Minimum p[i] = p[arr[i]] operations to regain the given Array

Given an array, A[] (1 – indexed) of size ‘N’ which contains a permutation of [1, N], the task is to find the minimum number of operations to be applied on any array P[] to get back the original array P[]. The operation must be applied at least once. In each operation, for every index of P[] we set P[i] = P[A[i]].

Examples:

Input: A[] = {1, 3, 2}
Output: 2
Explanation: Let P[] = {7, 4, 2}. 
After 1 operation, {P[1], P[3], P[2]} = {1, 2, 4}.
After 2 operations, {P[1], P[3], P[2]} = {7, 4, 2}
After 2 operation original array is reached.

Input: A[] = {5, 4, 2, 3, 1}
Output: 6
Explanation: Let P = {1, 2, 3, 4, 5},  
After 1 operation {P[5], P[4], P[2], P[3], P[1]} = {5, 4, 2, 3, 1}
After 2 operation {P[5], P[4], P[2], P[3], P[1]} = {1, 3, 4, 2, 5}
After 3 operation {P[5], P[4], P[2], P[3], P[1]} = {5, 2, 3, 4, 1}
After 4 operation {P[5], P[4], P[2], P[3], P[1]} = {1, 4, 2, 3, 5}
After 5 operation {P[5], P[4], P[2], P[3], P[1]} = {5, 3, 4, 2, 1}
After 6 operation {P[5], P[4], P[2], P[3], P[1]} = {1, 2, 3, 4, 5}
After 6 operation original array is reached.

Naive Approach:

A naive approach is to make any array and apply the given operation until the original array is reached again.

Below is the implementation of the approach.

C++

// C++ code for the naive approach
#include <bits/stdc++.h>
using namespace std;

// comparing 2 arrays
bool checkEqual(vector<int>& A, vector<int>& B)
{
  for (int i = 0; i < A.size(); i++) {
    if (A[i] != B[i])
      return false;
  }
  return true;
}

// function to find minm operations
int minOperations(vector<int>& A)
{
  int N = A.size();
  vector<int> P(N, 0);
  vector<int> originalArray(N, 0);

  // Let P be same as A
  for (int i = 0; i < N; i++) {
    P[i] = A[i];
    originalArray[i] = P[i];
  }

  // after applying operation 1 time
  for (int i = 0; i < N; i++) {
    P[i] = A[A[i] - 1];
  }

  int operations = 1;
  while (!checkEqual(originalArray, P)) {
    vector<int> temp(N);
    for (int i = 0; i < N; i++) {
      temp[i] = P[A[i] - 1];
    }
    P = temp;
    operations++;
  }
  return operations;
}

int main()
{

  // Given input
  vector<int> A = { 5, 4, 2, 3, 1 };

  // Function call
  cout << minOperations(A);

  return 0;
}

// This code is contributed by rakeshsahni

Java

// JAVA code for the naive approach

import java.util.*;

public class GFG {
    // comparing 2 arrays
    static boolean checkEqual(int[] A, int[] B)
    {
        for (int i = 0; i < A.length; i++) {
            if (A[i] != B[i])
                return false;
        }
        return true;
    }

    // function to find minm operations
    static int minOperations(int[] A)
    {
        int N = A.length;
        int[] P = new int[N];
        int[] originalArray = new int[N];

        // Let P be same as A
        for (int i = 0; i < N; i++) {
            P[i] = A[i];
            originalArray[i] = P[i];
        }

        // after applying operation 1 time
        for (int i = 0; i < N; i++) {
            P[i] = A[A[i] - 1];
        }

        int operations = 1;
        while (!checkEqual(originalArray, P)) {
            int[] temp = new int[N];
            for (int i = 0; i < N; i++) {
                temp[i] = P[A[i] - 1];
            }
            P = temp;
            operations++;
        }
        return operations;
    }
    public static void main(String[] args)
    {

        // Given input
        int[] A = { 5, 4, 2, 3, 1 };

        // Function call
        System.out.println(minOperations(A));
    }
}

Python3

# Python code for the above approach

# comparing 2 arrays
def checkEqual(A, B):
    for i in range(len(A)):
        if(A[i] is not B[i]):
            return False

    return True

# Function to find minimum operations
def minOperations(A):
    N = len(A)
    P = [0] * N
    originalArray = [0] * N

    # let P be same as A
    for i in range(N):
        P[i] = A[i]
        originalArray[i] = P[i]

    # after applying operation 1 time
    for i in range(N):
        P[i] = A[A[i]-1]

    operations = 1
    while(checkEqual(originalArray, P) is not True):
        temp = [0] * N
        for i in range(N):
            temp[i] = P[A[i]-1]
        P = temp
        operations += 1

    return operations


A = [5, 4, 2, 3, 1]

# Function call
print(minOperations(A))

# This code is contributed by lokesh.

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;

class GFG
{
  // comparing 2 arrays
  static bool checkEqual(int[] A, int[] B)
  {
    for (int i = 0; i < A.Length; i++) {
      if (A[i] != B[i])
        return false;
    }
    return true;
  }

