One [topic category]

sliding window

Two [question difficulty]

Simple

Three [topic number]

643. Maximum average number of subarrays I

Four [title description]

give you a by Integer array $of n elements$ $n u m s$ and an integer $k$ 。
Please find the largest average and length Contiguous subarrays of $k , and output that maximum mean.$
Any error less than $10^{-5}$ answers will be considered correct.

Five [topic example]

Example 1:
- Input: nums = [1,12,-5,-6,50,3], k = 4
- Output: 12.75
- Explanation: Maximum average (12-5-6+50)/4 = 51/4 = 12.75
Example 2:
- Input: nums = [5], k = 1
- Output: 5.00000

Six [topic prompt]

$n == n u m s . l e n g t h$
$1 <= k <= n <= 10^{5}$
$10^{4} <= nums[i] <= 10^{4}$

Seven [problem-solving ideas]

Using the idea of sliding windows
First calculate the sum of the first k element values
Then enter the loop to traverse the remaining elements, subtract the first element in the "window" every time an element is traversed, and then add the newly traversed elements to the "window", then compare and take the maximum value
The title requires us to return the duoble type, so we need to perform a forced type conversion at the end, and divide the sum of the largest "window" by the number of window elements
Finally return the result

Eight 【Time Frequency】

Time complexity: $O (n)$ , where $n$ is the length of the incoming array
Space complexity: $O (1)$

Nine [code implementation]

Java language version

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

C language version

double findMaxAverage(int* nums, int numsSize, int k)
{
    
    
    int tempSum = 0;
    for(int i = 0;i<k;i++)
    {
    
    
        tempSum += nums[i];
    }
    int sum = tempSum;
    for(int i = k;i<numsSize;i++)
    {
    
    
        sum = sum - nums[i - k] + nums[i];
        tempSum = fmax(tempSum,sum);
    }
    return (double)tempSum / k;
}

Python language version

class Solution:
    def findMaxAverage(self, nums: List[int], k: int) -> float:
        tempSum = resSum = sum(nums[:k])
        for i in range(k,len(nums)):
            tempSum = tempSum - nums[i - k] + nums[i]
            resSum = max(resSum,tempSum)
        return resSum / k

C++ language version

class Solution {
    
    
public:
    double findMaxAverage(vector<int>& nums, int k) {
    
    
        int tempSum = 0;
        for(int i = 0;i<k;i++)
        {
    
    
            tempSum += nums[i];
        }
        int sum = tempSum;
        for(int i = k;i<nums.size();i++)
        {
    
    
            sum = sum - nums[i - k] + nums[i];
            tempSum = max(tempSum,sum);
        }
        return (double)tempSum / k;
    }
};