Given integers N, K, A, and B, check if it is possible to reach B from A in a circular queue of integers from 1 to N placed sequentially, by jumps of K length. In each move, If it is possible, then print “Yes”. Otherwise, print “No”.
Examples:
Input: N = 5, A = 2, B = 1, K = 2
Output: Yes
Explanation: 2 -> 4 -> 1. Therefore, it is possible to reach B from A.Input: N = 9, A = 6, B = 5, K = 3
Output: No
Approach: The idea to solve the problem is based on the following observations:
- For position A, and after t steps, the position of A is (A + K*t)%N.
- For position B, and after t steps, the position of B is (A + K*t)%N.
- It can be written as:
(A + K*t) = (N*q + B), where q is any positive integer
(A – B) = N*q – K*t
On observing the above equation (N*q – K*t) is divisible by GCD of N and K. Therefore, (A – B) is also divisible by GCD of N and K. Therefore, to reach B from A, (A – B) must be divisible by GCD(N, K).
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to return GCD of two// numbers a and bint GCD(int a, int b){ // Base Case if (b == 0) return a; // Recursively Find the GCD return GCD(b, a % b);}// Function to check of B can be reaced// from A with a jump of K elements in// the circular queuevoid canReach(int N, int A, int B, int K){ // Find GCD of N and K int gcd = GCD(N, K); // If A - B is divisible by gcd // then print Yes if (abs(A - B) % gcd == 0) { cout << "Yes"; } // Otherwise else { cout << "No"; }}// Driver Codeint main(){ int N = 5, A = 2, B = 1, K = 2; // Function Call canReach(N, A, B, K); return 0;} |
Java
// Java program for the// above approachimport java.util.*;class solution{// Function to return GCD // of two numbers a and bstatic int GCD(int a, int b){ // Base Case if (b == 0) return a; // Recursively Find // the GCD return GCD(b, a % b);}// Function to check of B can// be reaced from A with a jump // of K elements in the circular // queuestatic void canReach(int N, int A, int B, int K){ // Find GCD of N and K int gcd = GCD(N, K); // If A - B is divisible // by gcd then print Yes if (Math.abs(A - B) % gcd == 0) { System.out.println("Yes"); } // Otherwise else { System.out.println("No"); }}// Driver Codepublic static void main(String args[]){ int N = 5, A = 2, B = 1, K = 2; // Function Call canReach(N, A, B, K);}}// This code is contributed by SURENDRA_GANGWAR |
Python3
# Python3 program for the# above approach# Function to return GCD # of two numbers a and bdef GCD(a, b): # Base Case if (b == 0): return a # Recursively Find # the GCD return GCD(b, a % b)# Function to check of B # can be reaced from A # with a jump of K elements # in the circular queuedef canReach(N, A, B, K): # Find GCD of N and K gcd = GCD(N, K) # If A - B is divisible # by gcd then prYes if (abs(A - B) % gcd == 0): print("Yes") # Otherwise else: print("No")# Driver Codeif __name__ == '__main__': N = 5 A = 2 B = 1 K = 2 # Function Call canReach(N, A, B, K)# This code is contributed by Mohit Kumar 29 |
C#
// C# program for the// above approachusing System;class GFG{ // Function to return GCD // of two numbers a and bstatic int GCD(int a, int b){ // Base Case if (b == 0) return a; // Recursively Find // the GCD return GCD(b, a % b);} // Function to check of B can// be reaced from A with a jump // of K elements in the circular // queuestatic void canReach(int N, int A, int B, int K){ // Find GCD of N and K int gcd = GCD(N, K); // If A - B is divisible // by gcd then print Yes if (Math.Abs(A - B) % gcd == 0) { Console.WriteLine("Yes"); } // Otherwise else { Console.WriteLine("No"); }} // Driver Codepublic static void Main(){ int N = 5, A = 2, B = 1, K = 2; // Function Call canReach(N, A, B, K);}}// This code is contributed by code_hunt |
Yes
Time Complexity: O(log(min(N, K))
Auxiliary Space: O(1)
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