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Largest Subarray Length K

Solution Explanation: Finding the Largest Subarray of Length K

This problem asks us to find the "largest" subarray of length k within a given array nums. The definition of "largest" is crucial: a subarray A is larger than subarray B if, at the first index where they differ, A's element is greater than B's element. This implies a lexicographical comparison.

Algorithm:

The most efficient approach leverages the fact that the integers in nums are distinct. We don't need to compare all possible subarrays of length k. Instead, we can directly locate the starting index of the largest subarray:

  1. Find the maximum element within the first nums.length - k + 1 elements: This is because any subarray of length k must start within this range. We only need to consider the first element of each possible subarray of length k for comparison.

  2. Extract the subarray: Once we find the index (maxIndex) of the maximum element within the range, we simply extract the subarray of length k starting at that index. This subarray will be the largest according to the problem's definition.

Time Complexity: O(n), where n is the length of nums. This is because finding the maximum element takes linear time, and extracting the subarray takes linear time in the worst case (when k is close to n).

Space Complexity: O(1) or O(k). If we don't consider the space used to store the result (the subarray of length k), then the space complexity is O(1). However, if the space used by the output is considered then it is O(k).

Code Examples (with explanations):

Python:

class Solution:
    def largestSubarray(self, nums: List[int], k: int) -> List[int]:
        # Find the index of the maximum element within the relevant range
        max_index = nums[:len(nums) - k + 1].index(max(nums[:len(nums) - k + 1]))  
 
        # Extract and return the subarray
        return nums[max_index: max_index + k]

Java:

import java.util.Arrays;
 
class Solution {
    public int[] largestSubarray(int[] nums, int k) {
        int maxIndex = 0;
        // Find the index of the maximum element
        for (int i = 1; i <= nums.length - k; ++i) {
            if (nums[i] > nums[maxIndex]) {
                maxIndex = i;
            }
        }
        //Extract the subarray using Arrays.copyOfRange
        return Arrays.copyOfRange(nums, maxIndex, maxIndex + k);
    }
}

C++:

#include <vector>
#include <algorithm>
 
class Solution {
public:
    std::vector<int> largestSubarray(std::vector<int>& nums, int k) {
        auto max_element_iter = std::max_element(nums.begin(), nums.begin() + nums.size() - k + 1);
        // Use iterators to get the subarray efficiently.
        return std::vector<int>(max_element_iter, max_element_iter + k);
    }
};

Go:

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

TypeScript:

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

Rust:

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

All these solutions follow the same algorithmic approach, differing only in their syntax and specific library functions used for finding the maximum element and extracting the subarray. They all achieve the optimal time complexity of O(n).