最小化递增或递减操作以使给定的 Array 元素连续

给定一个数组， arr[] ，大小为N .，任务是对数组的元素进行最小递增或递减操作，使所有元素连续，输出所有可能变化的最小和（加法和减法）需要做同样的事情。

例子：

Input: N = 5, arr[] = {13, 6, 11, 18, 4}
Output: 15
Explanation: Convert 4 to 9, 8 to 10, 13 to12 and 18to13, the new array becomes {9, 10, 11, 12, 13}. 
So the sum of changes are abs(9-4) + abs(10-8) + abs(12-13) + abs(13-18) = 15.

Input: N = 2, arr[] = {3, 8}
Output: 4

方法：可以使用观察来解决任务。数组元素的中位数应保持不变，其余元素应相应更改，以使所有元素变为连续。
请按照以下步骤解决问题：

• 对数组进行排序
• 取一个存储数组中位数的变量mid和一个存储其位置的变量pos 。
• 另外，取一个变量ele并用 这 结果数组的最小值 即(mid – pos)和一个变量sum = 0来存储数组元素中所有可能变化的总和。
• 遍历数组并在每次第 i次迭代中：
  • 用abs(arr[i]-ele)递增总和（添加原始元素和所需元素的差）
  • 将ele增加 1
• 输出总和。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to output sum of all the
// required changes in array
int minChanges(int arr[], int N)
{
    sort(arr, arr + N);
 
    // Stores median of array
    int mid = arr[N / 2];
 
    // Variable for position of median
    int pos = N / 2;
 
    // Smallest element of
    // the required array
    int ele = mid - pos;
    int sum = 0;
 
    // Loop to find  sum of changes
    for (int i = 0; i < N; i++) {
 
        // Adding difference of original
        // and required element to answer
        sum += abs(arr[i] - ele);
        ele++;
    }
    return sum;
}
 
// Driver code
int main()
{
    int N = 5;
    int arr[] = { 13, 6, 11, 18, 4 };
 
    cout << minChanges(arr, N);
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG
{
 
  // Function to output sum of all the
  // required changes in array
  static int minChanges(int arr[], int N)
  {
    Arrays.sort(arr);
 
    // Stores median of array
    int mid = arr[N / 2];
 
    // Variable for position of median
    int pos = N / 2;
 
    // Smallest element of
    // the required array
    int ele = mid - pos;
    int sum = 0;
 
    // Loop to find  sum of changes
    for (int i = 0; i < N; i++)
    {
 
      // Adding difference of original
      // and required element to answer
      sum += Math.abs(arr[i] - ele);
      ele++;
    }
    return sum;
  }
 
  // Driver code
  public static void main(String[] args)
  {
 
    int N = 5;
    int arr[] = { 13, 6, 11, 18, 4 };
    int ans = minChanges(arr, N);
    System.out.println(ans);
  }
}
 
// This code is contributed by hrithikgarg03188

Python3

# Python code for the above approach
import math as Math 
 
# Function to output sum of all the
# required changes in array
def minChanges(arr, N):
    arr.sort();
 
    # Stores median of array
    mid = arr[N // 2];
 
    # Variable for position of median
    pos = N // 2;
 
    # Smallest element of
    # the required array
    ele = mid - pos;
    sum = 0;
 
    # Loop to find  sum of changes
    for i in range(N):
 
        # Adding difference of original
        # and required element to answer
        sum += Math.fabs(arr[i] - ele);
        ele += 1
    return int(sum);
 
# Driver code
N = 5;
arr = [13, 6, 11, 18, 4];
print(minChanges(arr, N));
 
# This code is contributed by Saurabh Jaiswal

C#

// C# program for above approach
using System;
using System.Collections.Generic;
 
public class GFG
{
   
  // Function to output sum of all the
  // required changes in array
  static int minChanges(int[ ] arr, int N)
  {
    Array.Sort(arr);
 
    // Stores median of array
    int mid = arr[N / 2];
 
    // Variable for position of median
    int pos = N / 2;
 
    // Smallest element of
    // the required array
    int ele = mid - pos;
    int sum = 0;
 
    // Loop to find  sum of changes
    for (int i = 0; i < N; i++)
    {
       
      // Adding difference of original
      // and required element to answer
      sum += Math.Abs(arr[i] - ele);
      ele++;
    }
    return sum;
  }
 
  // Driver code
  public static void Main(String[] args)
  {
    int N = 5;
    int[ ] arr = { 13, 6, 11, 18, 4 };
 
    Console.WriteLine(minChanges(arr, N));
  }
}
 
// This code is contributed by hrithikgarg03188

Javascript

      // JavaScript code for the above approach
 
 
      // Function to output sum of all the
      // required changes in array
      function minChanges(arr, N) {
          arr.sort(function (a, b) { return a - b })
 
          // Stores median of array
          let mid = arr[Math.floor(N / 2)];
 
          // Variable for position of median
          let pos = Math.floor(N / 2);
 
          // Smallest element of
          // the required array
          let ele = mid - pos;
          let sum = 0;
 
          // Loop to find  sum of changes
          for (let i = 0; i < N; i++) {
 
              // Adding difference of original
              // and required element to answer
              sum += Math.abs(arr[i] - ele);
              ele++;
          }
          return sum;
      }
 
      // Driver code
 
      let N = 5;
      let arr = [13, 6, 11, 18, 4];
 
      document.write(minChanges(arr, N));
 
// This code is contributed by Potta Lokesh
  

输出

15

时间复杂度：O(N * logN)
辅助空间：O(1)