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

Difficulty Level : Hard
Last Updated : 12 Apr, 2019

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.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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.

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 
?> 

Output:

Time Complexity : O(N*LogN)

My Personal Notes

Related Articles

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP