LeetCode 643. Maximum Average Subarray I

You are given an integer array nums consisting of n elements, and an integer k.

Find a contiguous subarray whose length is equal to k that has the maximum average value and return this value. Any answer with a calculation error less than 10-5 will be accepted.

Example 1:

Input: nums = [1,12,-5,-6,50,3], k = 4
Output: 12.75000
Explanation: Maximum average is (12 - 5 - 6 + 50) / 4 = 51 / 4 = 12.75

Example 2:

Input: nums = [5], k = 1
Output: 5.00000

Constraints:

• n == nums.length
• 1 <= k <= n <= 105
• -104 <= nums[i] <= 104

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本质就是求数组里长度为k的字数组的最大值，是一道简单的prefix sum / sliding window题。刚开始又给想复杂了……其实可以直接算sum，不需要往window里加加减减。

sliding window就是一个window里放k个元素，每次remove掉第一个，加上新的，一个for loop遍历完所有。刚开始写的时候max初始化成了double.min，这时候第二个for loop就cover不到k==n的corner case，导致结果出错。

Runtime: 15 ms, faster than 23.03% of Java online submissions for Maximum Average Subarray I.

Memory Usage: 107.5 MB, less than 24.36% of Java online submissions for Maximum Average Subarray I.

class Solution {
    public double findMaxAverage(int[] nums, int k) {
        double sum = 0;
        for (int i = 0; i < k; i++) {
            sum += nums[i];
        }
        double max = sum;
        for (int i = k; i < nums.length; i++) {
            sum -= nums[i - k];
            sum += nums[i];
            max = Math.max(max, sum);
        }
        return max / k;
    }
}

prefix sum的做法就是先一个for loop计算出当前所有元素的和，然后再一个for loop对和进行相减，取最大。其中也有一些小坑需要注意，就是max需要初始化为sums[k - 1]，然后从i = k遍历到nums.length。

Runtime: 15 ms, faster than 23.03% of Java online submissions for Maximum Average Subarray I.

Memory Usage: 108 MB, less than 20.27% of Java online submissions for Maximum Average Subarray I.

class Solution {
    public double findMaxAverage(int[] nums, int k) {
        int[] sums = new int[nums.length];
        sums[0] = nums[0];
        for (int i = 1; i < nums.length; i++) {
            sums[i] = sums[i - 1] + nums[i];
        }
        double result = sums[k - 1];
        for (int i = k; i < nums.length; i++) {
            double sum = sums[i] - sums[i - k];
            result = Math.max(sum, result);
        }
        return result / k;
    }
}