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Count number of digits after decimal on dividing a number
  • Difficulty Level : Hard
  • Last Updated : 02 Feb, 2021

We are given two numbers A and B. We need to calculate the number of digits after decimal. If in case the numbers are irrational then print “INF”.
Examples: 

Input  : x = 5, y = 3
Output : INF
5/3 = 1.666....

Input :  x = 3, y = 6
Output : 1
3/6 = 0.5 

The idea is simple we follow school division and keep track of remainders while dividing one by one. If remainder becomes 0, we return count of digits seen after decimal. If remainder repeats, we return INF.  

C++




// CPP program to count digits after dot when a
// number is divided by another.
#include <bits/stdc++.h>
using namespace std;
 
int count(int x, int y)
{
    int ans = 0;  // Initialize result
     
    unordered_map<int, int> m;
     
    // calculating remainder
    while (x % y != 0) {
 
        x = x % y;
 
        ans++;
 
        // if this remainder appeared before then
        // the numbers are irrational and would not
        // converge to a solution the digits after
        // decimal will be infinite
        if (m.find(x) != m.end())
            return -1;
 
        m[x] = 1;
        x = x * 10;
    }
    return ans;
}
 
// Driver code
int main()
{
    int res = count(1, 2);
    (res == -1)? cout << "INF" : cout << res;
 
    cout << endl;
    res = count(5, 3);
    (res == -1)? cout << "INF" : cout << res;
 
    cout << endl;
    res = count(3, 5);
    (res == -1)? cout << "INF" : cout << res;
    return 0;
}

Java




// Java program to count digits after dot when a
// number is divided by another.
import java.util.*;
 
class GFG
{
 
static int count(int x, int y)
{
    int ans = 0; // Initialize result
     
    Map<Integer,Integer> m = new HashMap<>();
     
    // calculating remainder
    while (x % y != 0)
    {
 
        x = x % y;
 
        ans++;
 
        // if this remainder appeared before then
        // the numbers are irrational and would not
        // converge to a solution the digits after
        // decimal will be infinite
        if (m.containsKey(x))
            return -1;
 
        m.put(x, 1);
        x = x * 10;
    }
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    int res = count(1, 2);
    if((res == -1))
        System.out.println("INF");
    else
        System.out.println(res);
 
    res = count(5, 3);
    if((res == -1))
        System.out.println("INF");
    else
        System.out.println(res);
     
    res = count(3, 5);
    if((res == -1))
        System.out.println("INF");
    else
        System.out.println(res);
}
}
 
// This code is contributed by Rajput-Ji

Python3




# Python3 program to count digits after dot
# when a number is divided by another.
def count(x, y):
    ans = 0 # Initialize result
 
    m = dict()
 
    # calculating remainder
    while x % y != 0:
        x %= y
 
        ans += 1
 
        # if this remainder appeared before then
        # the numbers are irrational and would not
        # converge to a solution the digits after
        # decimal will be infinite
        if x in m:
            return -1
 
        m[x] = 1
        x *= 10
 
    return ans
 
# Driver Code
if __name__ == "__main__":
    res = count(1, 2)
 
    print("INF") if res == -1 else print(res)
 
    res = count(5, 3)
    print("INF") if res == -1 else print(res)
 
    res = count(3, 5)
    print("INF") if res == -1 else print(res)
 
# This code is contributed by
# sanjeev2552

C#




// C# program to count digits after dot when a
// number is divided by another.
using System;
using System.Collections.Generic;
     
class GFG
{
 
static int count(int x, int y)
{
    int ans = 0; // Initialize result
     
    Dictionary<int,int> m = new Dictionary<int,int>();
     
    // calculating remainder
    while (x % y != 0)
    {
 
        x = x % y;
 
        ans++;
 
        // if this remainder appeared before then
        // the numbers are irrational and would not
        // converge to a solution the digits after
        // decimal will be infinite
        if (m.ContainsKey(x))
            return -1;
 
        m.Add(x, 1);
        x = x * 10;
    }
    return ans;
}
 
// Driver code
public static void Main(String[] args)
{
    int res = count(1, 2);
    if((res == -1))
        Console.WriteLine("INF");
    else
        Console.WriteLine(res);
 
    res = count(5, 3);
    if((res == -1))
        Console.WriteLine("INF");
    else
        Console.WriteLine(res);
     
    res = count(3, 5);
    if((res == -1))
        Console.WriteLine("INF");
    else
        Console.WriteLine(res);
}
}
 
// This code is contributed by 29AjayKumar

Output: 

1
INF
1

Time Complexity: O(N * log(N) )

Auxiliary Space: O(N)

This article is contributed by Rahul Chawla. 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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