通过重复从一个元素中添加 K 并从其他元素中减去 K 来使 Array 相等的最少操作

给定一个数组arr[]和一个整数K，任务是找到使数组arr[] 的所有元素相等的最少操作次数。在一个操作中，从一个元素中减去K，并将这个K加到另一个元素中。如果不可能，则打印-1 。

例子：

Input: arr[] = {5, 8, 11}, K = 3
Output: 1
Explanation:
Operation 1: Subtract 3 from arr[2](=11) and add 3 to arr[0](=5).
Now, all the elements of the array arr[] are equal.

Input: arr[] = {1, 2, 3, 4}, K = 2
Output: -1

方法：这个问题可以使用贪心算法解决。首先检查数组arr[]的所有元素的总和是否可以被N整除。如果它不可整除，则意味着不可能使数组arr[] 的所有元素都相等。否则，尝试加减K的值，使每个元素等于arr[] / N 的总和。 请按照以下步骤解决此问题：

• 将变量sum初始化为0以存储数组arr[]的所有元素的总和。
• 使用变量i在范围[0, N-1] 中迭代并将sum更新为sum + arr[i]。
• 如果sum % N不等于0，则打印-1并返回。
• 将变量valueAfterDivision初始化为sum/N以存储这么多值，数组arr[] 的每个元素都被存储并计数为0以存储使所有数组元素相等所需的最小操作次数。
• 使用变量i在范围[0, N-1] 中迭代：
  • 如果 abs of valueAfterDivision – arr[i] % K不等于0，则打印-1并返回。
  • 更新计数为计数+ ABS valueAfterDivision – arr[i] / K。
• 完成以上步骤后，打印count/2作为答案。

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 minimum number of
// operations to make all the elements of
// the array equal
void miniOperToMakeAllEleEqual(int arr[], int n, int k)
{
    // Store the sum of the array arr[]
    int sum = 0;
 
    // Traverse through the array
    for (int i = 0; i < n; i++) {
 
        sum += arr[i];
    }
 
    // If it is not possible to make all
    // array element equal
    if (sum % n) {
        cout << -1;
        return;
    }
 
    int valueAfterDivision = sum / n;
 
    // Store the minimum number of operations needed
    int count = 0;
 
    // Traverse through the array
    for (int i = 0; i < n; i++) {
 
        if (abs(valueAfterDivision - arr[i]) % k != 0) {
            cout << -1;
            return;
        }
 
        count += abs(valueAfterDivision - arr[i]) / k;
    }
 
    // Finally, print the minimum number operation
    // to make array elements equal
    cout << count / 2 << endl;
}
 
// Driver Code
int main()
{
 
    // Given Input
    int n = 3, k = 3;
    int arr[3] = { 5, 8, 11 };
 
    // Function Call
    miniOperToMakeAllEleEqual(arr, n, k);
    // This code is contributed by Potta Lokesh
    return 0;
}

Java

// Java Program for the above approach
import java.io.*;
 
class GFG
{
 
  // Function to find the minimum number of
  // operations to make all the elements of
  // the array equal
  static void miniOperToMakeAllEleEqual(int arr[], int n, int k)
  {
    // Store the sum of the array arr[]
    int sum = 0;
 
    // Traverse through the array
    for (int i = 0; i < n; i++) {
 
      sum += arr[i];
    }
 
    // If it is not possible to make all
    // array element equal
    if (sum % n != 0) {
      System.out.println(-1);
      return;
    }
 
    int valueAfterDivision = sum / n;
 
    // Store the minimum number of operations needed
    int count = 0;
 
    // Traverse through the array
    for (int i = 0; i < n; i++) {
 
      if (Math.abs(valueAfterDivision - arr[i]) % k != 0) {
        System.out.println(-1);
        return;
      }
 
      count += Math.abs(valueAfterDivision - arr[i]) / k;
    }
 
    // Finally, print the minimum number operation
    // to make array elements equal
    System.out.println((int)count / 2);
  }
 
  // Driver Code
  public static void main (String[] args)
  {
 
    // Given Input
    int n = 3, k = 3;
    int arr[] = { 5, 8, 11 };
 
