Minimum Cost Path to visit all nodes situated at the Circumference of Circular Road

Given circumference of the circle and an array pos[] which marks the distance of N points on circle relative to a fixed point in the clockwise direction. We have to find a minimum distance through which we can visit all points. We can start with any point.

Examples:

Input: circumference = 20, pos = [3, 6, 9]
Output: min path cost =6
Explanation:
If we start from 3, we go to 6 and then we go to 9. Threfore, total path cost is 3 units for first movement and 3 units for second movement which sums up to 6.

Input:circumference=20 pos = [3, 6, 19]
Output: min path cost = 7
Explanation :
If we start from 19 and we go to 3 it will cost 4 units because we go from 19 -> 20 -> 1 -> 2 -> 3 which gives 4 units, and then 3 to 6 which gives 3 units. In total minimum cost will be 4 + 3 = 7.

Approach :



To solve the problem mentioned above we have to follow the steps given below:

  • Sort the array which marks the distance of N points on circle.
  • Make the array size twice by adding N element with value arr[i + n] = circumference + arr[i].
  • Find the minimum value (arr[i + (n-1)] – arr[i]) for all valid iterations of value i.

Below is the implementation of the above approach:

C++

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// C++ implementation to find the 
// Minimum Cost Path to visit all nodes
// situated at the Circumference of 
// Circular Road
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the minimum cost
int minCost(int arr[], int n, int circumference)
{
    // Sort the given array
    sort(arr, arr + n);
  
    // Initialise a new array of double size
    int arr2[2 * n];
  
    // Fill the array elements
    for (int i = 0; i < n; i++) {
        arr2[i] = arr[i];
        arr2[i + n] = arr[i] + circumference;
    }
  
    // Find the minimum path cost
    int res = INT_MAX;
  
    for (int i = 0; i < n; i++)
        res = min(res, arr2[i + (n - 1)] - arr2[i]);
  
    // Return the final result
    return res;
}
  
// Driver code
int main()
{
    int arr[] = { 19, 3, 6 };
  
    int n = sizeof(arr) / sizeof(arr[0]);
  
    int circumference = 20;
  
    cout << minCost(arr, n, circumference);
  
    return 0;
}

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Java

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// Java implementation to find the 
// Minimum Cost Path to visit all nodes
// situated at the Circumference of 
// Circular Road
import java.util.*; 
import java. util. Arrays;
  
class GFG { 
  
// Function to find the minimum cost
static int minCost(int arr[], int n,
                   int circumference)
{
    // Sort the given array
    Arrays.sort(arr);
  
    // Initialise a new array of double size
    int[] arr2 = new int[2 * n];
  
    // Fill the array elements
    for(int i = 0; i < n; i++)
    {
       arr2[i] = arr[i];
       arr2[i + n] = arr[i] + circumference;
    }
  
    // Find the minimum path cost
    int res = Integer.MAX_VALUE;
  
    for(int i = 0; i < n; i++)
       res = Math.min(res, 
                      arr2[i + (n - 1)] -
                      arr2[i]);
  
    // Return the final result
    return res;
}
  
// Driver code
public static void main(String args[])
{
    int arr[] = { 19, 3, 6 };
    int n = arr.length;
    int circumference = 20;
  
    System.out.println(minCost(arr, n, 
                               circumference));
}
  
}
  
// This code is contributed by ANKITKUMAR34

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Python3

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# Python3 implementation to find the 
# minimum cost path to visit all nodes
# situated at the circumference of 
# circular Road
  
# Function to find the minimum cost
def minCost(arr, n, circumference):
  
    # Sort the given array
    arr.sort()
  
    #Initialise a new array of double size
    arr2 = [0] * (2 * n)
  
    # Fill the array elements
    for i in range(n):
        arr2[i] = arr[i]
        arr2[i + n] = arr[i] + circumference
  
    # Find the minimum path cost
    res = 9999999999999999999;
  
    for i in range(n):
        res = min(res, 
                  arr2[i + (n - 1)] -
                  arr2[i]);
  
    # Return the final result
    return res;
  
# Driver code
arr = [ 19, 3, 6 ];
n = len(arr)
circumference = 20;
  
print(minCost(arr, n, circumference))
  
# This code is contributed by ANKITKUMAR34

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C#

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// C# implementation to find the 
// Minimum Cost Path to visit all nodes
// situated at the Circumference of 
// Circular Road
using System;
  
class GFG{ 
  
// Function to find the minimum cost
static int minCost(int []arr, int n,
                   int circumference)
{
    // Sort the given array
    Array.Sort(arr);
  
    // Initialise a new array of double size
    int[] arr2 = new int[2 * n];
  
    // Fill the array elements
    for(int i = 0; i < n; i++)
    {
       arr2[i] = arr[i];
       arr2[i + n] = arr[i] + circumference;
    }
  
    // Find the minimum path cost
    int res = int.MaxValue;
  
    for(int i = 0; i < n; i++)
       res = Math.Min(res, arr2[i + (n - 1)] -
                           arr2[i]);
  
    // Return the readonly result
    return res;
}
  
// Driver code
public static void Main(String []args)
{
    int []arr = { 19, 3, 6 };
    int n = arr.Length;
    int circumference = 20;
  
    Console.WriteLine(minCost(arr, n, circumference));
}
}
  
// This code is contributed by Princi Singh

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Output:

7

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