Maximize jump size to reach all given points from given starting point

Given an array, arr[] consisting of N integers representing coordinates on a number line and an integer S. Find the maximum size of jump required to visit all the coordinates at least once if the starting position is S. Also, Jumping in both directions is permitted.

Example:

Input: arr[]={1, 7, 3, 9, 5, 11}, S=3
Output: 2
Explanation: Jumps of size 2 is able to visit all the coordinates on the line, in the following way:
Jump -2: Reach 1 from 3.
Jump +2: Reach 3 from 1.
Jump +2: Reach 5 from 3.
Jump +2: Reach 7 from 5.
Jump +2: Reach 9 from 7.
Jump +2: Reach 11 from 9.

Input: arr[]={33, 105, 57}, S=81
Output: 24

Approach: An observation can be made after reading the problem, that the greatest common divisor of all the differences between two adjacent points on the line is the maximum size of Jump needed to visit all points. So, to solve this problem, follow the below steps:

1. Insert S, in the array arr, then sort it.
2. Find the gcd between the differences of all adjacent pairs in the array arr.
3. Return the final gcd as the answer to this problem.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find the maximum size of jump
// to reach all the coordinates
int minJumpSize(vector<int>& arr, int S)
{

    // Pushing S into the line
    arr.push_back(S);

    // Sorting arr, to get coordinates in
    // a sorted way
    sort(arr.begin(), arr.end());

    // Finding gcd between the differences of
    // all adjacent pairs
    int gcd = arr[1] - arr[0];
    for (int i = 1; i < arr.size(); i++) {
        gcd = __gcd(gcd,
                    __gcd(gcd,
                          arr[i] - arr[i - 1]));
    }

    return gcd;
}

// Driver Code
int main()
{
    vector<int> arr = { 1, 7, 3, 9, 5, 11 };
    int S = 3;

    cout << minJumpSize(arr, S);
}

Java

// Java program for the above approach
import java.util.ArrayList;
import java.util.Collections;

class GFG {

    public static int __gcd(int a, int b) {
        if (b == 0)
            return a;
        return __gcd(b, a % b);
    }

    // Function to find the maximum size of jump
    // to reach all the coordinates
    public static int minJumpSize(int[] t_arr, int S) {
        ArrayList<Integer> arr = new ArrayList<Integer>();

        for (int x : t_arr) {
            arr.add(x);
        }

        // Pushing S into the line
        arr.add(S);

        // Sorting arr, to get coordinates in
        // a sorted way
        Collections.sort(arr);

        // Finding gcd between the differences of
        // all adjacent pairs
        int gcd = arr.get(1) - arr.get(0);
        for (int i = 1; i < arr.size(); i++) {
            gcd = __gcd(gcd, __gcd(gcd, arr.get(i) - arr.get(i - 1)));
        }

        return gcd;
    }

    // Driver Code
    public static void main(String args[]) {
        int[] arr = { 1, 7, 3, 9, 5, 11 };
        int S = 3;

        System.out.println(minJumpSize(arr, S));
    }
}

// This code is contributed by saurabh_jaiswal.

Python3

# python program for the above approach
import math

# Function to find the maximum size of jump
# to reach all the coordinates
def minJumpSize(arr, S):

    # Pushing S into the line
    arr.append(S)

    # Sorting arr, to get coordinates in
    # a sorted way
    arr.sort()

    # Finding gcd between the differences of
    # all adjacent pairs
    gcd = arr[1] - arr[0]
    for i in range(1, len(arr)):
        gcd = math.gcd(gcd, math.gcd(gcd, arr[i] - arr[i - 1]))

    return gcd

# Driver Code
if __name__ == "__main__":

    arr = [1, 7, 3, 9, 5, 11]
    S = 3

    print(minJumpSize(arr, S))

# This code is contributed by rakeshsahni

C#

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG{

    public static int __gcd(int a, int b) {
        if (b == 0)
            return a;
        return __gcd(b, a % b);
    }

    // Function to find the maximum size of jump
    // to reach all the coordinates
    public static int minJumpSize(int[] t_arr, int S) {
        List<int> arr = new List<int>();

        foreach (int x in t_arr) {
            arr.Add(x);
        }

        // Pushing S into the line
        arr.Add(S);

        // Sorting arr, to get coordinates in
        // a sorted way
        arr.Sort();

        // Finding gcd between the differences of
        // all adjacent pairs
        int gcd = arr[1] - arr[0];
        for (int i = 1; i < arr.Count; i++) {
            gcd = __gcd(gcd, __gcd(gcd, arr[(i)] - arr[(i - 1)]));
        }

        return gcd;
    }


// Driver Code
public static void Main()
{
    int[] arr = { 1, 7, 3, 9, 5, 11 };
    int S = 3;

    Console.WriteLine(minJumpSize(arr, S));
}
}

// This code is contributed by sanjoy_62.

Javascript

<script>
        // JavaScript Program to implement
        // the above approach

        function __gcd(a, b) {
            // Everything divides 0
            if (a == 0)
                return b;
            if (b == 0)
                return a;

            // base case
            if (a == b)
                return a;

            // a is greater
            if (a > b)
                return __gcd(a - b, b);
            return __gcd(a, b - a);
        }


        // Function to find the maximum size of jump
        // to reach all the coordinates
        function minJumpSize(arr, S) {

            // Pushing S into the line
            arr.push(S);

            // Sorting arr, to get coordinates in
            // a sorted way
            arr.sort(function (a, b) { return a - b })

            // Finding gcd between the differences of
            // all adjacent pairs
            let gcd = arr[1] - arr[0];
            for (let i = 1; i < arr.length; i++) {
                gcd = __gcd(gcd,
                    __gcd(gcd,
                        arr[i] - arr[i - 1]));
            }

            return gcd;
        }

        // Driver Code

        let arr = [1, 7, 3, 9, 5, 11];
        let S = 3;

        document.write(minJumpSize(arr, S))


// This code is contributed by Potta Lokesh
    </script>

Output:

2

Time Complexity: O(NlogN)
Auxiliary Space: O(1).