Minimum length of jumps to avoid given array of obstacles

We are given coordinates of obstacles on a straight line. We start jumping from point 0, we need to reach the end, avoiding all obstacles. The length of every jump has to be the same (For example, if we jump from 0 to 4, then we must make the next jump from 4 to 8). We need to find the minimum length of the jump so that we can reach the end and we avoid all obstacles.

Examples:

Input : obs[] = [5, 3, 6, 7, 9]
Output : 4
Obstacles are at points 3, 5, 6, 7 and 9
We jump from 0 to 4, then 4 to 8, then 4
to 12. This is how we reach end with jumps
of length 4. If we try lower jump lengths,
we cannot avoid all obstacles.

Input : obs[] = [5, 8, 9, 13, 14]
Output : 6 

We insert the locations of all obstacles in a hash table. We also find the maximum value of the obstacle. Then we tried all possible jump sizes from 1 to maximum. If any jump size leads to an obstacle, we do not consider that jump. 

Implementation:

CPP

// C++ program to find length of a jump
// to reach end avoiding all obstacles
#include <bits/stdc++.h>
#define mod 1000000007
using namespace std;

int avoidObstacles(unordered_set<int> obs)
{

    // set jump distance to 1
    int jump_dist = 1;

    // flag to check if current jump distance
    // hits an obstacle
    bool obstacle_hit = true;

    while(obstacle_hit)
    {

        obstacle_hit = false;
        jump_dist += 1;

        // checking if jumping with current length
        // hits an obstacle
        for (auto i:obs)
        {
            if (i % jump_dist == 0)
            {

                // if obstacle is hit repeat process
                // after increasing jump distance
                obstacle_hit = true;
                break;
            }
            }
        }

    return jump_dist;
}

// Driver Code
int main()
{
    unordered_set<int> a = {5, 3, 6, 7, 9};
    int b = avoidObstacles(a);
    cout << b;
}

// This code is contributed by mohit kumar 29

Java

// Java program to find length of a jump
// to reach end avoiding all obstacles
import java.util.*;

public class obstacle {
    static int avoidObstacles(int[] obs)
    {
        // Insert all array elements in a hash table
        // and find the maximum value in the array
        HashSet<Integer> hs = new HashSet<Integer>();
        int max = obs[0];
        for (int i=0; i<obs.length; i++)
        {
            hs.add(obs[i]);
            max = Math.max(max, obs[i]);
        }

        // checking for every possible length which
        // yield us solution
        for (int i = 1; i <= max; i++) {
            int j;
            for (j = i; j <= max; j = j + i) {

                // if there is obstacle, break the loop.
                if (hs.contains(j))
                    break;
            }

            // If above loop did not break
            if (j > max)
                return i;
        }

        return max+1;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int a[] = new int[] { 5, 3, 6, 7, 9 };
        int b = avoidObstacles(a);
        System.out.println(b);
    }
}

Python3

# Python3 program to find length of a jump
# to reach end avoiding all obstacles
def avoidObstacles(obs):

    # sort the list in ascending order
    obs = sorted(obs)

    # set jump distance to 1
    jump_dist = 1

    # flag to check if current jump distance
    # hits an obstacle
    obstacle_hit = True

    while(obstacle_hit):

        obstacle_hit = False
        jump_dist += 1

        # checking if jumping with current length
        # hits an obstacle
        for i in range(0, len(obs)):
            if obs[i] % jump_dist == 0:

                # if obstacle is hit repeat process
                # after increasing jump distance
                obstacle_hit = True
                break

    return jump_dist

# Driver Code
a = [5, 3, 6, 7, 9]
b = avoidObstacles(a)
print(b)

# This code is contributed by ViratJoshi

C#

// C# program to find length of a jump
// to reach end avoiding all obstacles
using System;
using System.Collections.Generic;

public class obstacle
{
    static int avoidObstacles(int[] obs)
    {
        // Insert all array elements in a hash table
        // and find the maximum value in the array
        HashSet<int> hs = new HashSet<int>();
        int max = obs[0];
        for (int i = 0; i < obs.Length; i++)
        {
            hs.Add(obs[i]);
            max = Math.Max(max, obs[i]);
        }

        // checking for every possible length which
        // yield us solution
        for (int i = 1; i <= max; i++)
        {
            int j;
            for (j = i; j <= max; j = j + i)
            {

                // if there is obstacle, break the loop.
                if (hs.Contains(j))
                    break;
            }

            // If above loop did not break
            if (j > max)
                return i;
        }
        return max+1;
    }

    // Driver Code
    public static void Main()
    {
        int []a = new int[] { 5, 3, 6, 7, 9 };
        int b = avoidObstacles(a);
        Console.WriteLine(b);
    }
}

// This code is contributed by 29AjayKumar

Javascript

<script>
// Javascript program to find length of a jump
// to reach end avoiding all obstacles

    function avoidObstacles(obs)
    {
        // Insert all array elements in a hash table
        // and find the maximum value in the array
        let hs = new Set();
        let max = obs[0];
        for (let i=0; i<obs.length; i++)
        {
            hs.add(obs[i]);
            max = Math.max(max, obs[i]);
        }

        // checking for every possible length which
        // yield us solution
        for (let i = 1; i <= max; i++) {
            let j;
            for (j = i; j <= max; j = j + i) {

                // if there is obstacle, break the loop.
                if (hs.has(j))
                    break;
            }

            // If above loop did not break
            if (j > max)
                return i;
        }

        return max+1;
    }

    // Driver Code
    let a=[5, 3, 6, 7, 9 ];
    let b = avoidObstacles(a);
    document.write(b);


// This code is contributed by unknown2108
</script>

Output

4

Time Complexity: O(n)
Auxiliary Space: O(1)