LeetCode 1708: Largest Subarray Length K Solution

Master LeetCode problem 1708 (Largest Subarray Length K), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1708. Largest Subarray Length K

Easy

Array Greedy

Problem Explanation

Explanation

To find the largest subarray of length K in a given array, we can iterate through the array while keeping track of the maximum sum subarray of length K. We can use a sliding window approach to efficiently compute the sum of elements in each subarray of length K.

Here are the steps:

Initialize variables maxSum to store the maximum sum encountered so far, and currSum to store the sum of the current subarray of length K.
Calculate the sum of the first subarray of length K and store it in currSum.
Iterate through the array from index K to the end.
- Update currSum by subtracting the element at index i - K and adding the element at index i.
- Update maxSum if currSum is greater than maxSum.
Return the largest subarray of length K with sum maxSum.

The time complexity of this approach is O(N) where N is the number of elements in the input array. The space complexity is O(1) as we are using a constant amount of extra space.

Solution Code

class Solution {
    public int[] largestSubarray(int[] nums, int k) {
        int maxSum = 0;
        int currSum = 0;
        int startIdx = 0;

        for (int i = 0; i < k; i++) {
            currSum += nums[i];
        }
        maxSum = currSum;

        for (int i = k; i < nums.length; i++) {
            currSum = currSum - nums[i - k] + nums[i];
            if (currSum > maxSum) {
                maxSum = currSum;
                startIdx = i - k + 1;
            }
        }

        int[] result = new int[k];
        for (int i = 0; i < k; i++) {
            result[i] = nums[startIdx + i];
        }

        return result;
    }
}

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