Minimum number of given operations required to convert a permutation into an identity permutation

Given a permutation P (P1, P2, P3, … Pn) of first n natural numbers. Find the minimum number of operations to convert it into an identity permutation i.e. 1, 2, 3, …, n where each operation is defined as:
P[i] = P[P[P[i]]] $\forall$ i from 1 to n (1 based indexing). If there is no way to convert then print -1.
Examples:

Input: arr[] = {2, 3, 1}
Output: 1
After 1 operation:
P[1] = P[P[P[1]]] = P[P[2]] = P[3] = 1
P[2] = P[P[P[2]]] = P[P[3]] = P[1] = 2
P[3] = P[P[P[3]]] = P[P[1]] = P[2] = 3
Thus after 1 operation we obtain an identity permutation.
Input: arr[] = {2, 1, 3}
Output: -1
There is no way to obtain identity permutation
no matter how many operations we apply.

Approach: First, find all the cycles in the given permutation. Here, a cycle is a directed graph in which there is an edge from an element e to the element on position e.
For example, Here’s the graph for permutation {4, 6, 5, 3, 2, 1, 8, 7}

Now in one operation, each cycle of length 3k breaks into 3 cycles each of length k while cycles of length 3k+1 or 3k+2 do not break. Since in the end, we need all cycles of length 1, therefore, all cycles must be a power of 3 otherwise answer doesn’t exist. The answer would then be the maximum of log(base 3) of all cycle lengths.

Algorithm:

Step 1: Create the calculateCycleOperations function, which accepts an integer as an input named “len”.
Step 2: Initialize the cycle operations variable with a value of 0 within the function. Step 3: Start a while loop with the condition “len” is positive.
  a. divided “len” by three and update it every time
  b. divided “len” by three and update it every time    Step 4: Return the cycle operations value that has been reduced by 1.
Step 5: Create the function minimumOperations, which has two inputs: an integer array p[] and an integer n.
Step 6: Set all values in a boolean array with the name visited and a size of n+1 to 0.
Step 7: create the variable “ans” with the value 0.
Step 8: start a for loop and traverse through each element of the array.
   a. If the current element is not present in previous cycles, then only consider it.
   b. Mark the current element as visited so that it will not be considered in other cycles.
c. Initialize a variable named ele as the current element, and a variable named len as 1.
   d. Start a while loop with the condition element is not visited
   1. Update the value of ele with p[ele], and increment len by 1
     e. Calculate the number of operations needed for this cycle to reduce to length 1 (if possible) by calling the function calculateCycleOperations.
   f. Check if the length of the cycle is a power of 3, if not, then return -1.
   g. Take the maximum of the operations and ans.
Step 8: Return ans

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
int calculateCycleOperations(int len)
{
    int cycle_operations = 0;
    while (len) {
        len /= 3;
        ++cycle_operations;
    }
    return --cycle_operations;
}
 
// Function to return the minimum operations required
int minimumOperations(int p[], int n)
{
 
    // Array to keep track of visited elements
    bool visited[n + 1] = { 0 };
 
    // To store the final answer
    int ans = 0;
 
    // Looping through all the elements
    for (int i = 1; i <= n; i++) {
 
        // Current element
        int ele = p[i];
 
        // If current element is not present in the
        // previous cycles then only consider this
        if (!visited[ele]) {
 
            // Mark current element visited so that it
            // will not be considered in other cycles
            visited[ele] = 1;
 
            // To store the length of each cycle
            int len = 1;
            ele = p[ele];
 
            // Calculating cycle length
            while (!visited[ele]) {
                visited[ele] = 1;
                ++len;
                ele = p[ele];
            }
 
            // Operations needed for this cycle to reduce
            // to length 1 (if possible)
            int operations = calculateCycleOperations(len);
 
            // Checking cycle length to be power of 3
            // if not, then return -1
            int num = pow(3, operations);
            if (num != len) {
                return -1;
            }
 
            // Taking maximum of the operations
            ans = max(ans, operations);
        }
    }
    return ans;
}
 
// Driver code
int main()
{
    // 1-based indexing
    int P[] = { -1, 4, 6, 5, 3, 2, 7, 8, 9, 1 };
    int n = (sizeof(P) / sizeof(P[0])) - 1;
 
