查找使所有数组元素相等所需的操作数

给定一个由N个整数组成的数组，任务是找到使数组中的所有元素相等所需的操作数。在一个操作中，我们可以从最大元素到其余数组元素分配相等的权重。如果执行上述操作后无法使数组元素相等，则打印-1。

例子：

Input: arr = [1, 6, 1, 1, 1]; 
Output: 4 
Explanation: Since arr becomes [2, 2, 2, 2, 2] after distribution from max element.
Input : arr = [2, 2, 3]; 
Output : -1 
Explanation: Here arr becomes [3, 3, 1] after distribution.  

算法：

• 声明临时变量以存储执行操作的次数。
• 查找给定数组的最大元素并存储其索引值。
• 减去n后，检查所有元素是否等于最大元素。
• 再次检查每个元素是否等于其他元素，然后返回n。

下面是上述方法的实现：

C++

// C++ program to find the number
//of operations required to make
//all array elements Equal
#include
using namespace std;
 
//Function to find maximum
//element of the given array
int find_n(int a[],int n)
{
        int j = 0, k = 0, s = 0;
 
        int x = *max_element(a, a + n);
        int y = *min_element(a, a + n);
        for (int i = 0; i < n; i++)
        {
            if (a[i] == x)
            {
                s = i;
                break;
            }
 
        }
        for (int i =0;i

---

Java

// Java program to find the number
//of operations required to make
//all array elements Equal
 
import java.util.Arrays;
 
class GFG {
 
//Function to find maximum
//element of the given array
    static int find_n(int[] a) {
        int j = 0, k = 0, s = 0;
 
        int x = Arrays.stream(a).max().getAsInt();
        int y = Arrays.stream(a).min().getAsInt();
        for (int i : a) {
            if (a[i] == x) {
                s = i;
                break;
            }
 
        }
        for (int i : a) {
            if (i != x && i <= y && i != 0) {
                a[j] += 1;
                a[s] -= 1;
                x -= 1;
                k += 1;
                j += 1;
            } else if (i != 0) {
                j += 1;
            }
        }
 
        for (int i : a) {
            if (a[i] != x) {
                k = -1;
                break;
            }
        }
        return k;
    }
//Driver Code
 
    public static void main(String[] args) {
 
        int[] a = {1, 6, 1, 1, 1};
        System.out.println(find_n(a));
    }
 
}

Python3

# Python program to find the number
# of operations required to make
# all array elements Equal
 
# Function to find maximum
# element of the given array
def find_n(a):
    j, k = 0, 0
     
    x = max(a)
    for i in range(len(a)):
        if(a[i] == x):
            s = i
            break
     
    for i in a:
        if(i != x and i <= min(a) and i !='\0'):
            a[j] += 1
            a[s] -= 1
            x -= 1
            k += 1
            j += 1
        elif(i != '\0'):
            j += 1
             
    for i in range(len(a)):    
        if(a[i] != x):
            k = -1
        break
 
    return k
 
# Driver Code
a = [1, 6, 1, 1, 1]
print (find_n(a))

C#

// C# program to find the number
// of operations required to make
// all array elements Equal
using System;
using System.Linq;
 
class GFG
{
 
// Function to find maximum
// element of the given array
static int find_n(int []a)
{
    int j = 0, k = 0, s = 0;
 
    int x = a.Max();
    int y = a.Min();
    foreach(int i in a)
    {
        if (a[i] == x)
        {
            s = i;
            break;
        }
 
    }
     
    foreach (int i in a)
    {
        if (i != x && i <= y && i != 0)
        {
            a[j] += 1;
            a[s] -= 1;
            x -= 1;
            k += 1;
            j += 1;
        }
         
        else if (i != 0)
        {
            j += 1;
        }
    }
 
    foreach (int i in a)
    {
        if (a[i] != x)
        {
            k = -1;
            break;
        }
    }
    return k;
}
 
// Driver Code
public static void Main()
{
    int[] a = {1, 6, 1, 1, 1};
    Console.Write(find_n(a));
}
}
 
// This code contributed by 29AjayKumar

Javascript

 
// javascript program to find the number
//of operations required to make
//all array elements Equal
 
 
//Function to find maximum
//element of the given array
function find_n(a) {
    var j = 0, k = 0, s = 0;
 
    var x = Math.max.apply(Math, a);
    var y = Math.min.apply(Math, a);
    for (var i = 0; i < n; i++)
        {
            if (a[i] == x)
            {
                s = i;
                break;
            }
 
        }
        for (var i =0;i<n;i++)
        {
            if (a[i] != x && a[i] <= y && a[i] != 0)
            {
                a[j] += 1;
                a[s] -= 1;
                x -= 1;
                k += 1;
                j += 1;
            }
            else if (a[i] != 0)
            {
                j += 1;
            }
        }
 
        for (var i = 0; i < n; i++)
        {
            if (a[i] != x)
            {
                k = -1;
                break;
            }
        }
        return k;
}
//Driver Code
var a = [1, 6, 1, 1, 1];
var n = a.length;
document.write(find_n(a,n));
 
// This code is contributed by 29AjayKumar

输出：

4

时间复杂度：O(n)

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