# Steps to return to {1, 2, ..n} with specified movements

Given an array moves[] that contains a permutation of first n natural numbers, each element of this array represents a movement that is each element shows an index where the element goes after each step. Now following these steps we need to tell after how many steps array [1..n] returns back to [1..N]. A step is defined as follows, after one step each element will be moved to position defined by moves array indices. **Examples :**

Input : moves[] = [4, 5, 1, 3, 2] Output : 6 Explanation: We need to consider an array of first 5 natural numbers, i.e., arr[] = {1, 2, 3, 4, 5} as size of moves[] is 5. Now we one by one move elements of arr[] using given moves. moves[] = [4, 5, 1, 3, 2] arr[] = [1, 2, 3, 4, 5] In step 1, we move 1 to position 4, 2 to position 5, 3 to position 1, 4 to position 3 and 5 to position 2. After step 1: arr[] = [3, 5, 4, 1, 2] In step 2, we move 3 to position 4, 5 to position 5, 4 to position 1, 1 to position 3 and 2 to position 2 After step 2: arr[] = [4, 2, 1, 3, 5] After step 3: arr[] = [1, 5, 3, 4, 2] After step 4: arr[] = [3, 2, 4, 1, 5] After step 5: arr[] = [4, 5, 1, 3, 2] After step 6: arr[] = [1, 2, 3, 4, 5] So we can reach to initial array in 6 steps, this is the minimum steps for reverting to the initial configuration of array. Input : moves[] = {3, 2, 1} Output : 2

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We can solve this problem by observing a pattern among the sequences which are formed. A particular set of element of moves array, forms a cycle. As in above moves array example [4, 1, 3] and [5, 2] are two such sets. These two cycles are independent.

[4, 1, 3] causes [1, 3, 4] -> [3, 4, 1] -> [4, 1, 3] -> [1, 3, 4] -> [3, 4, 1] -> [4, 1, 3] -> [1, 3, 4] [5, 2] causes [2, 5] -> [5, 2] -> [2, 5] -> [5, 2] -> [2, 5] -> [5, 2] -> [2, 5]

We can see from above changes that a cycle of length 3, takes 3 steps to reach to same state and cycle of length 2, takes 2 steps to reach to same state. In general if a cycle has length N then after N steps we can reach to same state.

Now if given moves array has just one cycle, then we can reach to starting state in number of moves equal to total elements in array but if it has more than 1 cycle then all of them move their elements independently and all of them will reach to starting state after x number of moves where x should be divisible by all cycle lengths, as smallest x which divides all cycle length is their LCM, we remain with finding LCM of all cycle lengths.

Cycle lengths can be found by visiting elements one by one, starting from any element we will move until we reach to starting element and we will count number of elements in this process which will be the corresponding cycle length.

Below is example of cycles found for some moves arrays, [2, 3, 1, 5, 4] -> [2, 3, 1] and [5, 4] [1, 2, 3, 4, 5] -> [1] [2] [3] [4] [5] [2, 3, 4, 1, 5] -> [2, 3, 4, 1] and [5]

Only thing remain is to calculate LCM of these lengths which can be calculated easily using GCD.

## C++

`// C++ program to get minimum steps to return to` `// initial array with specified movement` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Utility method to get lcm of a and b` `int` `lcm(` `int` `a, ` `int` `b)` `{` ` ` `return` `(a * b) / __gcd(a, b);` `}` `// Method returns minimum number of steps to` `// return to initial array` `int` `getMinStepsToSort(` `int` `moves[], ` `int` `N)` `{` ` ` `// initially all cells are unvisited` ` ` `bool` `visit[N];` ` ` `memset` `(visit, ` `false` `, ` `sizeof` `(visit));` ` ` `// looping over all elements to get` ` ` `// various cycle` ` ` `int` `steps = 1;` ` ` `for` `(` `int` `i = 0; i < N; i++) {` ` ` `// if already visited, that means it` ` ` `// was a part of some cycle` ` ` `if` `(visit[i])` ` ` `continue` `;` ` ` `int` `cycleLen = 0;` ` ` `// Looping among cycle elements, -1 is` ` ` `// for converting value to 0-index based` ` ` `for` `(` `int` `j = i; !visit[j]; j = moves[j] - 1) {` ` ` `cycleLen++;` ` ` `visit[j] = ` `true` `;` ` ` `}` ` ` `// Take the lcm of current result and` ` ` `// new cycle length` ` ` `steps = lcm(steps, cycleLen);` ` ` `}` ` ` `return` `steps;` `}` `// Driver code to test above methods` `int` `main()` `{` ` ` `int` `moves[] = { 4, 5, 1, 3, 2 };` ` ` `int` `N = ` `sizeof` `(moves) / ` `sizeof` `(` `int` `);` ` ` `cout << getMinStepsToSort(moves, N);` ` ` `return` `0;` `}` |

