Java程序查找具有最小平均数的子数组

给定一个大小为 n 的数组 arr[] 和整数 k，使得 k <= n。

例子 ：

Input:  arr[] = {3, 7, 90, 20, 10, 50, 40}, k = 3
Output: Subarray between indexes 3 and 5
The subarray {20, 10, 50} has the least average 
among all subarrays of size 3.

Input:  arr[] = {3, 7, 5, 20, -10, 0, 12}, k = 2
Output: Subarray between [4, 5] has minimum average

我们强烈建议您单击此处并进行练习，然后再继续使用解决方案。

一个简单的解决方案是将每个元素视为大小为 k 的子数组的开始，并从该元素开始计算子数组的总和。该解决方案的时间复杂度为 O(nk)。

一个有效的解决方案是在 O(n) 时间和 O(1) 额外空间内解决上述问题。这个想法是使用大小为 k 的滑动窗口。跟踪当前 k 个元素的总和。要计算当前窗口的总和，请删除前一个窗口的第一个元素并添加当前元素（当前窗口的最后一个元素）。

1) Initialize res_index = 0 // Beginning of result index
2) Find sum of first k elements. Let this sum be 'curr_sum'
3) Initialize min_sum = sum
4) Iterate from (k+1)'th to n'th element, do following
   for every element arr[i]
      a) curr_sum = curr_sum + arr[i] - arr[i-k]
      b) If curr_sum < min_sum
           res_index = (i-k+1)
5) Print res_index and res_index+k-1 as beginning and ending
   indexes of resultant subarray.

下面是上述算法的实现。

Java

// A Simple Java program to find 
// minimum average subarray
  
class Test {
      
    static int arr[] = new int[] { 3, 7, 90, 20, 10, 50, 40 };
  
    // Prints beginning and ending indexes of subarray
    // of size k with minimum average
    static void findMinAvgSubarray(int n, int k)
    {
        // k must be smaller than or equal to n
        if (n < k)
            return;
  
        // Initialize beginning index of result
        int res_index = 0;
  
        // Compute sum of first subarray of size k
        int curr_sum = 0;
        for (int i = 0; i < k; i++)
            curr_sum += arr[i];
  
        // Initialize minimum sum as current sum
        int min_sum = curr_sum;
  
        // Traverse from (k+1)'th element to n'th element
        for (int i = k; i < n; i++) 
        {
            // Add current item and remove first
            // item of previous subarray
            curr_sum += arr[i] - arr[i - k];
  
            // Update result if needed
            if (curr_sum < min_sum) {
                min_sum = curr_sum;
                res_index = (i - k + 1);
            }
        }
  
        System.out.println("Subarray between [" +
                            res_index + ", " + (res_index + k - 1) +
                            "] has minimum average");
    }
  
    // Driver method to test the above function
    public static void main(String[] args)
    {
        int k = 3; // Subarray size
        findMinAvgSubarray(arr.length, k);
    }
}

输出：

Subarray between [3, 5] has minimum average

时间复杂度：O(n)
辅助空间：O(1)

有关更多详细信息，请参阅有关查找具有最低平均值的子数组的完整文章！