数组的最小和最大元素的总和与乘积

给定一个数组。任务是找到给定数组的最大和最小元素的和与乘积。

例子：

Input : arr[] = {12, 1234, 45, 67, 1}
Output : Sum = 1235
         Product = 1234

Input : arr[] = {5, 3, 6, 8, 4, 1, 2, 9}
Output : Sum = 10
         Product = 9

取变量min和max来存储数组的最小和最大元素。找到最小和最大元素并分别存储在这些变量中。最后，打印最小和最大元素的总和与乘积。

下面是说明上述方法的程序：

C++

// CPP program to find the sum  and product
// of minimum and maximum element in an array
  
#include 
using namespace std;
  
// Function to find minimum element
int getMin(int arr[], int n)
{
    int res = arr[0];
    for (int i = 1; i < n; i++)
        res = min(res, arr[i]);
    return res;
}
  
// Function to find maximum element
int getMax(int arr[], int n)
{
    int res = arr[0];
    for (int i = 1; i < n; i++)
        res = max(res, arr[i]);
    return res;
}
  
// Function to get Sum
int findSum(int arr[], int n)
{
    int min = getMin(arr, n);
    int max = getMax(arr, n);
  
    return min + max;
}
  
// Function to get product
int findProduct(int arr[], int n)
{
    int min = getMin(arr, n);
    int max = getMax(arr, n);
  
    return min * max;
}
  
// Driver Code
int main()
{
    int arr[] = { 12, 1234, 45, 67, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    // Sum of min and max element
    cout << "Sum = " << findSum(arr, n) << endl;
  
    // Product of min and max element
    cout << "Product = " << findProduct(arr, n);
  
    return 0;
}

Java

// Java  program to find the sum and product
// of minimum and maximum element in an array
  
import java.io.*;
  
class GFG {
      
    // Function to find minimum element
static int getMin(int arr[], int n)
{
    int res = arr[0];
    for (int i = 1; i < n; i++)
        res = Math.min(res, arr[i]);
    return res;
}
  
// Function to find maximum element
static int getMax(int arr[], int n)
{
    int res = arr[0];
    for (int i = 1; i < n; i++)
        res = Math.max(res, arr[i]);
    return res;
}
  
// Function to get Sum
static int findSum(int arr[], int n)
{
    int min = getMin(arr, n);
    int max = getMax(arr, n);
  
    return min + max;
}
  
// Function to get product
static int findProduct(int arr[], int n)
{
    int min = getMin(arr, n);
    int max = getMax(arr, n);
  
    return min * max;
}
  
// Driver Code
      
    public static void main (String[] args) {
    int arr[] = { 12, 1234, 45, 67, 1 };
    int n = arr.length;
  
    // Sum of min and max element
        System.out.println ("Sum = " + findSum(arr, n));
  
    // Product of min and max element
        System.out.println( "Product = " + findProduct(arr, n));
  
          
          
    }
}
//This Code is contributed by anuj_67....

Python 3

# Python 3 program to find the sum and product
# of minimum and maximum element in an array
  
# Function to find minimum element
def getMin(arr, n):
    res = arr[0]
    for i in range(1, n):
        res = min(res, arr[i])
    return res
  
# Function to find maximum element
def getMax(arr, n):
    res = arr[0]
    for i in range(1, n):
        res = max(res, arr[i])
    return res
  
# Function to get Sum
def findSum(arr, n):
    min = getMin(arr, n)
    max = getMax(arr, n)
  
    return min + max
  
# Function to get product
def findProduct(arr, n):
    min = getMin(arr, n)
    max = getMax(arr, n)
  
    return min * max
  
# Driver Code
if __name__ == "__main__":
      
    arr = [ 12, 1234, 45, 67, 1 ]
    n = len(arr)
  
    # Sum of min and max element
    print("Sum = " , findSum(arr, n))
  
    # Product of min and max element
    print("Product = " , findProduct(arr, n))
  
# This code is contributed 
# by ChitraNayal

C#

// C# program to find the sum and product
// of minimum and maximum element in an array
using System;
  
class GFG {
      
// Function to find minimum element
static int getMin(int []arr, int n)
{
    int res = arr[0];
    for (int i = 1; i < n; i++)
        res = Math.Min(res, arr[i]);
    return res;
}
  
// Function to find maximum element
static int getMax(int []arr, int n)
{
    int res = arr[0];
    for (int i = 1; i < n; i++)
        res = Math.Max(res, arr[i]);
    return res;
}
  
// Function to get Sum
static int findSum(int []arr, int n)
{
    int min = getMin(arr, n);
    int max = getMax(arr, n);
  
    return min + max;
}
  
// Function to get product
static int findProduct(int []arr, int n)
{
    int min = getMin(arr, n);
    int max = getMax(arr, n);
  
    return min * max;
}
  
    // Driver Code
    public static void Main() 
    {
        int []arr = { 12, 1234, 45, 67, 1 };
        int n = arr.Length;
      
        // Sum of min and max element
        Console.WriteLine("Sum = " + findSum(arr, n));
  
        // Product of min and max element
        Console.WriteLine( "Product = " + findProduct(arr, n));
    }
}
  
// This Code is contributed by anuj_67....

---

输出：

Sum = 1235
Product = 1234

最佳化
我们可以使用一个循环来找到最大值和最小值。这将只需要遍历数组一次。

在C++中，有直接函数可以找到最大值和最小值：max_element()和min_element()