从数组元素中减去以使所有数组元素相等的值的最小总和

给定一个由N个正整数组成的数组arr [] ，任务是找到需要从每个数组元素中减去的所有数组元素的总和，以使其余数组元素都相等。

例子：

Input: arr[] = {1, 2}
Output: 1
Explanation: Subtracting 1 from arr[1] modifies arr[] to {1, 1}. Therefore, the required sum is 1.

Input: arr[] = {1, 2, 3}
Output: 3
Explanation: Subtracting 1 and 2 from arr[1] and arr[2] modifies arr[] to {1, 1, 1}. Therefore, the required sum = 1 + 2 = 3.

方法：想法是将所有数组元素减少到数组中存在的最小元素。请按照以下步骤解决问题：

• 初始化一个变量，例如sum ，以存储所有相减值的和。
• 使用min_element()找到存在于数组中的最小元素，例如minimum 。
• 遍历数组，并为每个数组元素(例如arr [i] )加上(arr [i] –最小值)到所需的总和。
• 遍历数组后，打印获得的总和。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the sum of values
// removed to make all array elements equal
int minValue(int arr[], int n)
{
    // Stores the minimum of the array
    int minimum = *min_element(
        arr, arr + n);
 
    // Stores required sum
    int sum = 0;
 
    // Traverse the array
    for (int i = 0; i < n; i++) {
 
        // Add the value subtracted
        // from the current elemnt
        sum = sum + (arr[i] - minimum);
    }
 
    // Return the total sum
    return sum;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    cout << minValue(arr, N);
    return 0;
}

Java

// Java program for the above approach
import java.util.Arrays;
class GFG
{
     
// Function to find the sum of values
// removed to make all array elements equal
static int minValue(int []arr, int n)
{
    Arrays.sort(arr);
     
    // Stores the minimum of the array
    int minimum = arr[0];
 
    // Stores required sum
    int sum = 0;
 
    // Traverse the array
    for(int i = 0; i < n; i++)
    {
         
        // Add the value subtracted
        // from the current elemnt
        sum = sum + (arr[i] - minimum);
    }
     
    // Return the total sum
    return sum;
}
 
// Driver Code
static public void main(String args[])
{
    int []arr = { 1, 2, 3 };
    int N = arr.length;
     
    // Function Call
    System.out.println(minValue(arr, N));
}
}
 
// This code is contributed by AnkThon

Python3

# Python3 program for the above approach
 
# Function to find the sum of values
# removed to make all array elements equal
def minValue(arr, n):
     
    # Stores the minimum of the array
    minimum = min(arr)
 
    # Stores required sum
    sum = 0
 
    # Traverse the array
    for i in range(n):
         
        # Add the value subtracted
        # from the current elemnt
        sum = sum + (arr[i] - minimum)
         
    # Return the total sum
    return sum
 
# Driver Code
if __name__ == '__main__':
     
    arr = [ 1, 2, 3 ]
    N = len(arr)
     
    # Function Call
    print(minValue(arr, N))
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
 
class GFG{
     
// Function to find the sum of values
// removed to make all array elements equal
static int minValue(int []arr, int n)
{
    Array.Sort(arr);
     
    // Stores the minimum of the array
    int minimum = arr[0];
 
    // Stores required sum
    int sum = 0;
 
    // Traverse the array
    for(int i = 0; i < n; i++)
    {
         
        // Add the value subtracted
        // from the current elemnt
        sum = sum + (arr[i] - minimum);
    }
     
    // Return the total sum
    return sum;
}
 
// Driver Code
static public void Main ()
{
    int []arr = { 1, 2, 3 };
    int N = arr.Length;
     
    // Function Call
    Console.WriteLine(minValue(arr, N));
}
}
 
// This code is contributed by AnkThon

Javascript

 
// Javascript program for the above approach
 
// Function to find the sum of values
// removed to make all array elements equal
function minValue(arr, n)
{
    // Stores the minimum of the array
    var minimum = Math.min.apply(Math,arr);
 
    // Stores required sum
    var sum = 0;
     
    var i;
    // Traverse the array
    for (i = 0; i < n; i++) {
 
        // Add the value subtracted
        // from the current elemnt
        sum = sum + (arr[i] - minimum);
    }
 
    // Return the total sum
    return sum;
}
 
// Driver Code
 
 var arr = [1, 2, 3];
 var N = arr.length;
 
// Function Call
 document.write(minValue(arr, N));
 

输出：

3

时间复杂度： O(N)
辅助空间： O(1)