# 1708. Largest Subarray Length K

## Description

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.

Follow up: What if the integers in nums are not distinct?

## Solutions

Solution 1: Simulation

All integers in the array are distinct, so we can first find the index of the maximum element in the range $[0,..n-k]$, and then take $k$ elements starting from this index.

The time complexity is $O(n)$, where $n$ is the length of the array. Ignoring the space consumption of the answer, the space complexity is $O(1)$.

• class Solution {
public int[] largestSubarray(int[] nums, int k) {
int j = 0;
for (int i = 1; i < nums.length - k + 1; ++i) {
if (nums[j] < nums[i]) {
j = i;
}
}
return Arrays.copyOfRange(nums, j, j + k);
}
}

• class Solution {
public:
vector<int> largestSubarray(vector<int>& nums, int k) {
auto i = max_element(nums.begin(), nums.end() - k + 1);
return {i, i + k};
}
};

• class Solution:
def largestSubarray(self, nums: List[int], k: int) -> List[int]:
i = nums.index(max(nums[: len(nums) - k + 1]))
return nums[i : i + k]


• func largestSubarray(nums []int, k int) []int {
j := 0
for i := 1; i < len(nums)-k+1; i++ {
if nums[j] < nums[i] {
j = i
}
}
return nums[j : j+k]
}

• function largestSubarray(nums: number[], k: number): number[] {
let j = 0;
for (let i = 1; i < nums.length - k + 1; ++i) {
if (nums[j] < nums[i]) {
j = i;
}
}
return nums.slice(j, j + k);
}


• impl Solution {
pub fn largest_subarray(nums: Vec<i32>, k: i32) -> Vec<i32> {
let mut j = 0;
for i in 1..=nums.len() - (k as usize) {
if nums[i] > nums[j] {
j = i;
}
}
nums[j..j + (k as usize)].to_vec()
}
}