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SQP11 Sliding Window Maximum

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: 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


Notes

Note: Zoom for Better Understanding



Always start with the BruteForce Approch and then moved on the Optimal Approch

  • Now can we say that I start with Iteration from 0 and finding the all the possible subarray with the length ok k

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    Python Code
    C++ Code
    Java Code
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    Khud Bhi Kr le Khuch ..... Nalayk


    Time Complexity:O(N ^ 2) Becouse we used a nested Loop thats why
    Space Complexity:O(N) for the Answer

    Optimized Approch

  • Now here again we used the concept of the Next Greatest Element that we learned pevesolly in this Concept we store the element into the Increasing Fassion
  • In this problem we store the element into the Desciring order Fassion we required the deque data structure
  • we start Iteration from the Index = 0
  • And maintain the deque prpoperty that is decresing order and put that Index into the deque

  • Maintain Property
  • check if the deque is Empty() then put the Index into the deque
  • and check if the exisiting element is the < then arr[i] then removed all the element from the deque and then put Index into the deque
  • otherwise put Index into the deque

  • Code Zone!

    Optimized Python Code
    Optimized C++ Code
    Optimized Java Code
    Sb Mai He Kru ...

    Khud Bhi Kr le Khuch ..... Nalayk


    Time Complexity:O(N) + O(N) ~ O(N) for the linear Iteration, and for the checking the Window Size That is the Amotized Time Complexcity
    Space Complexity:O(K) for the Answer of window size


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