Minimum steps to come back to starting point in a circular tour
Consider circular track with n points marked as 1, 2, …n. A person is initially placed on point k. The person moves m > 0, slot forward (in circular way) in each step. Find the minimum number of steps required so that the person reaches initial point k.
Examples:
Input : n = 9, k = 2, m = 6 Output : 3 Explanation : Sequence of moves is 2 => 8 => 5 => 2 Input : n = 6, k = 3, m = 2 Output : 3
Naive Approach : Initialize a counter ‘i’ with ‘k’ and ‘count’ = 0. Further for each iteration increment ‘count’ add ‘m’ to ‘i’. Take its modulus with n i.e. i=((i+m)%n), if i > n. If i becomes equal to k then count will be our answer.
Time complexity: O(n).
Efficient Approach: We find GCD(n, m) and then divide n by GCD(n, m). That will be our answer. This can be explained as:
Think of n and m as per question now as we know that gcd(n, m) must divide n and the quotient tells us that after how many successive jumps(addition) of m numbers from starting position(say 0) we again reach the starting position.
Note: In circular arrangement of n numbers nth and 0th position are same.
C++
// C++ program to find minimum steps to reach // starting position in a circular tour. #include<bits / stdc++.h> using namespace std; // function for finding minimum steps int minStroke(int n, int m) { // return value n / gcd(n, m) return (n/__gcd(n, m)); } // Driver function int main() { int n = 12, k = 5, m = 8; cout << minStroke(n, m); return 0; } |
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Java
// Java program to find minimum steps to reach // starting position in a circular tour. class Test { // method for finding minimum steps static int minStroke(int n, int m) { // return value n / gcd(n, m) return (n/gcd(n, m)); } // method for gcd static int gcd(int n, int m) { if (n == 0 || m == 0) return 0; if (n == m) return n; if (n > m) return gcd(n-m, m); return gcd(n, m-n); } // Driver method public static void main(String args[]) { int n = 12, k = 5, m = 8; System.out.println(minStroke(n, m)); } } |
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Python3
# Python program to find minimum # steps to reach starting position # in a circular tour. # function for finding minimum steps def minStroke(n, m): # return value n / gcd(n, m) return (n / __gcd(n, m)) # method for gcd def __gcd(n, m): if(n == 0 or m == 0): return 0 if (n == m): return n if (n > m): return __gcd(n-m, m) return __gcd(n, m-n) # Driver code n = 12k = 5m = 8 print(minStroke(n, m)) # This code is contributed by Anant Agarwal. |
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C#
// C# program to find minimum steps to reach // starting position in a circular tour. using System; using System.Collections; class GFG { // method for finding minimum steps static int minStroke(int n, int m) { // return value n / gcd(n, m) return (n/gcd(n, m)); } // method for gcd static int gcd(int n, int m) { if (n == 0 || m == 0) return 0; if (n == m) return n; if (n > m) return gcd(n-m, m); return gcd(n, m-n); } // Driver method public static void Main() { int n = 12, m = 8; Console.WriteLine(minStroke(n, m)); } } // This code is contributed by Sam007 |
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PHP
<?php // PHP program to find minimum // steps to reach starting // position in a circular tour. // Recursive function to // return gcd of a and b function __gcd($a, $b) { // Everything divides 0 if($a == 0 || $b == 0) return 0 ; // base case if($a == $b) return $a ; // a is greater if($a > $b) return __gcd($a - $b , $b); return __gcd($a , $b - $a); } // function for // finding minimum steps function minStroke($n, $m) { // return value n / gcd(n, m) return ($n / __gcd($n, $m)); } // Driver Code $n = 12; $k = 5; $m = 8; echo minStroke($n, $m); // This code is contributed by anuj_67. ?> |
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Output:
3
Time Complexity: O(log(n))
This article is contributed by Shivam Pradhan (anuj_charm). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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