Skip to content
Related Articles

Related Articles

Improve Article

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 <bits/stdc++.h>
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[0]);
    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




<?php
// PHP implementation of the approach
 
 
// Function to return the number of moves
function minimumMoves($n, $a, $s, $t)
{
    $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
 
    $s = 2; $t = 1; $i;
    $a = array(-1, 2, 3, 4, 1);
    $n = count($a);
    echo minimumMoves($n, $a, $s, $t);
     // This code is contributed by inder_verma.
?>

Javascript




<script>
    // Javascript implementation of the approach  
     
    // Function to return the number of moves
    function minimumMoves(n, a, s, t)
    {
        let 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;
    }
     
    let s = 2, t = 1;
    let a = [-1, 2, 3, 4, 1];
    let n = a.length ;
    document.write(minimumMoves(n, a, s, t));
      
     // This code is contributed by suresh07.
</script>
Output: 
3

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :