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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?

Companies:
Google

Related Topics:
Array, Greedy

Solution 1.

// OJ: https://leetcode.com/problems/largest-subarray-length-k/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1) extra space
class Solution {
public:
    vector<int> largestSubarray(vector<int>& A, int k) {
        auto it = max_element(begin(A), end(A) - k + 1);
        return vector<int>(it, it + k);
    }
};