Minimum moves required to change position with the given operation

Given two integers S and T and an array arr that contains elements from 1 to N in unsorted fashion. The task is to find the minimum number of moves to move Sth element to the Tth place in the array with the following operation:
A single move consists of the following

// Initially b[] = {1, 2, 3, ..., N}
// arr[] is input array
for (i = 1..n)
temp[arr[i]] = b[i]
b = temp

If not possible then print -1 instead.

Examples:

Input: S = 2, T = 1, arr[] = {2, 3, 4, 1}
Output: 3
N is 4 (size of arr[])
Move 1: b[] = {4, 1, 2, 3}
Move 2: b[] = {3, 4, 1, 2}
Move 3: b[] = {2, 3, 4, 1}

Input: S = 3, T = 4, arr[] = {1, 2, 3, 4}
Output: -1
N is 4 (Size of arr[])
Regardless of how many moves are made, the permutation would remain the same.

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

Approach: The important observation here is that we are only concerned with the position of a single element, and not the entire array. So at each move we move the element at position S to the position arr[S], until we reach Tth position.
Since there are at most N distinct places that we can reach, if we don’t reach T within N moves, it would mean we can never reach it.

Below is the implementation of the above approach:

C++

 // C++ implementation of the approach #include using namespace std;    // Function to return the number of moves int minimumMoves(int n, int a[], int s, int t) {     int i, x;     x = s;     for (i = 1; i <= n; i++) {         if (x == t)             break;         x = a[x];     }        // Destination reached     if (x == t)         return i - 1;     else         return -1; }    // Driver Code int main() {     int s = 2, t = 1, i;     int a[] = {-1, 2, 3, 4, 1};     int n = sizeof(a) / sizeof(a);     cout << minimumMoves(n, a, s, t); }

Java

 // Java implementation of the approach  public class GFG{    // Function to return the number of moves  static int minimumMoves(int n, int a[], int s, int t)  {      int i, x;      x = s;      for (i = 1; i <= n; i++) {          if (x == t)              break;          x = a[x];      }         // Destination reached      if (x == t)          return i - 1;      else         return -1;  }         // Driver Code      public static void main(String []args){     int s = 2, t = 1, i;      int a[] = {-1, 2, 3, 4, 1};      int n = a.length ;     System.out.println(minimumMoves(n, a, s, t));       }     // This code is contributed by Ryuga }

Python3

 # Python3 implementation of the approach     # Function to return the number of moves  def minimumMoves(n, a, s, t):         x = s      for i in range(1, n+1):           # Destination reached         if x == t:             return i-1         x = a[x]              return -1       # Driver Code  if __name__ == "__main__":        s, t = 2, 1      a = [-1, 2, 3, 4, 1]      n = len(a)      print(minimumMoves(n, a, s, t))      # This code is contributed by Rituraj Jain

C#

 // C# implementation of the approach  using System; public class GFG{    // Function to return the number of moves  static int minimumMoves(int n, int []a, int s, int t)  {      int i, x;      x = s;      for (i = 1; i <= n; i++) {          if (x == t)              break;          x = a[x];      }         // Destination reached      if (x == t)          return i - 1;      else         return -1;  }         // Driver Code      public static void Main(){     int s = 2, t = 1;      int []a = {-1, 2, 3, 4, 1};      int n = a.Length ;     Console.WriteLine(minimumMoves(n, a, s, t));      }     // This code is contributed by inder_verma. }

PHP



Output:

3

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