## Java

`// Java program to get minimum steps to` `// return to initial array with specified` `// movement` `import` `java.util.Arrays;` `class` `GFG {` ` ` `// Recursive function to return gcd` ` ` `// of a and b` ` ` `static` `int` `__gcd(` `int` `a, ` `int` `b)` ` ` `{` ` ` `// Everything divides 0` ` ` `if` `(a == ` `0` `|| b == ` `0` `)` ` ` `return` `0` `;` ` ` `// base case` ` ` `if` `(a == b)` ` ` `return` `a;` ` ` `// a is greater` ` ` `if` `(a > b)` ` ` `return` `__gcd(a - b, b);` ` ` `return` `__gcd(a, b - a);` ` ` `}` ` ` `// Utility method to get lcm of a and b` ` ` `static` `int` `lcm(` `int` `a, ` `int` `b)` ` ` `{` ` ` `return` `(a * b) / __gcd(a, b);` ` ` `}` ` ` `// Method returns minimum number of steps` ` ` `// to return to initial array` ` ` `static` `int` `getMinStepsToSort(` `int` `moves[],` ` ` `int` `N)` ` ` `{` ` ` `// initially all cells are unvisited` ` ` `boolean` `visit[] = ` `new` `boolean` `[N];` ` ` `Arrays.fill(visit, ` `false` `);` ` ` `// looping over all elements to get` ` ` `// various cycle` ` ` `int` `steps = ` `1` `;` ` ` `for` `(` `int` `i = ` `0` `; i < N; i++) {` ` ` `// if already visited, that` ` ` `// means it was a part of some` ` ` `// cycle` ` ` `if` `(visit[i])` ` ` `continue` `;` ` ` `int` `cycleLen = ` `0` `;` ` ` `// Looping among cycle elements,` ` ` `// -1 is for converting value to` ` ` `// 0-index based` ` ` `for` `(` `int` `j = i; !visit[j];` ` ` `j = moves[j] - ` `1` `) {` ` ` `cycleLen++;` ` ` `visit[j] = ` `true` `;` ` ` `}` ` ` `// Take the lcm of current result` ` ` `// and new cycle length` ` ` `steps = lcm(steps, cycleLen);` ` ` `}` ` ` `return` `steps;` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `main(String arg[])` ` ` `{` ` ` `int` `moves[] = { ` `4` `, ` `5` `, ` `1` `, ` `3` `, ` `2` `};` ` ` `int` `N = moves.length;` ` ` `System.out.print(getMinStepsToSort(` ` ` `moves, N));` ` ` `}` `}` `// This code is contributed by Anant Agarwal.` |

## Python3

`# Python program to get` `# minimum steps to return to` `# initial array with` `# specified movement` `# Recursive function to` `# return gcd of a and b` `def` `__gcd(a, b):` ` ` `# Everything divides 0` ` ` `if` `(a ` `=` `=` `0` `or` `b ` `=` `=` `0` `):` ` ` `return` `0` ` ` ` ` `# base case` ` ` `if` `(a ` `=` `=` `b):` ` ` `return` `a` ` ` ` ` `# a is greater` ` ` `if` `(a > b):` ` ` `return` `__gcd(a` `-` `b, b)` ` ` `return` `__gcd(a, b` `-` `a)` `# Utility method to` `# get lcm of a and b` `def` `lcm(a, b):` ` ` `return` `(a ` `*` `b) ` `/` `/` `__gcd(a, b)` ` ` `# Method returns minimum` `# number of steps to` `# return to initial array` `def` `getMinStepsToSort(moves, N):` ` ` `# initially all cells are unvisited` ` ` `visit ` `=` `[` `False` `for` `i ` `in` `range` `(N ` `+` `1` `)]` ` ` ` ` `# looping over all` ` ` `# elements to get` ` ` `# various cycle` ` ` `steps ` `=` `1` ` ` `for` `i ` `in` `range` `(N):` ` ` ` ` `# if already visited,` ` ` `# that means it` ` ` `# was a part of some cycle` ` ` `if` `(visit[i]):` ` ` `continue` ` ` ` ` `cycleLen ` `=` `0` ` ` ` ` `# Looping among cycle` ` ` `# elements, -1 is` ` ` `# for converting value` ` ` `# to 0-index based` ` ` `j ` `=` `i` ` ` `while` `(` `not` `visit[j]):` ` ` ` ` `cycleLen` `+` `=` `1` ` ` `visit[j] ` `=` `True` ` ` `j ` `=` `moves[j] ` `-` `1` ` ` ` ` `# Take the lcm of` ` ` `# current result and` ` ` `# new cycle length` ` ` `steps ` `=` `lcm(steps, cycleLen)` ` ` `return` `steps` `# Driver code` `moves ` `=` `[` `4` `, ` `5` `, ` `1` `, ` `3` `, ` `2` `]` `N ` `=` `len` `(moves)` ` ` `print` `(getMinStepsToSort(moves, N))` `# This code is contributed` `# by Anant Agarwal.` |

