Number of digits before the decimal point in the division of two numbers

0 0 5 minutes read

Given two integers a and b. The task is to find the number of digits before the decimal point in a / b.
Examples:

Input: a = 100, b = 4
Output: 2
100 / 4 = 25 and number of digits in 25 = 2.
Input: a = 100000, b = 10
Output: 5

Naive approach: Divide the two numbers and then find the number of digits in the division. Take the absolute value of the division for finding the number of digits.
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the number of digits
// before the decimal in a / b
int countDigits(int a, int b)
{
    int count = 0;
 
    // Absolute value of a / b
    int p = abs(a / b);
 
    // If result is 0
    if (p == 0)
        return 1;
 
    // Count number of digits in the result
    while (p > 0) {
        count++;
        p = p / 10;
    }
 
    // Return the required count of digits
    return count;
}
 
// Driver code
int main()
{
    int a = 100;
    int b = 10;
    cout << countDigits(a, b);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG {
 
    // Function to return the number of digits
    // before the decimal in a / b
    static int countDigits(int a, int b)
    {
        int count = 0;
 
        // Absolute value of a / b
        int p = Math.abs(a / b);
 
        // If result is 0
        if (p == 0)
            return 1;
 
        // Count number of digits in the result
        while (p > 0) {
            count++;
            p = p / 10;
        }
 
        // Return the required count of digits
        return count;
    }
 
    // Driver code
    public static void main(String args[])
    {
        int a = 100;
        int b = 10;
        System.out.print(countDigits(a, b));
    }
}

Python

# Python 3 implementation of the approach
 
# Function to return the number of digits 
# before the decimal in a / b
def countDigits(a, b):
    count = 0
     
    # Absolute value of a / b
    p = abs(a // b)
     
    # If result is 0
    if (p == 0):
        return 1
     
    # Count number of digits in the result
    while (p > 0):
        count = count + 1
        p = p // 10
     
     
    # Return the required count of digits
    return count 
 
# Driver code
a = 100
b = 10
print(countDigits(a, b))

C#

// C# implementation of the approach
using System;
class GFG {
 
    // Function to return the number of digits
    // before the decimal in a / b
    static int countDigits(int a, int b)
    {
        int count = 0;
 
        // Absolute value of a / b
        int p = Math.Abs(a / b);
 
        // If result is 0
        if (p == 0)
            return 1;
 
        // Count number of digits in the result
        while (p > 0) {
            count++;
            p = p / 10;
        }
 
        // Return the required count of digits
        return count;
    }
 
    // Driver code
    public static void Main()
    {
        int a = 100;
        int b = 10;
        Console.Write(countDigits(a, b));
    }
}

PHP

<?php 
// PHP implementation of the approach
 
// Function to return the number of digits 
// before the decimal in a / b
function countDigits($a, $b) 
{ 
    $count = 0; 
     
    // Absolute value of a / b
    $p = abs($a / $b); 
     
    // If result is 0
    if ($p == 0) 
        return 1; 
     
    // Count number of digits in the result
    while ($p > 0) { 
        $count++; 
        $p = (int)($p / 10); 
    } 
     
    // Return the required count of digits
    return $count; 
} 
 
// Driver code 
$a = 100; 
$b = 10; 
echo countDigits($a, $b); 
?>

Javascript

<script>
// Javascript implementation of the approach
 
// Function to return the number of digits
// before the decimal in a / b
function countDigits(a, b)
{
    var count = 0;
 
    // Absolute value of a / b
    var p = Math.abs(parseInt(a / b));
 
    // If result is 0
    if (p == 0)
        return 1;
 
    // Count number of digits in the result
    while (p > 0) {
        count++;
        p = parseInt(p / 10);
    }
 
    // Return the required count of digits
    return count;
}
 
// Driver code
var a = 100;
var b = 10;
document.write(countDigits(a, b));
 
// This code is contributed by rrrtnx.
</script>

Output:

Time Complexity: O(log₁₀(a/ b))

Auxiliary Space: O(1)

Efficient approach: To count the number of digits in a / b, we can use the formula:

floor(log₁₀(a) – log₁₀(b)) + 1

Here both the numbers need to be positive integers. For this we can take the absolute values of a and b.
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the number of digits
// before the decimal in a / b
int countDigits(int a, int b)
{
    // Return the required count of digits
    return floor(log10(abs(a)) - log10(abs(b))) + 1;
}
 
// Driver code
int main()
{
    int a = 100;
    int b = 10;
    cout << countDigits(a, b);
 
    return 0;
}