最小化获得数组排序顺序所需的步骤

给定一个数组arr [] ，该数组由整数[1，N]的排列组成，通过重新排列排序的顺序[1，N]得出，任务是找到最小的步骤数，之后最小的步骤数将排序的顺序[1，N]通过重复在每个步骤中从排序后的序列中获得arr []的相同过程来重复该过程。
例子：

Input: arr[ ] = {3, 6, 5, 4, 1, 2} 
Output: 6 
Explanation: 
Increasing Permutation: {1, 2, 3, 4, 5, 6} 
Step 1 : arr[] = {3, 6, 5, 4, 1, 2} (Given array) 
Step 2 : arr[] = {5, 2, 1, 4, 3, 6} 
Step 3 : arr[] = {1, 6, 3, 4, 5, 2} 
Step 4 : arr[] = {3, 2, 5, 4, 1, 6} 
Step 5 : arr[] = {5, 6, 1, 4, 3, 2} 
Step 6 : arr[] = {1, 2, 3, 4, 5, 6} (Increasing Permutation) 
Therefore, the total number of steps required are 6.

Input: arr[ ] = [5, 1, 4, 3, 2] 
Output: 6 
 

方法：
通过使用直接寻址的概念可以简单地解决此问题。请按照以下步骤解决问题：

• 初始化数组dat []以直接寻址。
• 迭代[1，N]并按排序顺序从每个元素的索引计算当前索引的差。
• 计算数组dat []的LCM。
• 现在，将获得的LCM打印为获得排序顺序所需的最少步骤。

下面是上述方法的实现：

C++14

// C++ Program to implement 
// the above approach 
#include  
using namespace std; 
  
// Function to find 
// GCD of two numbers 
int gcd(int a, int b) 
{ 
    if (b == 0) 
        return a; 
  
    return gcd(b, a % b); 
} 
  
// Function to calculate the 
// LCM of array elements 
int findlcm(int arr[], int n) 
{ 
    // Initialize result 
    int ans = 1; 
  
    for (int i = 1; i <= n; i++) 
        ans = (((arr[i] * ans)) 
            / (gcd(arr[i], ans))); 
  
    return ans; 
} 
  
// Function to find minimum steps 
// required to obtain sorted sequence 
void minimumSteps(int arr[], int n) 
{ 
  
    // Inititalize dat[] array for 
    // Direct Address Table. 
    int i, dat[n + 1]; 
  
    for (i = 1; i <= n; i++) 
  
        dat[arr[i - 1]] = i; 
  
    int b[n + 1], j = 0, c; 
  
    // Calculating steps required 
    // for each element to reach 
    // its sorted position 
    for (i = 1; i <= n; i++) { 
        c = 1; 
        j = dat[i]; 
        while (j != i) { 
            c++; 
            j = dat[j]; 
        } 
        b[i] = c; 
    } 
  
    // Calculate LCM of the array 
    cout << findlcm(b, n); 
} 
  
// Driver Code 
int main() 
{ 
  
    int arr[] = { 5, 1, 4, 3, 2, 7, 6 }; 
  
    int N = sizeof(arr) / sizeof(arr[0]); 
  
    minimumSteps(arr, N); 
  
    return 0; 
}

Java

// Java program to implement
// the above approach
class GFG{
      
// Function to find
// GCD of two numbers
static int gcd(int a, int b)
{
    if (b == 0)
        return a;
  
    return gcd(b, a % b);
}
  
// Function to calculate the
// LCM of array elements
static int findlcm(int arr[], int n)
{
      
    // Initialize result
    int ans = 1;
  
    for(int i = 1; i <= n; i++)
        ans = (((arr[i] * ans)) / 
            (gcd(arr[i], ans)));
  
    return ans;
}
  
// Function to find minimum steps
// required to obtain sorted sequence
static void minimumSteps(int arr[], int n)
{
  
