Given two integers a and b. The problem is to find the number of digits in the product of these two integers.
Examples:
Input : a = 12, b = 4 Output : 2 12 * 4 = 48 (2 digits) Input : a = 33, b = -24 Output : 3 33 * -24 = -792 (3 digits)
Basic Approach: Multiply the two numbers and then by using looping construct find the number of digits in the product. Take the absolute value of the product for finding the number of digits.
C++
// C++ implementation to count number of digits// in the product of two numbers#include <bits/stdc++.h>using namespace std;// function to count number of digits // in the product of two numbersint countDigits(int a, int b){ int count = 0; // absolute value of the // product of two numbers int p = abs(a*b); // if product is 0 if (p == 0) { return 1; } // count number of digits in the product 'p' while (p > 0){ count++; p = p / 10; } // required count of digits return count;}// Driver program to test aboveint main(){ int a = 33; int b = -24; // Function call cout << "Number of digits = " << countDigits(a,b); return 0;} |
Java
// Java implementation to count // number of digits in the product // of two numbersimport java.io.*;import java.math.*;class GFG { // function to count number of digits // in the product of two numbers static int countDigits(int a, int b) { int count = 0; // absolute value of the // product of two numbers int p = Math.abs(a * b); // if product is 0 if (p == 0) { return 1; } // count number of digits in // the product 'p' while (p > 0) { count++; p = p / 10; } // required count of digits return count; } // Driver program to test above public static void main(String args[]) { int a = 33; int b = -24; // Function call System.out.println("Number of digits = " + countDigits(a, b)); }}/*This code is contributed by Nikita Tiwari.*/ |
Python3
# Python 3 implementation to count # number of digits in the product # of two numbers# function to count number of digits # in the product of two numbersdef countDigits(a, b) : count = 0 # absolute value of the # product of two numbers p = abs(a * b) # if product is 0 if (p == 0) : return 1 # count number of digits # in the product 'p' while (p > 0) : count = count + 1 p = p // 10 # required count of digits return count# Driver program to test abovea = 33b = -24# Function callprint("Number of digits = ", countDigits(a,b))# This code is contributed by Nikita Tiwari. |
C#
// C# implementation to count// number of digits in the product// of two numbersusing System;class GFG { // function to count number of digits // in the product of two numbers static int countDigits(int a, int b) { int count = 0; // absolute value of the // product of two numbers int p = Math.Abs(a * b); // if product is 0 if (p == 0) { return 1; } // count number of digits in // the product 'p' while (p > 0) { count++; p = p / 10; } // required count of digits return count; } // Driver program to test above public static void Main() { int a = 33; int b = -24; // Function call Console.WriteLine("Number of digits = " + countDigits(a, b)); }}// This code is contributed by Sam007 |
PHP
<?php// PHP implementation to count // number of digits in the // product of two numbers// function to count number // of digits in the product // of two numbersfunction countDigits($a, $b){ $count = 0; // absolute value of the // product of two numbers $p = abs($a * $b); // if product is 0 if ($p == 0) { return 1; } // count number of digits // in the product 'p' while ($p > 0) { $count++; $p = (int)($p / 10); } // required count of digits return $count;}// Driver Code$a = 33;$b = -24;echo "Number of digits = " . countDigits($a, $b);// This code is contributed by mits?> |
Javascript
<script> // Javascript implementation to count number of digits // in the product of two numbers // function to count number of digits // in the product of two numbers function countDigits(a, b) { let count = 0; // absolute value of the // product of two numbers let p = Math.abs(a*b); // if product is 0 if (p == 0) { return 1; } // count number of digits in the product 'p' while (p > 0) { count++; p = parseInt(p / 10, 10); } // required count of digits return count; } let a = 33; let b = -24; document.write("Number of digits = " + countDigits(a,b));// This code is contributed by divyeshrabadiya07.</script> |
Output:
Number of digits = 3
Time Complexity: O(log(abs(a*b))) where abs is absolute value of a*b.
Auxiliary Space: O(1)
Another Approach: To count the number of digits in the product of two numbers we can use the formula given below: Here time complexity will remain same.
count = 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.
C++
// C++ implementation to count number of digits// in the product of two numbers#include <bits/stdc++.h>using namespace std;// function to count number of digits // in the product of two numbersint countDigits(int a, int b){ // if either of the number is 0, then // product will be 0 if (a == 0 || b == 0){ return 1; } // required count of digits return floor(log10(abs(a)) + log10(abs(b))) + 1; }// Driver program to test aboveint main(){ int a = 33; int b = -24; // Function call cout << countDigits(a,b); return 0;} |
Java
// JAVA Code for Number of digits // in the product of two numbersclass GFG { // function to count number of digits // in the product of two numbers public static int countDigits(int a, int b) { // if either of the number is 0, then // product will be 0 if (a == 0 || b == 0) { return 1; } // required count of digits return (int)Math.floor(Math.log10(Math.abs(a)) + Math.log10(Math.abs(b))) + 1; } /* Driver program to test above function */ public static void main(String[] args) { int a = 33; int b = -24; // Function call System.out.print(countDigits(a,b)); }}// This code is contributed by Arnav Kr. Mandal. |
Python3
# Python 3 implementation to count# number of digits in the product# of two numbersimport math # function to count number of digits # in the product of two numbersdef countDigits(a, b) : # if either of the number is 0, # then product will be 0 if (a == 0 or b == 0) : return 1 # required count of digits return math.floor(math.log10(abs(a)) + math.log10(abs(b))) + 1# Driver program to test abovea = 33b = -24# Function callprint(countDigits(a, b))# This code is contributed by Nikita Tiwari. |
C#
// C# Code for Number of digits // in the product of two numbersusing System;class GFG { // function to count number of // digits in the product of two // numbers public static int countDigits(int a, int b) { // if either of the number is 0, // then product will be 0 if (a == 0 || b == 0) { return 1; } // required count of digits return (int)Math.Floor( Math.Log10(Math.Abs(a)) + Math.Log10(Math.Abs(b))) + 1; } // Driver code static void Main() { int a = 33; int b = -24; Console.Write(countDigits(a, b)); }}// This code is contributed by Sam007 |
PHP
<?php// PHP implementation to count// number of digits in the product// of two numbers// function to count number of digits // in the product of two numbersfunction countDigits($a, $b){ // if either of the number is // 0, then product will be 0 if ($a == 0 or $b == 0) { return 1; } // required count of digits return floor(log10(abs($a)) + log10(abs($b))) + 1;}// Driver Code$a = 33;$b = -24;echo countDigits($a, $b);// This code is contributed by mits?> |
Javascript
<script>// Javascript implementation to count number of digits// in the product of two numbers// function to count number of digits // in the product of two numbersfunction countDigits(a, b){ // if either of the number is 0, then // product will be 0 if (a == 0 || b == 0) { return 1; } // required count of digits return Math.floor(Math.log10(Math.abs(a)) + Math.log10(Math.abs(b))) + 1; }// Driver program to test above let a = 33; let b = -24; document.write(countDigits(a,b));// This code is contributed by Mayank Tyagi</script> |
Output:
3
Time Complexity: O(loga + logb) = O(log(a*b))
Auxiliary Space: O(1)
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