# Minimum Manhattan distance covered by visiting every coordinates from a source to a final vertex

Given an array arr[] of co-ordinate points and a source and final co-ordinate point, the task is to find the minimum manhattan distance covered from the source to the final vertex such that every point of the array is visited exactly once.

Manhattan Distance =

Examples:

Input: source = (0, 0), final = (100, 100)
arr[] = {(70, 40), (30, 10), (10, 5), (90, 70), (50, 20)}
Output: 200

Input: source = (0, 0), final = (5, 5)
arr[] = {(1, 1)}
Output: 10

Approach: The idea is to use permutation and combination to generate every possible permutation movements to the co-ordinates and then compute the total manhattan distance covered by moving from the first co-ordinate of the array to the final co-ordinate and If the final distance covered is less than the minimum distance covered till now. Then update the minimum distance covered.

Below is the implementation of the above approach:

## C++

 // C++ implementation to find the   // minimum manhattan distance   // covered by visiting N co-ordinates     #include  using namespace std;     // Class of co-ordinates  class pairs {  public:      int x;      int y;  };     // Function to calculate the   // manhattan distance between   // pair of points  int calculate_distance(pairs a,                          pairs b)  {      return abs(a.x - b.x) +              abs(a.y - b.y);  }     // Function to find the minimum   // distance covered for visiting   // every co-ordinate point  int findMinDistanceUtil(vector<int> nodes,              int noOfcustomer, int** matrix)  {      int mindistance = INT_MAX;             // Loop to compute the distance      // for every possible permutation      do {          int distance = 0;          int prev = 1;                     // Computing every total manhattan          // distance covered for the every           // co-ordinate points          for (int i = 0; i < noOfcustomer; i++) {              distance = distance +                          matrix[prev][nodes[i]];              prev = nodes[i];          }                     // Adding the final distance          distance = distance + matrix[prev][0];                     // if distance is less than           // minimum value than update it          if (distance < mindistance)              mindistance = distance;      }while (          next_permutation(              nodes.begin(), nodes.end()          ));      return mindistance;  }     // Function to intialize the input  // and find the minimum distance   // by visiting every cordinate  void findMinDistance()  {      int noOfcustomer = 1;      vector cordinate;      vector<int> nodes;      // filling the coordinates into vector      pairs office, home, customer;      office.x = 0;      office.y = 0;      cordinate.push_back(office);      home.x = 5;      home.y = 5;      cordinate.push_back(home);      customer.x = 1;      customer.y = 1;      cordinate.push_back(customer);             // make a 2d matrix which stores      // distance between two point      int** matrix = new int*[noOfcustomer + 2];             // Loop to compute the distance between      // every pair of points in the co-ordinate      for (int i = 0; i < noOfcustomer + 2; i++) {          matrix[i] = new int[noOfcustomer + 2];                     for (int j = 0; j < noOfcustomer + 2; j++) {              matrix[i][j] = calculate_distance(                      cordinate[i], cordinate[j]);          }                     // Condition to not move the           // index of the source or           // the final vertex          if (i != 0 && i != 1)              nodes.push_back(i);      }      cout << findMinDistanceUtil(          nodes, noOfcustomer, matrix);  }     // Driver Code  int main()  {      // Function Call      findMinDistance();      return 0;  }

Output:

10


Performance Analysis:

• Time Complexity: O(N! * N)
• Auxiliary Space: O(N2)

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