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

• Difficulty Level : Medium
• Last Updated : 01 Jun, 2021

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 =

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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-ordinatesclass pairs {public:    int x;    int y;}; // Function to calculate the// manhattan distance between// pair of pointsint 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 pointint 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 initialize the input// and find the minimum distance// by visiting every coordinatevoid findMinDistance(){    int noOfcustomer = 1;    vector coordinate;    vector<int> nodes;    // filling the coordinates into vector    pairs office, home, customer;    office.x = 0;    office.y = 0;    coordinate.push_back(office);    home.x = 5;    home.y = 5;    coordinate.push_back(home);    customer.x = 1;    customer.y = 1;    coordinate.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(                    coordinate[i], coordinate[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 Codeint main(){    // Function Call    findMinDistance();    return 0;}
Output:
10

Performance Analysis:

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

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