  // function to find minm operations
  static int minOperations(int[] A)
  {
    int N = A.Length;
    int[] P = new int[N];
    int[] originalArray = new int[N];

    // Let P be same as A
    for (int i = 0; i < N; i++) {
      P[i] = A[i];
      originalArray[i] = P[i];
    }

    // after applying operation 1 time
    for (int i = 0; i < N; i++) {
      P[i] = A[A[i] - 1];
    }

    int operations = 1;
    while (!checkEqual(originalArray, P)) {
      int[] temp = new int[N];
      for (int i = 0; i < N; i++) {
        temp[i] = P[A[i] - 1];
      }
      P = temp;
      operations++;
    }
    return operations;
  }

  // Driver code
  public static void Main(String[] args)
  {

    // Given input
    int[] A = { 5, 4, 2, 3, 1 };

    // Function call
    Console.WriteLine(minOperations(A));
  }
}

// This code is contributed by code_hunt.

Javascript

<script>
// JS code to implement the approach

    // comparing 2 arrays
    function checkEqual(A, B)
    {
        for (let i = 0; i < A.length; i++) {
            if (A[i] != B[i])
                return false;
        }
        return true;
    }

    // function to find minm operations
    function minOperations(A)
    {
        let N = A.length;
        let P = new Array(N);
        let originalArray = new Array(N);

        // Let P be same as A
        for (let i = 0; i < N; i++) {
            P[i] = A[i];
            originalArray[i] = P[i];
        }

        // after applying operation 1 time
        for (let i = 0; i < N; i++) {
            P[i] = A[A[i] - 1];
        }

        let operations = 1;
        while (!checkEqual(originalArray, P)) {
            let temp = new Array(N);
            for (let i = 0; i < N; i++) {
                temp[i] = P[A[i] - 1];
            }
            P = temp;
            operations++;
        }
        return operations;
    }

// Driver code

        // Given input
        let A = [ 5, 4, 2, 3, 1 ];

        // Function call
        document.write(minOperations(A));

// This code is contributed by sanjoy_62.
</script>

Output

6

Time Complexity: O(N * minOperations), for executing the operations until the original array is retrieved.
Auxiliary Space: O(N), for creating an additional array of size P.

Efficient Approach:

Use the following idea to solve the problem:

It can be observed that the elements form a cycle. When all the cycles are completed at the same operation for the first time that many moves are required.

Each cycle is completed after making moves same as their length. So all the cycles are completed at the same operation for the first time when LCM(all cycle lengths) number of moves are made.

Follow the below steps to solve the problem:

• Declare an array ‘cycleLengths[]’ to store the length of all cycles present.
• Declare a Boolean array ‘visited[]’ to check if the cycle length of corresponding element has already been calculated or not.
• For every unvisited index
  • traverse all the elements of corresponding cycle while updating the ‘visited[]’ and store its length in ‘cycleLength[]’.
• Return the LCM of all numbers present in ‘cycleLength[]’.

Code implementation of above approach:

C++

// C++ code to implement the approach

#include <iostream>
#include <vector>
using namespace std;

// Function to calculate GCD of two numbers
int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}

// Function to calculate LCM of two numbers
int lcm(int a, int b) { return (a * b) / gcd(a, b); }

// Traversing the cycle and returning
// the length of the cycle
int traverseCycle(int root, int A[], vector<int>& visited)
{
    if (visited[root])
        return 0;
    visited[root] = true;
    return 1 + traverseCycle(A[root - 1], A, visited);
}

// Function to find minm operations
int minOperations(int A[], int N)
{

    vector<int> cycleLength;
    vector<int> visited(N + 1, 0);

    // Detecting all cycles and storing
    // their length in cycleLength List
    for (int i = 1; i <= N; i++) {
        if (!visited[i]) {
            int len = traverseCycle(i, A, visited);
            cycleLength.push_back(len);
        }
    }

    // Finding lcm of all cycle lengths
    int res = 1;
    for (auto cycleLen : cycleLength) {
        res = lcm(res, cycleLen);
    }
    return res;
}

// Driver code
int main()
{
    int A[] = { 5, 4, 2, 3, 1 };

    // Function call
    cout << (minOperations(A, 5)) << endl;
    ;
}

// This code is contributed by garg28harsh

Java

// Java code to implement the approach

import java.util.*;

public class GFG {

    // Function to calculate GCD of two numbers
    static int gcd(int a, int b)
    {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }

    // Function to calculate LCM of two numbers
    static int lcm(int a, int b)
    {
        return (a * b) / gcd(a, b);
    }

    // Traversing the cycle and returning
    // the length of the cycle
    static int traverseCycle(int root, int[] A,
                             boolean[] visited)
    {
        if (visited[root])
            return 0;
        visited[root] = true;
        return 1 + traverseCycle(A[root - 1], A, visited);
    }

