LeetCode 1708. Largest Subarray Length K Solution in Java, C++, Python & More | Explanation + Code

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).

PythonJavaC++GoTypeScriptRust

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](code-box)

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); } }(code-box)

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}; } };(code-box)

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] }(code-box)

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); }(code-box)

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() } }(code-box)