Skip to content
Related Articles

Related Articles

Minimize Steps required to obtain Sorted Order of an Array

Improve Article
Save Article
  • Difficulty Level : Expert
  • Last Updated : 13 Dec, 2021
Improve Article
Save Article

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:
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:

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)
 


My Personal Notes arrow_drop_up
Related Articles

Start Your Coding Journey Now!