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.
CPP
// C++ program to find length of a jump// to reach end avoiding all obstacles#include <bits/stdc++.h>#define mod 1000000007using 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 Codeint 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 obstaclesimport 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 obstaclesdef 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 Codea = [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 obstaclesusing 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> |
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