    // Calling function
    cout << minimumOperations(P, n);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG {
 
    static int calculateCycleOperations(int len)
    {
        int cycle_operations = 0;
        while (len > 0) {
            len /= 3;
            ++cycle_operations;
        }
        return --cycle_operations;
    }
 
    // Function to return the minimum operations required
    static int minimumOperations(int p[], int n)
    {
 
        // Array to keep track of visited elements
        int[] visited = new int[n + 1];
        Arrays.fill(visited, 0);
 
        // To store the final answer
        int ans = 0;
 
        // Looping through all the elements
        for (int i = 1; i <= n; i++) {
 
            // Current element
            int ele = p[i];
 
            // If current element is not present in the
            // previous cycles then only consider this
            if (visited[ele] == 0) {
 
                // Mark current element visited so that it
                // will not be considered in other cycles
                visited[ele] = 1;
 
                // To store the length of each cycle
                int len = 1;
                ele = p[ele];
 
                // Calculating cycle length
                while (visited[ele] == 0) {
                    visited[ele] = 1;
                    ++len;
                    ele = p[ele];
                }
 
                // Operations needed for this cycle to
                // reduce to length 1 (if possible)
                int operations
                    = calculateCycleOperations(len);
 
                // Checking cycle length to be power of 3
                // if not, then return -1
                int num = (int)Math.pow(3, operations);
                if (num != len) {
                    return -1;
                }
 
                // Taking maximum of the operations
                ans = Math.max(ans, operations);
            }
        }
        return ans;
    }
 
    // Driver code
    public static void main(String args[])
    {
        // 1-based indexing
        int P[] = { -1, 4, 6, 5, 3, 2, 7, 8, 9, 1 };
        int n = P.length - 1;
 
        // Calling function
        System.out.println(minimumOperations(P, n));
    }
}
 
// This code is contributed by
// Surendra_Gangwar

Python3

# Python3 implementation of the approach
def calculateCycleOperations(length):
 
    cycle_operations = 0
    while length > 0:
        length //= 3
        cycle_operations += 1
 
    return cycle_operations - 1
 
# Function to return the minimum
# operations required
 
 
def minimumOperations(p, n):
 
    # Array to keep track of visited elements
    visited = [0] * (n + 1)
 
    # To store the final answer
    ans = 0
 
    # Looping through all the elements
    for i in range(1, n + 1):
 
        # Current element
        ele = p[i]
 
        # If current element is not present in the
        # previous cycles then only consider this
        if not visited[ele]:
 
            # Mark current element visited so that it
            # will not be considered in other cycles
            visited[ele] = 1
 
            # To store the length of each cycle
            length = 1
            ele = p[ele]
 
            # Calculating cycle length
            while not visited[ele]:
                visited[ele] = 1
                length += 1
                ele = p[ele]
 
            # Operations needed for this cycle to
            # reduce to length 1 (if possible)
            operations = calculateCycleOperations(length)
 
            # Checking cycle length to be power
            # of 3 if not, then return -1
            num = pow(3, operations)
            if num != length:
                return -1
 
            # Taking maximum of the operations
            ans = max(ans, operations)
 
    return ans
 
 
# Driver code
if __name__ == "__main__":
 
    # 1-based indexing
    P = [-1, 4, 6, 5, 3, 2, 7, 8, 9, 1]
    n = len(P) - 1
 
    # Calling function
    print(minimumOperations(P, n))
 
# This code is contributed by Rituraj Jain

C#

// C# implementation of the above approach
using System;
 
class GFG {
 
    static int calculateCycleOperations(int len)
    {
        int cycle_operations = 0;
        while (len > 0) {
            len /= 3;
            ++cycle_operations;
        }
        return --cycle_operations;
    }
 
    // Function to return the minimum operations required
    static int minimumOperations(int[] p, int n)
    {
 
        // Array to keep track of visited elements
        int[] visited = new int[n + 1];
 
        // To store the final answer
        int ans = 0;
 
        // Looping through all the elements
        for (int i = 1; i <= n; i++) {
 
            // Current element
            int ele = p[i];
 
            // If current element is not present in the
            // previous cycles then only consider this
            if (visited[ele] == 0) {
 
                // Mark current element visited so that it
                // will not be considered in other cycles
                visited[ele] = 1;
 
                // To store the length of each cycle
                int len = 1;
                ele = p[ele];
 
                // Calculating cycle length
                while (visited[ele] == 0) {
                    visited[ele] = 1;
                    ++len;
                    ele = p[ele];
                }
 
