给定大小为n且整数为k的数组arr []，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实践，然后再继续解决方案。
一个简单的解决方案是将每个元素视为大小为k的子数组的开头，并计算从该元素开始的子数组的总和。该解决方案的时间复杂度为O(nk)。
一个有效的解决方案是解决O(n)时间和O(1)额外空间的上述问题。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，请删除上一个窗口的第一个元素，然后添加当前元素(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。
C++
// A Simple C++ program to find minimum average subarray
#include
using namespace std;
// Prints beginning and ending indexes of subarray
// of size k with minimum average
void findMinAvgSubarray(int arr[], 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);
}
}
cout << "Subarray between [" << res_index << ", "
<< res_index + k - 1 << "] has minimum average";
}
// Driver program
int main()
{
int arr[] = { 3, 7, 90, 20, 10, 50, 40 };
int k = 3; // Subarray size
int n = sizeof arr / sizeof arr[0];
findMinAvgSubarray(arr, n, k);
return 0;
} 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);
}
}Python3
# Python3 program to find
# minimum average subarray
# Prints beginning and ending
# indexes of subarray of size k
# with minimum average
def findMinAvgSubarray(arr, n, k):
# k must be smaller than or equal to n
if (n < k): return 0
# Initialize beginning index of result
res_index = 0
# Compute sum of first subarray of size k
curr_sum = 0
for i in range(k):
curr_sum += arr[i]
# Initialize minimum sum as current sum
min_sum = curr_sum
# Traverse from (k + 1)'th
# element to n'th element
for i in range(k, n):
# 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)
print("Subarray between [", res_index,
", ", (res_index + k - 1),
"] has minimum average")
# Driver Code
arr = [3, 7, 90, 20, 10, 50, 40]
k = 3 # Subarray size
n = len(arr)
findMinAvgSubarray(arr, n, k)
# This code is contributed by Anant Agarwal.C#
// A Simple C# program to find
// minimum average subarray
using System;
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);
}
}
Console.Write("Subarray between [" + res_index + ", " +
(res_index + k - 1) +
"] has minimum average");
}
// Driver method to test the above function
public static void Main()
{
int k = 3; // Subarray size
findMinAvgSubarray(arr.Length, k);
}
}
// This code is contributed by nitin mittal.PHP
输出：
Subarray between [3, 5] has minimum average时间复杂度：O(n)
辅助空间：O(1)