# 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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