Minimize Steps required to obtain Sorted Order of an Array

Given an array arr[] consisting of a permutation of integers [1, N], derived by rearranging the sorted order [1, N], the task is to find the minimum number of steps after which the sorted order [1, N] is repeated, by repeating the same process by which arr[] is obtained from the sorted sequence at each step.

Examples:

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 

Approach:
This problem can be solved simply by using the concept of Direct Addressing. Follow the steps given below to solve the problem:  

• Initialize an array dat[] for direct addressing.
• Iterate over [1, N] and calculate the difference of the current index of every element from its index in the sorted sequence.
• Calculate the LCM of the array dat[].
• Now, print the obtained LCM as the minimum steps required to obtain the sorted order.

Below is the implementation of the above approach: 

C++14

// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
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)
{

    // Initialize 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)
{

    // Initialize 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):

    # Initialize 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)
{

    // Initialize 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

Javascript

<script>
// JavaScript program for the above approach

// Function to find
// GCD of two numbers
function gcd(a, b)
{
    if (b == 0)
        return a;

    return gcd(b, a % b);
}

// Function to calculate the
// LCM of array elements
function findlcm(arr, n)
{

    // Initialize result
    let ans = 1;

    for(let 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
function minimumSteps(arr, n)
{

    // Initialize dat[] array for
    // Direct Address Table.
    let i;
    let dat = Array.from({length: n+1}, (_, i) => 0);

    for(i = 1; i <= n; i++)
        dat[arr[i - 1]] = i;

    let b = Array.from({length: n+1}, (_, i) => 0);
    let 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
    document.write(findlcm(b, n));
}

// Driver Code

    let arr = [ 5, 1, 4, 3, 2, 7, 6 ];

    let N = arr.length;

    minimumSteps(arr, N);

</script>

Output:

6

Time Complexity: O(NlogN) 
Auxiliary Space: O(N)