Minimal distance such that for every customer there is at least one vendor at given distance

Given N and M number of points on the straight line, denoting the positions of the customers and vendors respectively. Each vendor provides service to all customers, which are located at a distance that is no more than R from the vendor. The task is to find minimum R such that for each customer there is at least one vendor at the distance which is no more than R.

Examples:

Input: N = 3, M = 2, customer[] = {-2, 2, 4}, vendor[] = {-3, 0}
Output: 4
Explanation: 4 is the minimum distance such that every customer is given service by at least one vendor. 

Input: N = 5, M = 3, customer [] = {1, 5, 10, 14, 17}, vendor[] = {4, 11, 15}
Output: 3
Explanation: 3 is the minimum distance such that every customer is given service by at least one vendor.

Approach: This problem can be solved by using the Two Pointer approach. Follow the steps below to solve the given problem. 

• Take two pointers, i for the customer array and j for the vendor array.
• Start moving the i pointer and for every customer i move the j index while customer[i] > vendor[j].
• Now when vendor[j] >= customer[i],
  • so check the right distance between them which is vendor[j] – customer[i] if j < m.
  • Check the left distance i.e. customer[i] – vendor[j – 1] if j > 0.
  • Find the minimum out of these two i.e the shortest range that the customer[i] can be covered with; achieved by the comparison of the distance of those two adjacent vendors.
  • Then maximize this property as much as possible to get the answer.
• Finally print the answer found.

Below is the implementation of the above approach.

C++

// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the minimal R.
int findMinDist(int customer[], int vendor[],
                int N, int M)
{
    // Variable to keep track of the minimal r
    int minR = 0;

    int i = 0, j = 0;

    // Two pointer approach
    while (i < N) {
        if (j < M and vendor[j] < customer[i])
            j++;
        else {
            int left_d = INT_MAX;
            int right_d = INT_MAX;

            // Find the distance of customer
            // from left vendor.
            if (j > 0)
                left_d = customer[i]
                         - vendor[j - 1];

            // Find the distance of customer
            // from right vendor.
            if (j < M)
                right_d = vendor[j]
                          - customer[i];

            // Find the minimum of
            // left_d and right_d.
            int mn_d = min(left_d, right_d);

            // Maximize the minimum distance.
            minR = max(minR, mn_d);

            // Go to the next customer.
            i++;
        }
    }
    return minR;
}

// Driver code
int main()
{
    int customer[] = { -2, 2, 4 };
    int vendor[] = { -3, 0 };

    int N = sizeof(customer)
            / sizeof(customer[0]);
    int M = sizeof(vendor)
            / sizeof(vendor[0]);

    // Function Call
    cout << findMinDist(customer, vendor,
                        N, M);

    return 0;
}

Java

// Java code for the above approach
import java.io.*;
class GFG
{
  static int INT_MAX = 2147483647;

  // Function to find the minimal R.
  static int findMinDist(int customer[], int vendor[],
                         int N, int M)
  {

    // Variable to keep track of the minimal r
    int minR = 0;

    int i = 0, j = 0;

    // Two pointer approach
    while (i < N) {
      if (j < M && vendor[j] < customer[i])
        j++;
      else {
        int left_d = INT_MAX;
        int right_d = INT_MAX;

        // Find the distance of customer
        // from left vendor.
        if (j > 0)
          left_d = customer[i] - vendor[j - 1];

        // Find the distance of customer
        // from right vendor.
        if (j < M)
          right_d = vendor[j] - customer[i];

        // Find the minimum of
        // left_d and right_d.
        int mn_d = Math.min(left_d, right_d);

        // Maximize the minimum distance.
        minR =Math.max(minR, mn_d);

        // Go to the next customer.
        i++;
      }
    }
    return minR;
  }

  // Driver code
  public static void main(String[] args)
  {
    int customer[] = { -2, 2, 4 };
    int vendor[] = { -3, 0 };

    int N = customer.length;
    int M = vendor.length;

