最小化操作次数以使所有元素在给定条件下相等

给定一个数组arr[]。任务是最小化使arr[]中的所有元素相等所需的操作数。几乎可以用任何其他元素替换arr[]中的任何元素。找到这样做所需的最小操作数，在一个操作中采用arr[] 的任何后缀并将该后缀中的值递增/递减1 。

例子

Input: arr[] = {-1, 0, 2}
Output: 1
Explanation: Following are the operations done to make all the elements in arr[] to be equal. 
Initially, change the last element of array to 0, so arr[] = {-1, 0, 0}
Now, using the operation once on the suffix starting at arr₂, which means arr₂ and arr₃ are decreased by 1 . Thus, making all elements of array -1.
Hence, the number of operations is 1. 

Input: arr[] = {-3, -5, -2, 1 }
Output : 4

方法：这个问题是基于实现的。请按照以下步骤解决给定的问题。

• 因为，不需要对从 arr ₁开始的后缀执行任何操作，因为这可以更改数组中的所有整数。
• 因此，使arr _i等于arr _i-1的唯一方法 是从a _i开始对 suffix 进行运算， abs(a _i −a _i-1 )次。
• 现在，最初更改数组中的值的最佳方法是最小化操作。
• 为了使arr ₁等于arr ₂ ，最小操作减少了 abs (arr ₂ – arr ₁ ) 。
• 类似地，为了使arr _n等于 arr _n-1 ，操作 = abs(arr _n – arr _n-1 ) 。
• 对于左元素，更改任何元素arr _i , 影响abs(a _i -a _i-1 ) 和 abs(a _i+1 -a _i ) 。
• 此外，请注意这一重要事实，即当 a _i介于 a _i-1和 a _i+1之间时，该值最小化。
• 因此，运算次数从abs(a _i -a _i-1 )+abs(a _i+1 -a _i ) 减少到 abs(a _i+1 -a _i-1 ) 。
• 返回最终答案。

下面是上述方法的实现：

C++

// C++ program for above approach
#include 
using namespace std;
 
void findMinOperations(vector ar, int n)
{
 
    // Initializing vector to avoid overflows
    vector arr(n + 5);
 
    for (int i = 1; i <= n; i++) {
        arr[i] = ar[i - 1];
    }
    int result = 0;
 
    // Calculating minimum operations to be
    // performed on initial array
    for (int i = 2; i <= n; i++) {
        result += abs(arr[i] - arr[i - 1]);
    }
 
    // Way to change a value to make
    // a1 equal to a2 or a(n)
    // equal to a(n-1)
    int max_operations
        = max(abs(arr[1] - arr[2]),
              abs(arr[n] - arr[n - 1]));
    for (int i = 2; i < n; i++) {
 
        // For the rest of elements
        // taking the max of
        // operations already done +
        // the ones performed here
        max_operations
            = max(
                max_operations,
                abs(arr[i] - arr[i - 1])
                    + abs(arr[i + 1] - arr[i])
                    - abs(arr[i + 1] - arr[i - 1]));
    }
 
    // Print the final result
    cout << result - max_operations << "\n";
}
 
// Driver Code
int main()
{
 
    int N = 3;
    vector arr = { -1, 0, 2 };
    findMinOperations(arr, N);
    return 0;
}

Java

// Java program for above approach
class GFG {
 
  static void findMinOperations(int[] ar, int n) {
 
    // Initializing vector to avoid overflows
    int[] arr = new int[n + 5];
 
    for (int i = 1; i <= n; i++) {
      arr[i] = ar[i - 1];
    }
    int result = 0;
 
    // Calculating minimum operations to be
    // performed on initial array
    for (int i = 2; i <= n; i++) {
      result += Math.abs(arr[i] - arr[i - 1]);
    }
 
    // Way to change a value to make
    // a1 equal to a2 or a(n)
    // equal to a(n-1)
    int max_operations = Math.max(Math.abs(arr[1] - arr[2]),
                                  Math.abs(arr[n] - arr[n - 1]));
    for (int i = 2; i < n; i++) {
 
      // For the rest of elements
      // taking the max of
      // operations already done +
      // the ones performed here
      max_operations = Math.max(
        max_operations,
        Math.abs(arr[i] - arr[i - 1])
        + Math.abs(arr[i + 1] - arr[i])
        - Math.abs(arr[i + 1] - arr[i - 1]));
    }
 
    // Print the final result
    System.out.println(result - max_operations);
  }
 
  // Driver Code
  public static void main(String args[]) {
 
    int N = 3;
    int[] arr = { -1, 0, 2 };
    findMinOperations(arr, N);
  }
}
 
// This code is contributed by saurabh_jaiswal.

