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:** [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

Naive Approach §

• Set: i=0, j=k
• Slice the array into i -> k
  • sort in descending
  • push[0] into the max array
• i++; j++

let nums = [1,3,-1,-3,5,3,6,7]
function window(nums, k){
    let j = k
    let i = 0
    let max = []
    while (j <= nums.length){
        let section = nums.slice(i, j)
        section.sort((a,b)=>{return b-a})
        max.push(section[0])
        i++
        j++
    }
    return max
}
 
console.log(window(nums, 3))

This shit is hella slow for larger arrays, So organizing for larger arrays is not really possilbe

Time Complexity §

The time complexity of the given function can be analyzed step by step:

1. Initializing variables j, i, and max takes constant time, O(1).

2. The while loop runs nums.length - k + 1 times. This is because the loop starts from index i = 0 and goes up to index i = nums.length - k, since you’re considering subarrays of length k. So, the loop runs approximately nums.length - k times.

3. Inside the loop, you’re using the slice method, which takes O(k) time, and then you’re using the sort method, which takes O(k * log(k)) time (since you’re sorting a subarray of length k).

4. Pushing an element into the max array is a constant time operation.

Overall, the time complexity of each iteration of the loop is O(k * log(k)) due to the sorting step.

Since the loop runs approximately nums.length - k times, and each iteration has a time complexity of O(k * log(k)), the total time complexity of the function can be approximated as follows: Total time complexity ≈ (nums.length - k) * (O(k * log(k)))

Asymptotically, when k is smaller than nums.length, the O(k * log(k)) term dominates, so you can simplify the expression to: Total time complexity ≈ O((nums.length - k) * k * log(k))

This can be further simplified to: Total time complexity ≈ O(nums.length * k * log(k))

So, the time complexity of the provided function is approximately O(nums.length * k * log(k)) in the worst case.