    // Function Call
    miniOperToMakeAllEleEqual(arr, n, k);
  }
}
 
// This code is contributed by Potta Lokesh

Python3

# Python3 program for the above approach
 
# Function to find the minimum number of
# operations to make all the elements of
# the array equal
def miniOperToMakeAllEleEqual(arr, n, k):
     
    # Store the sum of the array arr[]
    sum = 0
 
    # Traverse through the array
    for i in range(n):
        sum += arr[i]
 
    # If it is not possible to make all
    # array element equal
    if (sum % n):
        print(-1)
        return
 
    valueAfterDivision = sum // n
 
    # Store the minimum number of operations needed
    count = 0
 
    # Traverse through the array
    for i in range(n):
        if (abs(valueAfterDivision - arr[i]) % k != 0):
            print(-1)
            return
 
        count += abs(valueAfterDivision - arr[i]) // k
 
    # Finally, print the minimum number operation
    # to make array elements equal
    print(count // 2)
 
# Driver Code
if __name__ == '__main__':
     
    # Given Input
    n = 3
    k = 3
    arr = [ 5, 8, 11 ]
 
    # Function Call
    miniOperToMakeAllEleEqual(arr, n, k)
 
# This code is contributed by ipg2016107

C#

// C# program for the above approach
using System;
 
class GFG
{
   
    // Function to find the minimum number of
    // operations to make all the elements of
    // the array equal
    static void miniOperToMakeAllEleEqual(int[] arr, int n,
                                          int k)
    {
        // Store the sum of the array arr[]
        int sum = 0;
 
        // Traverse through the array
        for (int i = 0; i < n; i++) {
 
            sum += arr[i];
        }
 
        // If it is not possible to make all
        // array element equal
        if (sum % n != 0) {
            Console.WriteLine(-1);
            return;
        }
 
        int valueAfterDivision = sum / n;
 
        // Store the minimum number of operations needed
        int count = 0;
 
        // Traverse through the array
        for (int i = 0; i < n; i++) {
 
            if (Math.Abs(valueAfterDivision - arr[i]) % k
                != 0) {
                Console.WriteLine(-1);
                return;
            }
 
            count += Math.Abs(valueAfterDivision - arr[i])
                     / k;
        }
 
        // Finally, print the minimum number operation
        // to make array elements equal
        Console.WriteLine((int)count / 2);
    }
 
    static void Main()
    {
        // Given Input
        int n = 3, k = 3;
        int[] arr = { 5, 8, 11 };
 
        // Function Call
        miniOperToMakeAllEleEqual(arr, n, k);
    }
}
 
// This code is contributed by abhinavjain194

Javascript

       // JavaScript program for the above approach
 
       // Function to find the minimum number of
       // operations to make all the elements of
       // the array equal
       function miniOperToMakeAllEleEqual(arr, n, k)
       {
        
           // Store the sum of the array arr[]
           let sum = 0;
 
           // Traverse through the array
           for (let i = 0; i < n; i++) {
 
               sum += arr[i];
           }
 
           // If it is not possible to make all
           // array element equal
           if (sum % n) {
               document.write(-1);
               return;
           }
 
           let valueAfterDivision = sum / n;
 
           // Store the minimum number of operations needed
           let count = 0;
 
           // Traverse through the array
           for (let i = 0; i < n; i++) {
 
               if (Math.abs(valueAfterDivision - arr[i]) % k != 0) {
                   document.write(-1);
                   return;
               }
 
               count += Math.abs(valueAfterDivision - arr[i]) / k;
           }
 
           // Finally, print the minimum number operation
           // to make array elements equal
           document.write(Math.floor(count / 2));
       }
 
       // Driver Code
 
 
       // Given Input
       let n = 3, k = 3;
       let arr = [5, 8, 11];
 
       // Function Call
       miniOperToMakeAllEleEqual(arr, n, k);
 
   // This code is contributed by Potta Lokesh
 
   

输出

1

时间复杂度： O(N)
辅助空间： O(1)

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