Given integers k, d1, d2 and an integer array arr[]. Starting from number k you are allowed to make jumps of size d1 and d2 i.e. all the possible moves from k are:
- k + d1 and k – d1
- k + d2 and k – d2
The task is to find which of the numbers from the array are reachable from k making any number of jumps and any given jump is either of size d1 or d2. For every number print 1 if its reachable else print 0.
Examples:
Input: k = 10, d1 = 4, d2 = 6, arr[] = {10, 15, 20}
Output: 1 0 1
10 can be reached from k with no extra move.
20 can be reached with k + d1 + d2 = 10 + 4 + 6 = 20Input: k = 8, d1 = 3, d2 = 2, arr[] = {9, 4}
Output: 1 1
Approach:
Any number x that is reachable from k with jumps d1 or d2 will be of the form x = k + (i * d1) + (j * d2) where i and j are integers.
Let the GCD of d1 and d2 be gcd. Since, gcd divides both d1 and d2. Therefore we can write d1 = m1 * gcd and d2 = m2 * gcd where m1 and m2 are integers
And x = k + gcd * (i * m1 + j * m2) = k + M * gcd.
So, any number x that is reachable from k should satisfy (x – k) % gcd = 0.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <algorithm>#include <iostream>using namespace std;// Function that returns the vector containing the// result for the reachability of the required numbersvoid reachTheNums(int nums[], int k, int d1, int d2, int n){ int i, ans[n] = { 0 }; int gcd = __gcd(d1, d2); for (i = 0; i < n; i++) { int x = nums[i] - k; // If distance x is coverable if (x % gcd == 0) ans[i] = 1; else ans[i] = 0; } for (i = 0; i < n; i++) cout << ans[i] << " ";}// Driver codeint main(){ // Numbers to be checked for reachability int nums[] = { 9, 4 }; int n = sizeof(nums) / sizeof(nums[0]); // Starting number K int k = 8; // Sizes of jumps d1 and d2 int d1 = 3, d2 = 2; reachTheNums(nums, k, d1, d2, n); return 0;} |
Java
// Java implementation of the approachimport java.io.*;class GFG {// Recursive function to return gcd of a and b static int __gcd(int a, int 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 that returns the vector containing the// result for the reachability of the required numbersstatic void reachTheNums(int nums[], int k, int d1, int d2, int n){ int i; int ans[] = new int[n]; int gcd = __gcd(d1, d2); for (i = 0; i < n; i++) { int x = nums[i] - k; // If distance x is coverable if (x % gcd == 0) ans[i] = 1; else ans[i] = 0; } for (i = 0; i < n; i++) System.out.print(ans[i] + " ");}// Driver code public static void main (String[] args) { // Numbers to be checked for reachability int nums[] = { 9, 4 }; int n =nums.length; // Starting number K int k = 8; // Sizes of jumps d1 and d2 int d1 = 3, d2 = 2; reachTheNums(nums, k, d1, d2, n); }}// This code is contributed by inder_verma.. |
Python3
# Python3 implementation of the approachimport math as mt# Function that returns the vector # containing the result for the reachability # of the required numbersdef reachTheNums(nums, k, d1, d2, n): ans = [0 for i in range(n)] gcd = mt.gcd(d1, d2) for i in range(n): x = nums[i] - k # If distance x is coverable if (x % gcd == 0): ans[i] = 1 else: ans[i] = 0 for i in range(n): print(ans[i], end = " ") # Driver code# Numbers to be checked for# reachabilitynums = [9, 4]n = len(nums)# Starting number Kk = 8# Sizes of jumps d1 and d2d1, d2 = 3, 2reachTheNums(nums, k, d1, d2, n)# This code is contributed # by mohit kumar 29 |
C#
// C# implementation of the above approach using System ;class GFG { // Recursive function to return gcd of a and b static int __gcd(int a, int 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 that returns the vector containing the // result for the reachability of the required numbers static void reachTheNums(int []nums, int k, int d1, int d2, int n) { int i; int []ans = new int[n]; int gcd = __gcd(d1, d2); for (i = 0; i < n; i++) { int x = nums[i] - k; // If distance x is coverable if (x % gcd == 0) ans[i] = 1; else ans[i] = 0; } for (i = 0; i < n; i++) Console.Write(ans[i] + " "); } // Driver code public static void Main () { // Numbers to be checked for reachability int []nums = { 9, 4 }; int n =nums.Length; // Starting number K int k = 8; // Sizes of jumps d1 and d2 int d1 = 3, d2 = 2; reachTheNums(nums, k, d1, d2, n); } // This code is contributed by Ryuga } |
PHP
<?php// PHP implementation of the approach// gcd functionfunction GCD($a, $b){ if ($b == 0) return $a; return GCD($b, $a % $b); }// Function that returns the vector // containing the result for the// reachability of the required numbersfunction reachTheNums($nums, $k, $d1, $d2, $n){ $i = 0; $ans = array(0, 0); $gcd = GCD($d1, $d2); for ($i = 0; $i < $n; $i++) { $x = $nums[$i] - $k; // if distance x is coverable if ($x % $gcd == 0) $ans[$i] = 1; else $ans[$i] = 0; } for ($i = 0; $i < $n; $i++) echo $ans[$i] . " ";}// Driver Code// Numbers to be checked for reachability$nums = array(9, 4);$n = 2;// Starting number $K$k = 8;// Sizes of jumps $d1 and $d2$d1 = 3; $d2 = 2;reachTheNums($nums, $k, $d1, $d2, $n); // This code is contributed by Adesh Singh1 ?> |
Javascript
<script>// javascript implementation of the approach // Recursive function to return gcd of a and b 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 that returns the vector containing the // result for the reachability of the required numbers function reachTheNums(nums , k , d1 , d2 , n) { var i; var ans = Array(n).fill(0); var gcd = __gcd(d1, d2); for (let i = 0; i < n; i++) { var x = nums[i] - k; // If distance x is coverable if (x % gcd == 0) ans[i] = 1; else ans[i] = 0; } for (let i = 0; i < n; i++) document.write(ans[i] + " "); } // Driver code // Numbers to be checked for reachability var nums = [ 9, 4 ]; var n = nums.length; // Starting number K var k = 8; // Sizes of jumps d1 and d2 var d1 = 3, d2 = 2; reachTheNums(nums, k, d1, d2, n);// This code is contributed by aashish1995. </script> |
Output
1 1
Complexity Analysis:
- Time Complexity: O(N), since there runs a loop from 0 to (n – 1).
- Auxiliary Space: O(N), since N extra space has been taken.
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