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.
C++
#include <iostream>
using namespace std;
int minStroke(int n, int m, int k)
{
int i = k;
int count = 0;
do {
i = (i + m) % n;
count++;
} while (i != k);
return count;
}
int main()
{
int n = 12, k = 5, m = 8;
cout << minStroke(n, m, k);
return 0;
}
|
Java
public class Main
{
public static int minStroke(int n, int m, int k)
{
int i = k;
int count = 0;
do {
i = (i + m) % n;
count++;
} while (i != k);
return count;
}
public static void main(String[] args)
{
int n = 12;
int k = 5;
int m = 8;
System.out.print(minStroke(n, m, k));
}
}
|
Python3
def minStroke( n, m, k):
i = k;
count = 0;
i = (i + m) % n;
count+=1;
while (i != k):
i = (i + m) % n;
count += 1;
return count;
n = 12;
k = 5;
m = 8;
print(minStroke(n, m, k));
|
C#
using System;
public static class GFG
{
public static int minStroke(int n, int m, int k)
{
int i = k;
int count = 0;
do {
i = (i + m) % n;
count++;
} while (i != k);
return count;
}
public static void Main()
{
int n = 12;
int k = 5;
int m = 8;
Console.Write(minStroke(n, m, k));
}
}
|
Javascript
function minStroke(n, m, k)
{
let i = k;
let count = 0;
do {
i = (i + m) % n;
count++;
} while (i != k);
return count;
}
let n = 12, k = 5, m = 8;
console.log(minStroke(n, m, k));
|
Time complexity: O(n)
Auxiliary Space: O(1)
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++
#include <bits/stdc++.h>
using namespace std;
int minStroke(int n, int m)
{
return (n / __gcd(n, m));
}
int main()
{
int n = 12, k = 5, m = 8;
cout << minStroke(n, m);
return 0;
}
|
Java
class Test {
static int minStroke(int n, int m)
{
return (n / gcd(n, m));
}
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);
}
public static void main(String args[])
{
int n = 12, k = 5, m = 8;
System.out.println(minStroke(n, m));
}
}
|
Python3
def minStroke(n, m):
return (n / __gcd(n, m))
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)
n = 12
k = 5
m = 8
print(minStroke(n, m))
|
C#
using System;
using System.Collections;
class GFG {
static int minStroke(int n, int m)
{
return (n / gcd(n, m));
}
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);
}
public static void Main()
{
int n = 12, m = 8;
Console.WriteLine(minStroke(n, m));
}
}
|
PHP
<?php
function __gcd($a, $b)
{
if($a == 0 || $b == 0)
return 0 ;
if($a == $b)
return $a ;
if($a > $b)
return __gcd($a - $b , $b);
return __gcd($a , $b - $a);
}
function minStroke($n, $m)
{
return ($n / __gcd($n, $m));
}
$n = 12; $k = 5; $m = 8;
echo minStroke($n, $m);
?>
|
Javascript
<script>
function minStroke(n, m)
{
return (n/gcd(n, m));
}
function gcd(n, 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);
}
let n = 12, k = 5, m = 8;
document.write(minStroke(n, m));
</script>
|
Time Complexity: O(log(n))
Auxiliary space: O(1)
This article is contributed by Shivam Pradhan (anuj_charm). If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
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