将给定数组重新排列为 2 的幂序列所需的最小步骤

给定一个由N 个正整数组成的数组arr[] ，任务是找到将给定的整数数组变成2的幂序列所需的最小步骤 通过以下操作：

• 重新排序给定的数组。这不算是一步。
• 对于每一步，从数组中选择任何索引i并将arr[i]更改为arr[i] − 1或arr[i] + 1 。

 A sequence is called power sequence of 2, if for every i^th index (0 ≤i ≤ N − 1), 
arr[i] = 2ⁱ , where N is length of the given array.

例子：

Input: arr[] = { 1, 8, 2, 10, 6 }
Output: 8
Explanation: 
Reorder the array arr[] to { 1, 2, 6, 8, 10 }
Step 1: Decrement arr[2] to 5
Step 2: Decrement arr[2] to 4
Step 3 – 8: Increment arr[4] by 1. Final value of arr[4] becomes 16.
Therefore, arr[] = {1, 2, 4, 8, 16}
Hence, the minimum number of steps required to obtain the power sequence of 2 is 8.

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

方法：解决给定的问题，思路是对数组进行升序排序，对于排序后的数组的每个^第i^个索引，计算arr[i]之间的绝对差  和2 ⁱ 。绝对差异的总和为我们提供了所需的答案。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to calculate the minimum
// steps required to convert given
// array into a power sequence of 2
int minsteps(int arr[], int n)
{
 
    // Sort the array in
    // ascending order
    sort(arr, arr + n);
 
    int ans = 0;
 
    // Calculate the absolute difference
    // between arr[i] and 2^i for each index
    for (int i = 0; i < n; i++) {
        ans += abs(arr[i] - pow(2, i));
    }
 
    // Return the answer
    return ans;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 8, 2, 10, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << minsteps(arr, n) << endl;
    return 0;
}

Java

// Java Program to implement
// the above approach
 
import java.util.*;
import java.lang.Math;
 
class GFG {
 
    // Function to calculate the minimum
    // steps required to convert given
    // array into a power sequence of 2
    static int minsteps(int arr[], int n)
    {
        // Sort the array in ascending order
        Arrays.sort(arr);
        int ans = 0;
 
        // Calculate the absolute difference
        // between arr[i] and 2^i for each index
        for (int i = 0; i < n; i++) {
            ans += Math.abs(arr[i] - Math.pow(2, i));
        }
        return ans;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = { 1, 8, 2, 10, 6 };
        int n = arr.length;
        System.out.println(minsteps(arr, n));
    }
}

Python3

# Python 3 program for the above approach
 
# Function to calculate the minimum
# steps required to convert given
# array into a power sequence of 2
def minsteps(arr, n):
 
    # Sort the array in ascending order
    arr.sort()
    ans = 0
    for i in range(n):
        ans += abs(arr[i]-pow(2, i))
    return ans
 
 
# Driver Code
arr = [1, 8, 2, 10, 6]
n = len(arr)
print(minsteps(arr, n))

C#

// C# Program to the above approach
 
using System;
 
class GFG {
 
    // Function to calculate the minimum
    // steps required to convert given
    // array into a power sequence of 2
    static int minsteps(int[] arr, int n)
    {
 
        // Sort the array in ascending order
        Array.Sort(arr);
        int ans = 0;
 
        // Calculate the absolute difference
        // between arr[i] and 2^i for each index
        for (int i = 0; i < n; i++) {
            ans += Math.Abs(arr[i]
                            - (int)(Math.Pow(2, i)));
        }
        return ans;
    }
 
    // Driver Code
    public static void Main()
    {
 
        int[] arr = { 1, 8, 2, 10, 6 };
        int n = arr.Length;
        Console.WriteLine(minsteps(arr, n));
    }
}

Javascript

// Javascript program to implement
// the above approach
 
// Function to calculate the minimum
// steps required to convert given
// array into a power sequence of 2
function minsteps(arr, n)
{
 
    // Sort the array in
    // ascending order
    arr.sort((a,b)=>a-b)
 
    var ans = 0;
 
    // Calculate the absolute difference
    // between arr[i] and 2^i for each index
    for (var i = 0; i < n; i++) {
        ans += Math.abs(arr[i] - Math.pow(2, i));
    }
 
    // Return the answer
    return ans;
}
 
// Driver Code
var arr = [ 1, 8, 2, 10, 6 ];
var n = arr.length;
document.write( minsteps(arr, n));
 
// This code is contributed by noob2000.

输出：

8

时间复杂度： O(NlogN)
辅助空间： O(1)

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