    // Inititalize dat[] array for
    // Direct Address Table.
    int i;
    int dat[] = new int[n + 1];
  
    for(i = 1; i <= n; i++)
        dat[arr[i - 1]] = i;
  
    int b[] = new int[n + 1];
    int j = 0, c;
  
    // Calculating steps required
    // for each element to reach
    // its sorted position
    for(i = 1; i <= n; i++)
    {
        c = 1;
        j = dat[i];
          
        while (j != i) 
        {
            c++;
            j = dat[j];
        }
        b[i] = c;
    }
  
    // Calculate LCM of the array
    System.out.println(findlcm(b, n));
}
  
// Driver code    
public static void main(String[] args)
{
    int arr[] = { 5, 1, 4, 3, 2, 7, 6 };
  
    int N = arr.length;
  
    minimumSteps(arr, N);
}
}
  
// This code is contributed by rutvik_56

Python3

# Python3 program to implement 
# the above approach 
  
# Function to find 
# GCD of two numbers 
def gcd(a, b): 
  
    if(b == 0): 
        return a 
  
    return gcd(b, a % b) 
  
# Function to calculate the 
# LCM of array elements 
def findlcm(arr, n): 
  
    # Initialize result 
    ans = 1
  
    for i in range(1, n + 1): 
        ans = ((arr[i] * ans) //
            (gcd(arr[i], ans))) 
  
    return ans 
  
# Function to find minimum steps 
# required to obtain sorted sequence 
def minimumSteps(arr, n): 
  
    # Inititalize dat[] array for 
    # Direct Address Table. 
    dat = [0] * (n + 1) 
  
    for i in range(1, n + 1): 
        dat[arr[i - 1]] = i 
  
    b = [0] * (n + 1) 
    j = 0
  
    # Calculating steps required 
    # for each element to reach 
    # its sorted position 
    for i in range(1, n + 1): 
        c = 1
        j = dat[i] 
        while(j != i): 
            c += 1
            j = dat[j] 
  
        b[i] = c 
  
    # Calculate LCM of the array 
    print(findlcm(b, n)) 
  
# Driver Code 
arr = [ 5, 1, 4, 3, 2, 7, 6 ] 
  
N = len(arr) 
  
minimumSteps(arr, N) 
  
# This code is contributed by Shivam Singh

C#

// C# program to implement
// the above approach
using System;
  
class GFG{
      
// Function to find
// GCD of two numbers
static int gcd(int a, int b)
{
    if (b == 0)
        return a;
  
    return gcd(b, a % b);
}
  
// Function to calculate the
// LCM of array elements
static int findlcm(int []arr, int n)
{
      
    // Initialize result
    int ans = 1;
  
    for(int i = 1; i <= n; i++)
        ans = (((arr[i] * ans)) / 
            (gcd(arr[i], ans)));
  
    return ans;
}
  
// Function to find minimum steps
// required to obtain sorted sequence
static void minimumSteps(int []arr, int n)
{
  
    // Inititalize dat[] array for
    // Direct Address Table.
    int i;
    int []dat = new int[n + 1];
  
    for(i = 1; i <= n; i++)
        dat[arr[i - 1]] = i;
  
    int []b = new int[n + 1];
    int j = 0, c;
  
    // Calculating steps required
    // for each element to reach
    // its sorted position
    for(i = 1; i <= n; i++)
    {
        c = 1;
        j = dat[i];
          
        while (j != i) 
        {
            c++;
            j = dat[j];
        }
        b[i] = c;
    }
  
    // Calculate LCM of the array
    Console.WriteLine(findlcm(b, n));
}
  
// Driver code 
public static void Main(String[] args)
{
    int []arr = { 5, 1, 4, 3, 2, 7, 6 };
  
    int N = arr.Length;
  
    minimumSteps(arr, N);
}
}
  
// This code is contributed by gauravrajput1

输出：

6

时间复杂度： O(NlogN)
辅助空间： O(N)