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

• Difficulty Level : Hard
• Last Updated : 26 May, 2021

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[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.
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[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

 ``

## Javascript

 ``

Output:

`2`

Time Complexity : O(N*LogN)

My Personal Notes arrow_drop_up
Recommended Articles
Page :