    // Function Call

    System.out.println(
      findMinDist(customer, vendor, N, M));
  }
}

// This code is contributed by Potta Lokesh

Python3

# Python program for above approach
INT_MAX = 2147483647

# Function to find the minimal R.
def findMinDist(customer, vendor, N, M):

    # Variable to keep track of the minimal r
    minR = 0

    i = 0
    j = 0

    # Two pointer approach
    while (i < N):
        if (j < M and vendor[j] < customer[i]):
            j += 1
        else:
            left_d = 10 ** 9
            right_d = 10 ** 9

            # Find the distance of customer
            # from left vendor.
            if (j > 0):
                left_d = customer[i] - vendor[j - 1]

            # Find the distance of customer
            # from right vendor.
            if (j < M):
                right_d = vendor[j] - customer[i]

            # Find the minimum of
            # left_d and right_d.
            mn_d = min(left_d, right_d)

            # Maximize the minimum distance.
            minR = max(minR, mn_d)

            # Go to the next customer.
            i += 1
    return minR

# Driver code
customer = [-2, 2, 4]
vendor = [-3, 0]

N = len(customer)
M = len(vendor)

# Function Call
print(findMinDist(customer, vendor, N, M))

# This code is contributed by Saurabh Jaiswal

Javascript

<script>
    // JavaScript program for above approach
    const INT_MAX = 2147483647;

    // Function to find the minimal R.
    const findMinDist = (customer, vendor, N, M) => {

        // Variable to keep track of the minimal r
        let minR = 0;

        let i = 0, j = 0;

        // Two pointer approach
        while (i < N) {
            if (j < M && vendor[j] < customer[i])
                j++;
            else {
                let left_d = INT_MAX;
                let right_d = INT_MAX;

                // Find the distance of customer
                // from left vendor.
                if (j > 0)
                    left_d = customer[i]
                        - vendor[j - 1];

                // Find the distance of customer
                // from right vendor.
                if (j < M)
                    right_d = vendor[j]
                        - customer[i];

                // Find the minimum of
                // left_d and right_d.
                let mn_d = Math.min(left_d, right_d);

                // Maximize the minimum distance.
                minR = Math.max(minR, mn_d);

                // Go to the next customer.
                i++;
            }
        }
        return minR;
    }

    // Driver code
    let customer = [-2, 2, 4];
    let vendor = [-3, 0];

    let N = customer.length;
    let M = vendor.length;

    // Function Call
    document.write(findMinDist(customer, vendor, N, M));

    // This code is contributed by rakeshsahni

</script>

C#

// C# program to implement
// the above approach
using System;
class GFG
{
  static int INT_MAX = 2147483647;

  // Function to find the minimal R.
  static int findMinDist(int []customer, int []vendor,
                         int N, int M)
  {

    // Variable to keep track of the minimal r
    int minR = 0;

    int i = 0, j = 0;

    // Two pointer approach
    while (i < N) {
      if (j < M && vendor[j] < customer[i])
        j++;
      else {
        int left_d = INT_MAX;
        int right_d = INT_MAX;

        // Find the distance of customer
        // from left vendor.
        if (j > 0)
          left_d = customer[i] - vendor[j - 1];

        // Find the distance of customer
        // from right vendor.
        if (j < M)
          right_d = vendor[j] - customer[i];

        // Find the minimum of
        // left_d and right_d.
        int mn_d = Math.Min(left_d, right_d);

        // Maximize the minimum distance.
        minR =Math.Max(minR, mn_d);

        // Go to the next customer.
        i++;
      }
    }
    return minR;
  }

  // Driver code
  public static void Main()
  {
    int []customer = { -2, 2, 4 };
    int []vendor = { -3, 0 };

    int N = customer.Length;
    int M = vendor.Length;

    // Function Call

    Console.WriteLine(
      findMinDist(customer, vendor, N, M));
  }
}

// This code is contributed by Samim Hossain Mondal.

Output

4

Time Complexity: O(N + M)
Auxiliary Space: O(1)