Find Recurring Sequence in a Fraction

Given a fraction, find recurring sequence of digits if exists, otherwise print “No recurring sequence”.

Examples:

Input  : Numerator = 8, Denominator = 3
Output : Recurring sequence is 6 
Explanation : 8/3 = 2.66666666.......  

Input : Numerator = 50, Denominator = 22
Output : Recurring sequence is 27
Explanation : 50/22 = 2.272727272..... 

Input : Numerator = 11, Denominator = 2
Output : No recurring sequence
Explanation : 11/2 = 5.5 

We strongly recommend that you click here and practice it, before moving on to the solution.

When does the fractional part repeat ?

Let us simulate the process of converting fraction to decimal. Let us look at the part where we have already figured out the integer part which is floor(numerator/denominator). Now we are left with ( remainder = numerator%denominator ) / denominator.
If you remember the process of converting to decimal, at each step we do the following :

  1. Multiply the remainder by 10.
  2. Append remainder / denominator to result.
  3. Remainder = remainder % denominator.

At any moment, if remainder becomes 0, we are done.

However, when there is a recurring sequence, remainder never becomes 0. For example if you look at 1/3, the remainder never becomes 0.

Below is one important observation :
If we start with remainder ‘rem’ and if the remainder repeats at any point of time, the digits between the two occurrence of ‘rem’ keep repeating.

So the idea is to store seen remainders in a map. Whenever a remainder repeats, we return the sequence before the next occurrence. Below is C++ implementation of above idea.

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// C++ program to find repeating sequence in a fraction
#include <bits/stdc++.h>
using namespace std;
  
// This function returns repeating sequence of
// a fraction.  If repeating sequence doesn't
// exits, then returns empty string
string fractionToDecimal(int numr, int denr)
{
    string res; // Initialize result
  
    // Create a map to store already seen remainders
    // remainder is used as key and its position in
    // result is stored as value. Note that we need
    // position for cases like 1/6.  In this case,
    // the recurring sequence doesn't start from first
    // remainder.
    map <int, int> mp;
    mp.clear();
  
    // Find first remainder
    int rem = numr%denr;
  
    // Keep finding remainder until either remainder
    // becomes 0 or repeats
    while ( (rem!=0) && (mp.find(rem) == mp.end()) )
    {
        // Store this remainder
        mp[rem] = res.length();
  
        // Multiply remainder with 10
        rem = rem*10;
  
        // Append rem / denr to result
        int res_part = rem / denr;
        res += to_string(res_part);
  
        // Update remainder
        rem = rem % denr;
    }
  
    return (rem == 0)? "" : res.substr(mp[rem]);
}
  
// Driver code
int main()
{
    int numr = 50, denr = 22;
    string res = fractionToDecimal(numr, denr);
    if (res == "")
        cout << "No recurring sequence";
    else
        cout << "Recurring sequence is " << res;
    return 0;
}

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Output :

Recurring sequence is 27

This article is contributed by Dhruv Mahajan. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above



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