给定一个由n个整数和一个数字m组成的数组,请找到从给定数组中选择的两组m个元素之间的最大可能差。
例子:
Input : arr[] = 1 2 3 4 5
m = 4
Output : 4
The maximum four elements are 2, 3,
4 and 5. The minimum four elements are
1, 2, 3 and 4. The difference between
two sums is (2 + 3 + 4 + 5) - (1 + 2
+ 3 + 4) = 4
Input : arr[] = 5 8 11 40 15
m = 2
Output : 42
The difference is (40 + 15) - (5 + 8)
想法是先对数组排序,然后找到前m个元素的和与后m个元素的和。最后返回两个和之间的差。
CPP
// C++ program to find difference
// between max and min sum of array
#include
using namespace std;
// utility function
int find_difference(int arr[], int n, int m)
{
int max = 0, min = 0;
// sort array
sort(arr, arr + n);
for (int i = 0, j = n - 1;
i < m; i++, j--) {
min += arr[i];
max += arr[j];
}
return (max - min);
}
// Driver code
int main()
{
int arr[] = { 1, 2, 3, 4, 5 };
int n = sizeof(arr) / sizeof(arr[0]);
int m = 4;
cout << find_difference(arr, n, m);
return 0;
}
Java
// Java program to find difference
// between max and min sum of array
import java.util.Arrays;
class GFG {
// utility function
static int find_difference(int arr[], int n,
int m)
{
int max = 0, min = 0;
// sort array
Arrays.sort(arr);
for (int i = 0, j = n - 1;
i < m; i++, j--) {
min += arr[i];
max += arr[j];
}
return (max - min);
}
// Driver program
public static void main(String arg[])
{
int arr[] = { 1, 2, 3, 4, 5 };
int n = arr.length;
int m = 4;
System.out.print(find_difference(arr, n, m));
}
}
// This code is contributed by Anant Agarwal.
Python3
# Python program to
# find difference
# between max and
# min sum of array
def find_difference(arr, n, m):
max = 0; min = 0
# sort array
arr.sort();
j = n-1
for i in range(m):
min += arr[i]
max += arr[j]
j = j - 1
return (max - min)
# Driver code
if __name__ == "__main__":
arr = [1, 2, 3, 4, 5]
n = len(arr)
m = 4
print(find_difference(arr, n, m))
# This code is contributed by
# Harshit Saini
C#
// C# program to find difference
// between max and min sum of array
using System;
class GFG {
// utility function
static int find_difference(int[] arr, int n,
int m)
{
int max = 0, min = 0;
// sort array
Array.Sort(arr);
for (int i = 0, j = n - 1;
i < m; i++, j--) {
min += arr[i];
max += arr[j];
}
return (max - min);
}
// Driver program
public static void Main()
{
int[] arr = { 1, 2, 3, 4, 5 };
int n = arr.Length;
int m = 4;
Console.Write(find_difference(arr, n, m));
}
}
// This code is contributed by nitin mittal
PHP
Javascript
输出:
4
我们可以使用下文中讨论的更有效的方法来优化上述解决方案。
数组中的k个最大(或最小)元素|添加了最小堆方法