查找数组的N-1个元素的最小和最大和

给定大小为N的未排序数组A ，任务是找到可以通过恰好相加N-1个元素来计算出的最小值和最大值。

例子：

Input: a[] = {13, 5, 11, 9, 7} 
Output: 32 40 
Explanation: Minimum sum is 5 + 7 + 9 + 11 = 32 and maximum sum is 7 + 9 + 11 + 13 = 40.
Input: a[] = {13, 11, 45, 32, 89, 21} 
Output: 122 200 
Explanation: Minimum sum is 11 + 13 + 21 + 32 + 45 = 122 and maximum sum is 13 + 21 + 32 + 45 + 89 = 200.
Input: a[] = {6, 3, 15, 27, 9} 
Output: 33 57 
Explanation: Minimum sum is 3 + 6 + 9 + 15 = 33 and maximum sum is 6 + 9 + 15 + 27 = 57.

简单方法：

1. 以升序对数组进行排序。
2. 数组中前N-1个元素的总和给出了最小可能的总和。
3. 数组中最后N-1个元素的总和给出最大可能的总和。

下面是上述方法的实现：

Java

// Java Implementation of the above approach
import java.util.*;
 
class GFG {
 
    static void minMax(int[] arr)
    {
        // Initialize the min_value
        // and max_value to 0
        long min_value = 0;
        long max_value = 0;
        int n = arr.length;
       
        // Sort array before calculating
        // min and max value
        Arrays.sort(arr);
                           
        for (int i = 0, j = n - 1;
             i < n - 1; i++, j--)
        {
            // All elements except
            // rightmost will be added
            min_value += arr[i];
           
            // All elements except
            // leftmost will be added
            max_value += arr[j];
        }
        
        // Output: min_value and max_value
        System.out.println(
            min_value + " "
            + max_value);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        Scanner sc = new Scanner(System.in);
 
        // Initialize your array elements here
        int[] arr = { 10, 9, 8, 7, 6, 5 };
        int[] arr1 = { 100, 200, 300, 400, 500 };
        minMax(arr);
        minMax(arr1);
    }
}

输出：

35 40
1000 1400

时间复杂度： O(NlogN)

高效方法：

1. 查找数组的最小和最大元素。
2. 计算数组中所有元素的总和。
3. 从总和中排除最大元素将得出最小的总和。
4. 从总和中排除最小的元素可以得到最大的总和。

下面是上述方法的实现：

C++

// C++ program to find the minimum and maximum
// sum from an array.
#include 
using namespace std;
 
// Function to calculate minimum and maximum sum
static void miniMaxSum(int arr[], int n)
{
 
    // Initialize the minElement, maxElement
    // and sum by 0.
    int minElement = 0, maxElement = 0, sum = 0;
 
    // Assigning maxElement, minElement
    // and sum as the first array element
    minElement = arr[0];
    maxElement = minElement;
    sum = minElement;
 
    // Traverse the entire array
    for(int i = 1; i < n; i++)
    {
         
        // Calculate the sum of
        // array elements
        sum += arr[i];
 
        // Keep updating the
        // minimum element
        if (arr[i] < minElement)
        {
            minElement = arr[i];
        }
 
        // Keep updating the
        // maximum element
        if (arr[i] > maxElement)
        {
            maxElement = arr[i];
        }
    }
 
    // print the minimum and maximum sum
    cout << (sum - maxElement) << " "
         << (sum - minElement) << endl;
}
 
// Driver Code
int main()
{
     
    // Test Case 1:
    int a1[] = { 13, 5, 11, 9, 7 };
    int n = sizeof(a1) / sizeof(a1[0]);
     
    // Call miniMaxSum()
    miniMaxSum(a1, n);
 
    // Test Case 2:
    int a2[] = { 13, 11, 45, 32, 89, 21 };
    n = sizeof(a2) / sizeof(a2[0]);
    miniMaxSum(a2, n);
 
    // Test Case 3:
    int a3[] = { 6, 3, 15, 27, 9 };
    n = sizeof(a3) / sizeof(a3[0]);
    miniMaxSum(a3, n);
}
 
// This code is contributed by chitranayal

Java

// Java program to find the minimum and maximum
// sum from an array.
class GFG {
 
    // Function to calculate minimum and maximum sum
    static void miniMaxSum(int[] arr)
    {
 
