1708. Largest Subarray Length K

An array A is larger than some array B if for the first index i where A[i] != B[i], A[i] > B[i].

For example, consider 0-indexing:

• [1,3,2,4] > [1,2,2,4], since at index 1, 3 > 2.
• [1,4,4,4] < [2,1,1,1], since at index 0, 1 < 2.

A subarray is a contiguous subsequence of the array.

Given an integer array nums of distinct integers, return the largest subarray of nums of length k.

 

Example 1:

Input: nums = [1,4,5,2,3], k = 3
Output: [5,2,3]
Explanation: The subarrays of size 3 are: [1,4,5], [4,5,2], and [5,2,3].
Of these, [5,2,3] is the largest.

Example 2:

Input: nums = [1,4,5,2,3], k = 4
Output: [4,5,2,3]
Explanation: The subarrays of size 4 are: [1,4,5,2], and [4,5,2,3].
Of these, [4,5,2,3] is the largest.

Example 3:

Input: nums = [1,4,5,2,3], k = 1
Output: [5]

 

Constraints:

• 1 <= k <= nums.length <= 105
• 1 <= nums[i] <= 109
• All the integers of nums are unique.

---

class Solution:
    def largestSubarray(self, nums: List[int], k: int) -> List[int]:
        nindex,nmax=0,0
        for i in range(len(nums)-k+1):
            if nums[i]>nmax:
                nmax=nums[i]
                nindex=i
        return (nums[nindex:nindex+k])