LeetCode 3118: Friday Purchase III Solution

Problem Explanation

Explanation

To solve this problem, we can use a sliding window approach. We maintain a window of size n and calculate the maximum sum of elements within this window for each possible starting position. We then return the maximum sum found.

Algorithm:

Initialize a variable maxSum to store the maximum sum found.
Iterate through the array from 0 to len(arr) - n to define the starting index of the window.
For each starting index, calculate the sum of elements within the window of size n.
Update maxSum if the current sum is greater.
Return maxSum at the end.

Time complexity: O(n), where n is the length of the input array. Space complexity: O(1), as we are using only a constant amount of extra space.

Solution Code

public int maxSum(int[] arr, int n) {
    int maxSum = Integer.MIN_VALUE;
    for (int i = 0; i <= arr.length - n; i++) {
        int sum = 0;
        for (int j = i; j < i + n; j++) {
            sum += arr[j];
        }
        maxSum = Math.max(maxSum, sum);
    }
    return maxSum;
}

Frequently Asked Questions

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3118. Friday Purchase III