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 / bint 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 codeint main(){ int a = 100; int b = 10; cout << countDigits(a, b); return 0;} |
Java
// Java implementation of the approachclass 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 / bdef 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 codea = 100b = 10print(countDigits(a, b)) |
C#
// C# implementation of the approachusing 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 / bfunction 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 / bfunction 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 codevar a = 100;var b = 10;document.write(countDigits(a, b));// This code is contributed by rrrtnx.</script> |
Output:
2
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 / bint countDigits(int a, int b){ // Return the required count of digits return floor(log10(abs(a)) - log10(abs(b))) + 1;}// Driver codeint main(){ int a = 100; int b = 10; cout << countDigits(a, b); return 0;} |
Java
// Java implementation of the approachclass GFG { // Function to return the number of digits // before the decimal in a / b public static int countDigits(int a, int b) { double digits = Math.log10(Math.abs(a)) - Math.log10(Math.abs(b)) + 1; // Return the required count of digits return (int)Math.floor(digits); } // Driver code public static void main(String[] args) { int a = 100; int b = 10; System.out.print(countDigits(a, b)); }} |
Python
# Python3 implementation of the approachimport math # Function to return the number of digits # before the decimal in a / bdef countDigits(a, b): # Return the required count of digits return math.floor(math.log10(abs(a)) - math.log10(abs(b))) + 1# Driver codea = 100b = 10print(countDigits(a, b)) |
C#
// C# implementation of the approachusing System;class GFG { // Function to return the number of digits // before the decimal in a / b public static int countDigits(int a, int b) { double digits = Math.Log10(Math.Abs(a)) - Math.Log10(Math.Abs(b)) + 1; // Return the required count of digits return (int)Math.Floor(digits); } // Driver code 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 / bfunction countDigits($a, $b) { // Return the required count of digits return floor(log10(abs($a)) - log10(abs($b))) + 1; } // Driver code$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 / bfunction countDigits(a, b){ // Return the required count of digits return Math.floor((Math.log(Math.abs(a))/Math.log(10)) - (Math.log(Math.abs(b))/Math.log(10))) + 1;}// Driver codevar a = 100;var b = 10;document.write(countDigits(a, b));// This code is contributed by rutvik_56.</script> |
Output:
2
Time Complexity: O(log10(a/ b))
Auxiliary Space: O(1)
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