通过 k 次操作使所有元素相等的最小增量

给定一个包含 n 个元素的数组，您必须找到使数组中的所有元素相等所需的操作数。其中单个操作可以将元素增加 k。如果不可能使所有元素都相等，则打印 -1。
例子 ：

Input : arr[] = {4, 7, 19, 16},  k = 3
Output : 10

Input : arr[] = {4, 4, 4, 4}, k = 3
Output : 0

Input : arr[] = {4, 2, 6, 8}, k = 3
Output : -1

为了解决这个问题，我们需要检查所有元素是否可以变得相等，并且也只能通过从元素值增加 k 来实现。为此，我们必须检查任何两个元素的差值应该总是可以被 k 整除。如果是这样，那么所有元素都可以变得相等，否则它们在任何情况下都不能通过从它们增加 k 来变得相等。此外，可以通过找到所有元素的 (max – Ai)/k 值来计算所需的操作次数。其中 max 是数组的最大元素。
算法 ：

// iterate for all elements
for (int i=0; i

C++

// Program to make all array equal
#include 
using namespace std;
 
// function for calculating min operations
int minOps(int arr[], int n, int k)
{
    // max elements of array
    int max = *max_element(arr, arr + n);
    int res = 0;
 
    // iterate for all elements
    for (int i = 0; i < n; i++) {
 
        // check if element can make equal to
        // max or not if not then return -1
        if ((max - arr[i]) % k != 0)
            return -1;
 
        // else update res for required operations
        else
            res += (max - arr[i]) / k;
    }
 
    // return result
    return res;
}
 
// driver program
int main()
{
    int arr[] = { 21, 33, 9, 45, 63 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 6;
    cout << minOps(arr, n, k);
    return 0;
}

Java

// Program to make all array equal
import java.io.*;
import java.util.Arrays;
 
class GFG {
    // function for calculating min operations
    static int minOps(int arr[], int n, int k)
    {
        // max elements of array
        Arrays.sort(arr);
        int max = arr[arr.length - 1];
        int res = 0;
 
        // iterate for all elements
        for (int i = 0; i < n; i++) {
 
            // check if element can make equal to
            // max or not if not then return -1
            if ((max - arr[i]) % k != 0)
                return -1;
 
            // else update res for required operations
            else
                res += (max - arr[i]) / k;
        }
 
        // return result
        return res;
    }
 
    // Driver program
    public static void main(String[] args)
    {
        int arr[] = { 21, 33, 9, 45, 63 };
        int n = arr.length;
        int k = 6;
        System.out.println(minOps(arr, n, k));
    }
}
 
// This code is contributed by vt_m

Python3

# Python3 Program to make all array equal
 
# function for calculating min operations
def minOps(arr, n, k):
 
    # max elements of array
    max1 = max(arr)
    res = 0
 
    # iterate for all elements
    for i in range(0, n):
 
        # check if element can make equal to
        # max or not if not then return -1
        if ((max1 - arr[i]) % k != 0):
            return -1
 
        # else update res fo
        # required operations
        else:
            res += (max1 - arr[i]) / k
     
    # return result
    return int(res)
 
# driver program
arr = [21, 33, 9, 45, 63]
n = len(arr)
k = 6
print(minOps(arr, n, k))
 
# This code is contributed by
# Smitha Dinesh Semwal

C#

// C# program Minimum increment by
// k operations to make all elements equal.
using System;
 
class GFG {
     
    // function for calculating min operations
    static int minOps(int[] arr, int n, int k)
    {
         
        // max elements of array
        Array.Sort(arr);
        int max = arr[arr.Length - 1];
        int res = 0;
 
        // iterate for all elements
        for (int i = 0; i < n; i++) {
 
            // check if element can make
            // equal to max or not if not
            // then return -1
            if ((max - arr[i]) % k != 0)
                return -1;
 
            // else update res for required
            // operations
            else
                res += (max - arr[i]) / k;
        }
 
        // return result
        return res;
    }
 
    // Driver program
    public static void Main()
    {
        int[] arr = { 21, 33, 9, 45, 63 };
        int n = arr.Length;
        int k = 6;
 
        Console.Write(minOps(arr, n, k));
    }
}
 
// This code is contributed by nitin mittal.

Javascript

 
// Program to make all array equal
 
// function for calculating min operations
function minOps(arr, n, k)
{
    // max elements of array
    var max = arr[0];
     
    for(var i=0; i<arr.length; i++)
    {
        if(arr[i]>max)
            max = arr[i];
    }
    var res = 0;
 
    // iterate for all elements
    for (var i = 0; i < n; i++) {
 
        // check if element can make equal to
        // max or not if not then return -1
        if ((max - arr[i]) % k != 0)
            return -1;
 
        // else update res for required operations
        else
            res += (max - arr[i]) / k;
    }
 
    // return result
    return res;
}
 
// driver program
var arr = [ 21, 33, 9, 45, 63 ];
var n = arr.length;
var k = 6;
document.write( minOps(arr, n, k));
 
// This code is contributed by rutvik_56.

输出 ：

24

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