# Maximum Average Subarray I

Easy

## Question

Given an array consisting of n integers, find the contiguous subarray of given length k that has the maximum average value. And you need to output the maximum average value.

Example 1:

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

Note
1 <= k <= n <= 30,000.
Elements of the given array will be in the range [-10,000, 10,000].

## Thinking

1. Using loop in loop to get the max value. Max initial value is Integer.MIN_VALUE (Note: Not Dobule.MIN_VALUE, Time complexity is O(n^2) and timeout)
2. Summery first and then find max average
3. Using previous sum result to find max Average

## Solution

### Java

#### solution 2

class Solution {
public double findMaxAverage(int[] nums, int k) {
if (nums == null || nums.length < k || k < 1) {
throw new IllegalArgumentException("Invalid input. There is no solution");
}
double sums[] = new double[nums.length];
sums[0] = nums[0];
for (int i = 1; i < nums.length; i++) {
sums[i] = sums[i - 1] + nums[i];
}
double maxAvg = sums[k -1] / k;
for (int i = k; i < nums.length; i++) {
double sum = sums[i] - sums[i - k];
maxAvg = Math.max(maxAvg, sum / k);
}
return maxAvg;
}
}

#### Solution 3

class Solution {
public double findMaxAverage(int[] nums, int k) {
if (nums == null || nums.length < k || k < 1) {
throw new IllegalArgumentException("Invalid input. There is no solution");
}
double sum = 0;
for (int i = 0; i < k; i++) {
sum += nums[i];
}
double maxAvg = sum / k;
for (int i = k; i < nums.length; i++) {
sum += (nums[i] - nums[i - k]);
maxAvg = Math.max(maxAvg, sum / k);
}
return maxAvg;
}
}