Shortest path to traverse all the elements of a circular array in increasing order

There are N distinct integers arranged on a circle. The distance between any two adjacent numbers is 1. The task is to travel on this circle starting with the smallest number, then moving to the second smallest, third smallest, and so on until the largest number and print the minimum travel distance.

Examples:

Input: arr[] = {3, 6, 5, 1, 2, 4}
Output: 8
1 -> 2 (Left to right)
2 -> 3 (Left to right)
3 -> 4 (Right to left)
4 -> 5 (Left to right or right to left)
5 -> 6 (Right to left)

Input: arr[] = {14, 16, 8, 17, 12, 10, 4, 13, 11, 20}
Output: 27

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

Approach: If we want to travel between two elements with indices i and j, we have two possible paths, one of length |i – j| and the other of length n – |i – j| and we will take the one with the minimum distance.
Initially, we build an array of pairs of (element, index) and we sort it in increasing order of the elements. Then we can take every two consecutive pairs and find the shortest way to travel between these two elements.

Below is the implementation of the above approach:

Java

// Java implementation of the approach 
import java.util.*; 
  
class GFG { 
  
    // Function to return the minimum distance 
    static int min_distance(int n, int[] arr) 
    { 
  
        // Declare a Pair[] array of size n 
        Pair[] val = new Pair[n]; 
  
        // Store all the (element, index) pairs 
        for (int i = 0; i < n; i++) 
            val[i] = new Pair(arr[i], i); 
  
        // Sort the pairs in ascending order of the value 
        Arrays.sort(val, com); 
  
        int min_dist = 0; 
        for (int i = 1; i < n; i++) { 
  
            // Choose the minimum distance path 
            // of the two available 
            min_dist 
                += Math.min(Math.abs(val[i].y - val[i - 1].y), 
                            n - Math.abs(val[i].y - val[i - 1].y)); 
        } 
  
        return min_dist; 
    } 
  
    // Comparator to be used in the 
    // sorting of pairs based on the values 
    static final Comparator<Pair> com = new Comparator<Pair>() { 
        public int compare(Pair a, Pair b) 
        { 
            if (Integer.compare(a.x, b.x) != 0) 
                return Integer.compare(a.x, b.x); 
            else
                return Integer.compare(a.y, b.y); 
        } 
    }; 
  
    // Class to represent a pair 
    static class Pair { 
        int x, y; 
        Pair(int p, int q) 
        { 
            x = p; 
            y = q; 
        } 
    } 
  
    // Driver code 
    public static void main(String[] args) 
    { 
        int[] arr = new int[] { 3, 6, 5, 1, 2, 4 }; 
        int n = arr.length; 
  
        System.out.println(min_distance(n, arr)); 
    } 
} 

Python

# Python implementation of the approach  
  
# Function to return the minimum distance  
def min_distance(n, arr):  
      
    # Declare a Pair[] array of size n  
    val = [None] * n  
  
    # Store all the (element, index) pairs  
    for i in range(0, n):  
        val[i] = (arr[i], i)  
  
    # Sort the pairs in ascending order of the value  
    val.sort()  
  
    min_dist = 0
    for i in range(1, n):  
  
        # Choose the minimum distance path  
        # of the two available  
        min_dist += min(abs(val[i][1] - val[i - 1][1]),  
                        n - abs(val[i][1] - val[i - 1][1]))  
          
    return min_dist  
  
# Driver code  
if __name__ == "__main__": 
      
    arr = [3, 6, 5, 1, 2, 4]  
    n = len(arr)  
  
    print(min_distance(n, arr)) 
  
# This code is contributed by Rituraj Jain 

Output:

My Personal Notes

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

Java

Python

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