                // Operations needed for this cycle to
                // reduce to length 1 (if possible)
                int operations
                    = calculateCycleOperations(len);
 
                // Checking cycle length to be power of 3
                // if not, then return -1
                int num = (int)Math.Pow(3, operations);
                if (num != len) {
                    return -1;
                }
 
                // Taking maximum of the operations
                ans = Math.Max(ans, operations);
            }
        }
        return ans;
    }
 
    // Driver code
    public static void Main()
    {
        // 1-based indexing
        int[] P = { -1, 4, 6, 5, 3, 2, 7, 8, 9, 1 };
        int n = P.Length - 1;
 
        // Calling function
        Console.WriteLine(minimumOperations(P, n));
    }
}
 
// This code is contributed by Ryuga

PHP

<?php
// PHP implementation of the approach
function calculateCycleOperations($len)
{
    $cycle_operations = 0;
    while ($len)
    {
        $len = (int)($len / 3);
        ++$cycle_operations;
    }
    return --$cycle_operations;
}
 
// Function to return the minimum
// operations required
function minimumOperations($p, $n)
{
 
    // Array to keep track of visited elements
    $visited[$n + 1] = array(0);
 
    // To store the final answer
    $ans = 0;
 
    // Looping through all the elements
    for ($i = 1; $i <= $n; $i++)
    {
 
        // Current element
        $ele = $p[$i];
 
        // If current element is not present in the
        // previous cycles then only consider this
        if (!$visited[$ele])
        {
 
            // Mark current element visited so that it
            // will not be considered in other cycles
            $visited[$ele] = 1;
 
            // To store the length of each cycle
            $len = 1;
            $ele = $p[$ele];
 
            // Calculating cycle length
            while (!$visited[$ele])
            {
                $visited[$ele] = 1;
                ++$len;
                $ele = $p[$ele];
            }
 
            // Operations needed for this cycle to reduce
            // to length 1 (if possible)
            $operations = calculateCycleOperations($len);
 
            // Checking cycle length to be power of 3
            // if not, then return -1
            $num = pow(3, $operations);
            if ($num != $len)
            {
                return -1;
            }
 
            // Taking maximum of the operations
            $ans = max($ans, $operations);
        }
    }
    return $ans;
}
 
// Driver code
 
// 1-based indexing
$P = array(-1, 4, 6, 5, 3, 2, 7, 8, 9, 1);
$n = sizeof($P) - 1;
 
// Calling function
echo minimumOperations($P, $n);
 
// This code is contributed by Akanksha Rai
?>

Javascript

<script>
 
// JavaScript implementation of the approach
 
function calculateCycleOperations(len)
{
    let cycle_operations = 0;
    while (len > 0)
    {
        len = Math.floor(len / 3);
        ++cycle_operations;
    }
    return --cycle_operations;
}
   
// Function to return the minimum operations required
function minimumOperations(p, n)
{
   
    // Array to keep track of visited elements
    let visited = Array.from({length: n + 1}, (_, i) => 0);
   
    // To store the final answer
    let ans = 0;
   
    // Looping through all the elements
    for (let i = 1; i <= n; i++)
    {
   
        // Current element
        let ele = p[i];
   
        // If current element is not present in the
        // previous cycles then only consider this
        if (visited[ele] == 0)
        {
   
            // Mark current element visited so that it
            // will not be considered in other cycles
            visited[ele] = 1;
   
            // To store the length of each cycle
            let len = 1;
            ele = p[ele];
   
            // Calculating cycle length
            while (visited[ele] == 0)
            {
                visited[ele] = 1;
                ++len;
                ele = p[ele];
            }
   
            // Operations needed for this cycle to reduce
            // to length 1 (if possible)
            let operations = calculateCycleOperations(len);
   
            // Checking cycle length to be power of 3
            // if not, then return -1
            let num = Math.floor(Math.pow(3, operations));
            if (num != len) {
                return -1;
            }
   
            // Taking maximum of the operations
            ans = Math.max(ans, operations);
        }
    }
    return ans;
}
     
    // Driver code
     
    // 1-based indexing
    let P = [ -1, 4, 6, 5, 3, 2, 7, 8, 9, 1 ];
    let n = P.length - 1;
   
    // Calling function
    document.write(minimumOperations(P, n));
                 
</script>