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# Minimum moves required to change position with the given operation

• Last Updated : 02 Jun, 2021

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:
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.

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

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## Javascript

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Output:
`3`

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