    // Function to find minm operations
    static int minOperations(int[] A)
    {
        int N = A.length;
        ArrayList<Integer> cycleLength = new ArrayList<>();
        boolean[] visited = new boolean[N + 1];

        // Detecting all cycles and storing
        // their length in cycleLength List
        for (int i = 1; i <= N; i++) {
            if (!visited[i]) {
                int len = traverseCycle(i, A, visited);
                cycleLength.add(len);
            }
        }

        // Finding lcm of all cycle lengths
        int res = 1;
        for (Integer cycleLen : cycleLength) {
            res = lcm(res, cycleLen);
        }
        return res;
    }

    // Driver code
    public static void main(String[] args)
    {
        int[] A = { 5, 4, 2, 3, 1 };

        // Function call
        System.out.println(minOperations(A));
    }
}

Python3

class GFG:
    # Function to calculate GCD of two numbers
    @staticmethod
    def gcd(a,  b):
        if (a == 0):
            return b
        return GFG.gcd(b % a, a)
    # Function to calculate LCM of two numbers

    @staticmethod
    def lcm(a,  b):
        return int((a * b) / GFG.gcd(a, b))
    # Traversing the cycle and returning
    # the length of the cycle

    @staticmethod
    def traverseCycle(root,  A,  visited):
        if (visited[root]):
            return 0
        visited[root] = True
        return 1 + GFG.traverseCycle(A[root - 1], A, visited)
    # Function to find minm operations

    @staticmethod
    def minOperations(A):
        N = 5
        cycleLength = []
        visited = [False] * (N + 1)
        # Detecting all cycles and storing
        # their length in cycleLength List
        i = 1
        while (i <= N):
            if (not visited[i]):
                len = GFG.traverseCycle(i, A, visited)
                cycleLength.append(len)
            i += 1
        # Finding lcm of all cycle lengths
        res = 1
        for cycleLen in cycleLength:
            res = GFG.lcm(res, cycleLen)
        return res
    # Driver code

    @staticmethod
    def main(args):
        A = [5, 4, 2, 3, 1]
        # Function call
        print(GFG.minOperations(A))


if __name__ == "__main__":
    GFG.main([])

C#

// C# code to implement the approach
using System;
using System.Collections.Generic;

public class GFG
{

  // Function to calculate GCD of two numbers
  static int gcd(int a, int b)
  {
    if (a == 0)
      return b;
    return gcd(b % a, a);
  }

  // Function to calculate LCM of two numbers
  static int lcm(int a, int b)
  {
    return (a * b) / gcd(a, b);
  }

  // Traversing the cycle and returning
  // the length of the cycle
  static int traverseCycle(int root, int[] A,
                           Boolean[] visited)
  {
    if (visited[root])
      return 0;
    visited[root] = true;
    return 1 + traverseCycle(A[root - 1], A, visited);
  }

  // Function to find minm operations
  static int minOperations(int[] A)
  {
    int N = A.Length;
    List<int> cycleLength = new List<int>();
    Boolean[] visited = new Boolean[N + 1];

    // Detecting all cycles and storing
    // their length in cycleLength List
    for (int i = 1; i <= N; i++)
    {
      if (!visited[i])
      {
        int len = traverseCycle(i, A, visited);
        cycleLength.Add(len);
      }
    }

    // Finding lcm of all cycle lengths
    int res = 1;
    foreach (int cycleLen in cycleLength)
    {
      res = lcm(res, cycleLen);
    }
    return res;
  }

  // Driver code
  public static void Main()
  {
    int[] A = { 5, 4, 2, 3, 1 };

    // Function call
    Console.Write(minOperations(A));
  }
}

// This code is contributed by saurabh_jaiswal.

Javascript

// Javascript code to implement the approach

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

// Function to calculate LCM of two numbers
function lcm(a, b)
{
    let c = a * b;
    return (c / gcd(a, b));
}

// Traversing the cycle and returning
// the length of the cycle
function traverseCycle(root, A, visited)
{
    if (visited[root])
        return 0;
    visited[root] = true;
    return 1 + traverseCycle(A[root - 1], A, visited);
}

// Function to find minm operations
function minOperations(A, N)
{

    let cycleLength = [];

    let visited = [];
    for (let i = 0; i <= N; i++) {
        visited.push(0);
    }

    // Detecting all cycles and storing
    // their length in cycleLength List
    for (let i = 1; i <= N; i++) {
        if (visited[i] == false) {
            let len = traverseCycle(i, A, visited);
            cycleLength.push(len);
        }
    }

    // Finding lcm of all cycle lengths
    let res = 1;
    for (let i = 0; i < cycleLength.length; i++) {
        res = lcm(res, cycleLength[i]);
    }
    return res;
}

// Driver code
let A = [ 5, 4, 2, 3, 1 ];

// Function call
console.log(minOperations(A, 5));

// This code is contributed by garg28harsh.

Output

6

Time Complexity: O(N*log(Arr[i])), where N is the size of the given array.
Auxiliary Space: O(N), for creating an additional array of size N + 1.