        // Initialize the minElement, maxElement
        // and sum by 0.
        int minElement = 0, maxElement = 0, sum = 0;
 
        // Assigning maxElement, minElement
        // and sum as the first array element
        minElement = arr[0];
        maxElement = minElement;
        sum = minElement;
 
        // Traverse the entire array
        for (int i = 1; i < arr.length; i++) {
 
            // calculate the sum of
            // array elements
            sum += arr[i];
 
            // Keep updating the
            // minimum element
            if (arr[i] < minElement) {
                minElement = arr[i];
            }
 
            // Keep updating the
            // maximum element
            if (arr[i] > maxElement) {
                maxElement = arr[i];
            }
        }
 
        // print the minimum and maximum sum
        System.out.println((sum - maxElement) + " "
                        + (sum - minElement));
    }
 
    // Driver Code
    public static void main(String args[])
    {
 
        // Test Case 1:
        int a1[] = { 13, 5, 11, 9, 7 };
        // Call miniMaxSum()
        miniMaxSum(a1);
 
        // Test Case 2:
        int a2[] = { 13, 11, 45, 32, 89, 21 };
        miniMaxSum(a2);
 
        // Test Case 3:
        int a3[] = { 6, 3, 15, 27, 9 };
        miniMaxSum(a3);
    }
}

Python3

# Python3 program to find the minimum and
# maximum sum from a list.
 
# Function to calculate minimum and maximum sum
def miniMaxSum(arr, n):
 
    # Initialize the minElement, maxElement
    # and sum by 0.
    minElement = 0
    maxElement = 0
    sum = 0
 
    # Assigning maxElement, minElement
    # and sum as the first list element
    minElement = arr[0]
    maxElement = minElement
    sum = minElement
 
    # Traverse the entire list
    for i in range(1, n):
 
        # Calculate the sum of
        # list elements
        sum += arr[i]
 
        # Keep updating the
        # minimum element
        if (arr[i] < minElement):
            minElement = arr[i]
 
        # Keep updating the
        # maximum element
        if (arr[i] > maxElement):
            maxElement = arr[i]
 
    # Print the minimum and maximum sum
    print(sum - maxElement,
          sum - minElement)
 
# Driver Code
 
# Test Case 1:
a1 = [ 13, 5, 11, 9, 7 ]
n = len(a1)
 
# Call miniMaxSum()
miniMaxSum(a1, n)
 
# Test Case 2:
a2 = [ 13, 11, 45, 32, 89, 21 ]
n = len(a2)
miniMaxSum(a2, n)
 
# Test Case 3:
a3 = [ 6, 3, 15, 27, 9 ]
n = len(a3)
miniMaxSum(a3, n)
 
# This code is contributed by vishu2908

C＃

// C# program to find the minimum and maximum
// sum from an array.
using System;
 
class GFG{
 
// Function to calculate minimum and maximum sum
static void miniMaxSum(int[] arr)
{
     
    // Initialize the minElement, maxElement
    // and sum by 0.
    int minElement = 0, maxElement = 0, sum = 0;
 
    // Assigning maxElement, minElement
    // and sum as the first array element
    minElement = arr[0];
    maxElement = minElement;
    sum = minElement;
 
    // Traverse the entire array
    for(int i = 1; i < arr.Length; i++)
    {
         
        // Calculate the sum of
        // array elements
        sum += arr[i];
 
        // Keep updating the
        // minimum element
        if (arr[i] < minElement)
        {
            minElement = arr[i];
        }
 
        // Keep updating the
        // maximum element
        if (arr[i] > maxElement)
        {
            maxElement = arr[i];
        }
    }
 
    // Print the minimum and maximum sum
    Console.WriteLine((sum - maxElement) + " " +
                      (sum - minElement));
}
 
// Driver Code
public static void Main()
{
     
    // Test Case 1:
    int[] a1 = new int[]{ 13, 5, 11, 9, 7 };
     
    // Call miniMaxSum()
    miniMaxSum(a1);
 
    // Test Case 2:
    int[] a2 = new int[]{ 13, 11, 45, 32, 89, 21 };
    miniMaxSum(a2);
 
    // Test Case 3:
    int[] a3 = new int[]{ 6, 3, 15, 27, 9 };
    miniMaxSum(a3);
}
}
 
// This code is contributed by sanjoy_62

输出

32 40
122 200
33 57

时间复杂度： O(N)