---
tags: data_structure_python
---

# Largest Subarray Length K <img src="https://img.shields.io/badge/-easy-brightgreen">

- https://leetcode.com/discuss/interview-question/352459/


# Solution

def largest_subarray_length_k(L, k):
    # Time complexity: O(nk).
    # Space complexity: O().
    n = len(L)
    if n == k:
        return L
    else:
        nbSubarray = n - k
        largestSubarray = L[0:k]
        for i in range(1, nbSubarray):
            for a, b in zip(largestSubarray, L[i:i+k]):
                if b > a:
                    largestSubarray = L[i:i+k]
                    break 
                elif b < a:
                    break
        return largestSubarray 

print(largest_subarray_length_k([1, 4, 3, 2, 5], 4))