## C#

`// C# program to get minimum steps to return` `// to initial array with specified movement` `using` `System;` `class` `GFG {` ` ` `// Recursive function to return gcd` ` ` `// of a and b` ` ` `static` `int` `__gcd(` `int` `a, ` `int` `b)` ` ` `{` ` ` `// Everything divides 0` ` ` `if` `(a == 0 || b == 0)` ` ` `return` `0;` ` ` `// base case` ` ` `if` `(a == b)` ` ` `return` `a;` ` ` `// a is greater` ` ` `if` `(a > b)` ` ` `return` `__gcd(a - b, b);` ` ` `return` `__gcd(a, b - a);` ` ` `}` ` ` `// Utility method to get lcm of a and b` ` ` `static` `int` `lcm(` `int` `a, ` `int` `b)` ` ` `{` ` ` `return` `(a * b) / __gcd(a, b);` ` ` `}` ` ` `// Method returns minimum number of steps` ` ` `// to return to initial array` ` ` `static` `int` `getMinStepsToSort(` `int` `[] moves,` ` ` `int` `N)` ` ` `{` ` ` `// initially all cells are unvisited` ` ` `bool` `[] visit = ` `new` `bool` `[N];` ` ` `// looping over all elements to get` ` ` `// various cycle` ` ` `int` `steps = 1;` ` ` `for` `(` `int` `i = 0; i < N; i++) {` ` ` `// if already visited, that` ` ` `// means it was a part of some cycle` ` ` `if` `(visit[i])` ` ` `continue` `;` ` ` `int` `cycleLen = 0;` ` ` `// Looping among cycle elements,` ` ` `// -1 is for converting value to` ` ` `// 0-index based` ` ` `for` `(` `int` `j = i; !visit[j];` ` ` `j = moves[j] - 1) {` ` ` `cycleLen++;` ` ` `visit[j] = ` `true` `;` ` ` `}` ` ` `// Take the lcm of current result` ` ` `// and new cycle length` ` ` `steps = lcm(steps, cycleLen);` ` ` `}` ` ` `return` `steps;` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `int` `[] moves = { 4, 5, 1, 3, 2 };` ` ` `int` `N = moves.Length;` ` ` `Console.WriteLine(getMinStepsToSort(moves, N));` ` ` `}` `}` `// This code is contributed by vt_m.` |

## Javascript

`<script>` ` ` `// Javascript program to get minimum steps to return` ` ` `// to initial array with specified movement` ` ` ` ` `// Recursive function to return gcd` ` ` `// of a and b` ` ` `function` `__gcd(a, b)` ` ` `{` ` ` `// Everything divides 0` ` ` `if` `(a == 0 || b == 0)` ` ` `return` `0;` ` ` ` ` `// base case` ` ` `if` `(a == b)` ` ` `return` `a;` ` ` ` ` `// a is greater` ` ` `if` `(a > b)` ` ` `return` `__gcd(a - b, b);` ` ` ` ` `return` `__gcd(a, b - a);` ` ` `}` ` ` ` ` `// Utility method to get lcm of a and b` ` ` `function` `lcm(a, b)` ` ` `{` ` ` `return` `parseInt((a * b) / __gcd(a, b), 10);` ` ` `}` ` ` ` ` `// Method returns minimum number of steps` ` ` `// to return to initial array` ` ` `function` `getMinStepsToSort(moves, N)` ` ` `{` ` ` `// initially all cells are unvisited` ` ` `let visit = ` `new` `Array(N);` ` ` `visit.fill(` `false` `);` ` ` ` ` `// looping over all elements to get` ` ` `// various cycle` ` ` `let steps = 1;` ` ` `for` `(let i = 0; i < N; i++) {` ` ` ` ` `// if already visited, that` ` ` `// means it was a part of some cycle` ` ` `if` `(visit[i])` ` ` `continue` `;` ` ` ` ` `let cycleLen = 0;` ` ` ` ` `// Looping among cycle elements,` ` ` `// -1 is for converting value to` ` ` `// 0-index based` ` ` `for` `(let j = i; !visit[j]; j = moves[j] - 1) {` ` ` ` ` `cycleLen++;` ` ` `visit[j] = ` `true` `;` ` ` `}` ` ` ` ` `// Take the lcm of current result` ` ` `// and new cycle length` ` ` `steps = lcm(steps, cycleLen);` ` ` `}` ` ` ` ` `return` `steps;` ` ` `}` ` ` ` ` `let moves = [ 4, 5, 1, 3, 2 ];` ` ` `let N = moves.length;` ` ` `document.write(getMinStepsToSort(moves, N));` `</script>` |

Output:

6

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