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

• Difficulty Level : Expert
• Last Updated : 12 Nov, 2021

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. Therefore, 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++

 // C++ implementation to find the// Minimum Cost Path to visit all nodes// situated at the Circumference of// Circular Road #include using namespace std; // Function to find the minimum costint 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 codeint main(){    int arr[] = { 19, 3, 6 };     int n = sizeof(arr) / sizeof(arr);     int circumference = 20;     cout << minCost(arr, n, circumference);     return 0;}

Java

 // Java implementation to find the// Minimum Cost Path to visit all nodes// situated at the Circumference of// Circular Roadimport java.util.*;import java. util. Arrays; class GFG { // Function to find the minimum coststatic 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 codepublic 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

Python3

 # Python3 implementation to find the# minimum cost path to visit all nodes# situated at the circumference of# circular Road # Function to find the minimum costdef minCost(arr, n, circumference):     # Sort the given array    arr.sort()     #Initialise a new array of double size    arr2 =  * (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 codearr = [ 19, 3, 6 ];n = len(arr)circumference = 20; print(minCost(arr, n, circumference)) # This code is contributed by ANKITKUMAR34

C#

 // C# implementation to find the// Minimum Cost Path to visit all nodes// situated at the Circumference of// Circular Roadusing System; class GFG{ // Function to find the minimum coststatic 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 codepublic 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

Javascript


Output:
7

Time Complexity: O(n * log n)

Auxiliary Space: O(n)

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