];
        for(int i=0; i<k; i++)
            heap.offer(new Tuple(nums[i], i));
        Tuple tup = null;
        for(int p=0; p<res.length; p++){
            tup = heap.peek();
            while(tup.idx<p) {
                heap.poll();
                tup = heap.peek();
            }
            res[p] = tup.val;
            if(p+k < nums.length)
                heap.offer(new Tuple(nums[p+k], p+k));
        }
        return res;
    }
    class Tuple implements Comparable<Tuple>{
        int val;
        int idx;
        public Tuple(int value, int index){
            val = value;
            idx = index;
        }
        @Override
        public int compareTo(Tuple tup){
            return (val - tup.val);
        }
    }