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 <bits/stdc++.h>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[0]); 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 = [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 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
<script>// Javascript implementation to find the // Minimum Cost Path to visit all nodes// situated at the Circumference of // Circular Road// Function to find the minimum costfunction minCost(arr, n,circumference){ // Sort the given array arr.sort((a,b)=>a-b) // Initialise a new array of double size var arr2 = Array(2* n).fill(0); // Fill the array elements for (var i = 0; i < n; i++) { arr2[i] = arr[i]; arr2[i + n] = arr[i] + circumference; } // Find the minimum path cost var res = 1000000000; for (var i = 0; i < n; i++) res = Math.min(res, arr2[i + (n - 1)] - arr2[i]); // Return the final result return res;}// Driver codevar arr = [19, 3, 6 ];var n = arr.length;var circumference = 20;document.write( minCost(arr, n, circumference));</script> |
Output:
7
Time Complexity: O(n * log n)
Auxiliary Space: O(n)
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