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# 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]]] 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:
After 1 operation:
P = P[P[P]] = P[P] = P = 1
P = P[P[P]] = P[P] = P = 2
P = P[P[P]] = P[P] = P = 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 ``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)) - 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

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## Javascript

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Output:

`2`

Time Complexity: O(N*LogN)
Auxiliary Space: O(N)