数组包含从1到N的所有元素的最小增/减操作数

给定一个由N个元素组成的数组，任务是通过使用以下最少次数的操作将其转换为一个排列(从1到N的每个数字恰好发生一次)：

• 递增任何数字。
• 减少任何数字。

例子：

Input: arr[] = {1, 1, 4}
Output: 2
The array can be converted into [1, 2, 3]
by adding 1 to the 1st index i.e. 1 + 1 = 2
and decrementing 2nd index by 1 i.e. 4- 1 = 3

Input: arr[] = {3, 0}
Output: 2

The array can be converted into [2, 1]

方法：为了最大程度地减少移动/操作的次数，请对给定的数组进行排序，并使a [i] = i + 1(从0开始)，这将采用abs(i + 1-a [i])否。每个元素的操作数。

下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include 
using namespace std;
  
// Function to find the minimum operations
long long minimumMoves(int a[], int n)
{
  
    long long operations = 0;
  
    // Sort the given array
    sort(a, a + n);
  
    // Count operations by assigning a[i] = i+1
    for (int i = 0; i < n; i++)
        operations += abs(a[i] - (i + 1));
  
    return operations;
}
  
// Driver Code
int main()
{
    int arr[] = { 5, 3, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << minimumMoves(arr, n);
  
    return 0;
}

Java

// Java implementation of the above approach
  
import java.util.*;
class solution
{
// Function to find the minimum operations
static long minimumMoves(int a[], int n)
{
   
    long operations = 0;
   
    // Sort the given array
    Arrays.sort(a);
   
    // Count operations by assigning a[i] = i+1
    for (int i = 0; i < n; i++)
        operations += (long)Math.abs(a[i] - (i + 1));
   
    return operations;
}
   
// Driver Code
public static void main(String args[])
{
    int arr[] = { 5, 3, 2 };
    int n = arr.length;
   
    System.out.print(minimumMoves(arr, n));
  
}
  
}
//contributed by Arnab Kundu

Python3

# Python 3 implementation of the above approach
  
# Function to find the minimum operations
def minimumMoves(a, n):
      
    operations = 0
    # Sort the given array
    a.sort(reverse = False)
      
    # Count operations by assigning a[i] = i+1
    for i in range(0,n,1):
        operations = operations + abs(a[i] - (i + 1))
  
    return operations
  
# Driver Code
if __name__ == '__main__':
    arr = [ 5, 3, 2 ]
    n = len(arr)
  
    print(minimumMoves(arr, n))
  
# This code is contributed by 
# Surendra_Gangwar

C#

// C# implementation of the above approach 
using System;
  
class GFG
{
// Function to find the minimum operations 
static long minimumMoves(int []a, int n) 
{ 
  
    long operations = 0; 
  
    // Sort the given array 
    Array.Sort(a); 
  
    // Count operations by assigning 
    // a[i] = i+1 
    for (int i = 0; i < n; i++) 
        operations += (long)Math.Abs(a[i] - (i + 1)); 
  
    return operations; 
} 
  
// Driver Code 
static public void Main ()
{
    int []arr = { 5, 3, 2 }; 
    int n = arr.Length; 
      
    Console.WriteLine(minimumMoves(arr, n)); 
}
}
  
// This code is contributed by Sach_Code

输出：

4

时间复杂度： O(NlogN)