使所有数组元素相等的最小增量-其他操作数。

我们得到一个由 n 个元素组成的数组。在每个操作中，您可以选择任何一个元素并将其余 n-1 个元素增加 1。您必须使所有元素相等，执行您希望多次的此类操作。找到为此所需的最少操作数。
例子：

Input : arr[] = {1, 2, 3}
Output : Minimum Operation = 3
Explanation : 
operation | increased elements | after increment
    1     |    1, 2            | 2, 3, 3
    2     |    1, 2            | 3, 4, 3
    3     |    1, 3            | 4, 4, 4

Input : arr[] = {4, 3, 4}
Output : Minimum Operation = 2
Explanation : 
operation | increased elements | after increment
     1    |    1, 2           | 5, 4, 4
     2    |    2, 3           | 5, 5, 5

蛮力：使所有元素相等的一个简单方法是在每一步找到最大的元素，然后将所有其余的 n-1 个元素加 1。我们应该继续执行此操作，直到所有元素相等。时间复杂度：O(n^2)
更好的方法：如果我们仔细研究每个操作以及问题陈述，我们会发现增加除最大元素之外的所有 n-1 元素类似于仅减少最大元素。因此，最小的元素不再需要减少，其余元素将减少到最小的元素。这样，使所有元素相等所需的操作总数将是arraySum – n * (smallestElement)。时间复杂度将与寻找最小元素和数组总和的时间复杂度相同，即 O(n)。

// find array sum
sum = arraySum (int arr[], int n);

// find the smallest element from array
small = smallest (int arr, int n);

// calculate min operation required
minOperation = sum - (n * small);

// return result
return minOperation;

C++

// CPP for finding minimum operation required
#include
using namespace std;
 
// function for finding array sum
int arraySum (int arr[], int n)
{
    int sum = 0;
    for (int i=0; iC++
// [STL] - CPP for finding minimum operation required
#include
using namespace std;
 
// function for finding min operation
int minOp (int arr[], int n)
{
    // find array sum
    int sum = accumulate(arr,arr+n,0);
 
    // find the smallest element from array
    int small = *min_element(arr,arr+n);
 
    // calculate min operation required
    int minOperation = sum - (n * small);
 
    // return result
    return minOperation;
}
 
//driver function
int main()
{
    int arr[] = {5, 6, 2, 4, 3};
    int n = sizeof(arr)/ sizeof(arr[0]);
    cout << "Minimum Operation = " << minOp (arr, n);
    return 0;
}
// This code is contributed by Shubham Singh

Java
// Java program to find min operation
import java.io.*;
  
class minOperation
{
    /* Function to print minimum operation required
       to make all elements of an array equal */
    static void printMinOp(int arr[])
    {
        int arraySum, smallest, arr_size = arr.length;
        arraySum = 0;
        smallest = arr[0];
        for (int i = 0; i < arr_size ; i ++)
        {
            /* If current element is smaller than
               update smallest */
            if (arr[i] < smallest)           
                smallest = arr[i];           
  
            /*find array sum */
            arraySum += arr[i];
        }
 
        int minOperation = arraySum - arr_size * smallest;
 
        /* Print min operation required */ 
        System.out.println("Minimum Operation = " +
                               minOperation);
 
    }
  
    /* Driver program to test above functions */
    public static void main (String[] args)
    {
        int arr[] = {5, 6, 2, 4, 3};
        printMinOp(arr);
    }
}

Python3
# Python 3 for finding minimum
# operation required
 
# function for finding min
# operation
def minOp (arr, n) :
     
    # find array sum
    sm = sum(arr)
 
    # find the smallest element from
    # array
    small = min(arr)
 
    # calculate min operation required
    minOperation = sm - (n * small)
 
    # return result
    return minOperation
     
# Driver function
arr = [5, 6, 2, 4, 3]
n = len(arr)
print( "Minimum Operation = ", minOp (arr, n))
 
# This code is contributed by Shubham Singh

C#
// C# program to find min operation
using System;
 
class GFG {
     
    /* Function to print minimum
    operation required to make all
    elements of an array equal */
    static void printMinOp(int []arr)
    {
        int arraySum, smallest,
        arr_size = arr.Length;
        arraySum = 0;
        smallest = arr[0];
         
        for (int i = 0; i < arr_size ; i ++)
        {
            /* If current element is
            smaller than update smallest */
            if (arr[i] < smallest)        
                smallest = arr[i];        
 
            /*find array sum */
            arraySum += arr[i];
        }
 
        int minOperation = arraySum - 
                        arr_size * smallest;
 
        /* Print min operation required */
        Console.Write("Minimum Operation = " +
                            minOperation);
    }
 
    /* Driver program to test above
    functions */
    public static void Main ()
    {
        int []arr = {5, 6, 2, 4, 3};
         
        printMinOp(arr);
    }
}
 
// This code is contributed by nitin mittal.


Javascript
// javascript program to find min operationclass minOeration{
    /*
     * Function to print minimum operation
     required to make all elements of an array
     * equal
     */
    function printMinOp(arr)
    {
        var arraySum, smallest, arr_size = arr.length;
        arraySum = 0;
        smallest = arr[0];
        for (i = 0; i < arr_size; i++)
        {
         
            /*
             * If current element is smaller than update smallest
             */
            if (arr[i] < smallest)
                smallest = arr[i];
 
            /* find array sum */
            arraySum += arr[i];
        }
 
        var minOperation = arraySum - arr_size * smallest;
 
        /* Print min operation required */
        document.write("Minimum Operation = " + minOperation);
 
    }
 
    /* Driver program to test above functions */
        var arr = [ 5, 6, 2, 4, 3 ];
        printMinOp(arr);
 
// This code is contributed by aashish1995

输出

Minimum Operation = 10