Python3

# Python code for the above approach
def findMinOperations(ar, n):
 
    # Initializing vector to avoid overflows
    arr = [0] * (n + 5)
 
    for i in range(1, n + 1):
        arr[i] = ar[i - 1]
 
    result = 0
 
    # Calculating minimum operations to be
    # performed on initial array
    for i in range(2, n + 1):
        result += abs(arr[i] - arr[i - 1])
 
    # Way to change a value to make
    # a1 equal to a2 or a(n)
    # equal to a(n-1)
    max_operations = max(abs(arr[1] - arr[2]), abs(arr[n] - arr[n - 1]))
    for i in range(2, n):
 
        # For the rest of elements
        # taking the max of
        # operations already done +
        # the ones performed here
        max_operations = max(
            max_operations,
            abs(arr[i] - arr[i - 1])
            + abs(arr[i + 1] - arr[i])
            - abs(arr[i + 1] - arr[i - 1]))
 
    # Print the final result
    print((result - max_operations))
 
# Driver Code
N = 3
arr = [-1, 0, 2]
findMinOperations(arr, N)
 
# This code is contributed by gfgking

C#

// C# program to implement
// the above approach
using System;
 
class GFG
{
 
  static void findMinOperations(int[] ar, int n) {
 
    // Initializing vector to avoid overflows
    int[] arr = new int[n + 5];
 
    for (int i = 1; i <= n; i++) {
      arr[i] = ar[i - 1];
    }
    int result = 0;
 
    // Calculating minimum operations to be
    // performed on initial array
    for (int i = 2; i <= n; i++) {
      result += Math.Abs(arr[i] - arr[i - 1]);
    }
 
    // Way to change a value to make
    // a1 equal to a2 or a(n)
    // equal to a(n-1)
    int max_operations = Math.Max(Math.Abs(arr[1] - arr[2]),
                                  Math.Abs(arr[n] - arr[n - 1]));
    for (int i = 2; i < n; i++) {
 
      // For the rest of elements
      // taking the max of
      // operations already done +
      // the ones performed here
      max_operations = Math.Max(
        max_operations,
        Math.Abs(arr[i] - arr[i - 1])
        + Math.Abs(arr[i + 1] - arr[i])
        - Math.Abs(arr[i + 1] - arr[i - 1]));
    }
 
    // Print the final result
    Console.Write(result - max_operations);
  }
 
  // Driver Code
  public static void Main()
  {
    int N = 3;
    int[] arr = { -1, 0, 2 };
    findMinOperations(arr, N);
  }
}
 
// This code is contributed by sanjoy_62.

Javascript

       // JavaScript code for the above approach
       function findMinOperations(ar, n) {
 
           // Initializing vector to avoid overflows
           let arr = new Array(n + 5);
 
           for (let i = 1; i <= n; i++) {
               arr[i] = ar[i - 1];
           }
           let result = 0;
 
           // Calculating minimum operations to be
           // performed on initial array
           for (let i = 2; i <= n; i++) {
               result += Math.abs(arr[i] - arr[i - 1]);
           }
 
           // Way to change a value to make
           // a1 equal to a2 or a(n)
           // equal to a(n-1)
           let max_operations
               = Math.max(Math.abs(arr[1] - arr[2]),
                   Math.abs(arr[n] - arr[n - 1]));
           for (let i = 2; i < n; i++) {
 
               // For the rest of elements
               // taking the max of
               // operations already done +
               // the ones performed here
               max_operations
                   = Math.max(
                       max_operations,
                       Math.abs(arr[i] - arr[i - 1])
                       + Math.abs(arr[i + 1] - arr[i])
                       - Math.abs(arr[i + 1] - arr[i - 1]));
           }
 
           // Print the final result
           document.write((result - max_operations) + "</br>");
       }
 
       // Driver Code
       let N = 3;
       let arr = [-1, 0, 2];
       findMinOperations(arr, N);
 
      // This code is contributed by Potta Lokesh
   

输出

1

时间复杂度：O(N)
辅助空间：O(1)