Given a number, the task is to find the product of the digits of a number.
Examples:
Input: n = 4513
Output: 60
Input: n = 5249
Output: 360
General Algorithm for product of digits in a given number:
Get the number
Declare a variable to store the product and set it to 1
Repeat the next two steps till the number is not 0
Get the rightmost digit of the number with help of remainder ‘%’ operator by dividing it with 10 and multiply it with product.
Divide the number by 10 with help of ‘/’ operator
Print or return the product.
Below is the solution to get the product of the digits:
C++
// C++ program to compute
// product of digits in the number.
#include<bits/stdc++.h>
usingnamespacestd;
/* Function to get product of digits */
intgetProduct(intn)
{
intproduct = 1;
while(n != 0)
{
product = product * (n % 10);
n = n / 10;
}
returnproduct;
}
// Driver program
intmain()
{
intn = 4513;
cout << (getProduct(n));
}
// This code is contributed by
// Surendra_Gangwar
Java
// Java program to compute
// product of digits in the number.
importjava.io.*;
classGFG {
/* Function to get product of digits */
staticintgetProduct(intn)
{
intproduct = 1;
while(n != 0) {
product = product * (n % 10);
n = n / 10;
}
returnproduct;
}
// Driver program
publicstaticvoidmain(String[] args)
{
intn = 4513;
System.out.println(getProduct(n));
}
}
Python3
# Python3 program to compute
# product of digits in the number.
# Function to get product of digits
defgetProduct(n):
product =1
while(n !=0):
product =product *(n %10)
n =n //10
returnproduct
# Driver Code
n =4513
print(getProduct(n))
# This code is contributed
# by mohit kumar
C#
// C# program to compute
// product of digits in the number.
usingSystem;
classGFG
{
/* Function to get product of digits */
staticintgetProduct(intn)
{
intproduct = 1;
while(n != 0)
{
product = product * (n % 10);
n = n / 10;
}
returnproduct;
}
// Driver program
publicstaticvoidMain()
{
intn = 4513;
Console.WriteLine(getProduct(n));
}
}
// This code is contributed by Ryuga
PHP
<?php
<?php
// PHP program to compute
// $product of digits in the number.
/* Function to get $product of digits */
functiongetProduct($n)
{
$product= 1;
while($n!= 0)
{
$product= $product* ( $n% 10);
$n= intdiv($n, 10);
}
return$product;
}
// Driver code
$n= 4513;
echogetProduct($n);
// This code is contributed by
// ihritik
?>
Javascript
<script>
// JavaScript program to compute
// product of digits in the number.
// Function to get product of digits
functiongetProduct(n)
{
let product = 1;
while(n != 0)
{
product = product * (n % 10);
n = Math.floor(n / 10);
}
returnproduct;
}
// Driver code
let n = 4513;
document.write(getProduct(n));
// This code is contributed by Manoj.
</script>
Output
60
Time Complexity: O(log10N) Auxiliary Space: O(1)
Method #2:Using string() method:
Convert the integer to string
Traverse the string and multiply the characters by converting them to integer
When this method can be used?: When the number of digits of a number exceeds , we can’t take that number as an integer since the range of long long int doesn’t satisfy the given number. So take input as a string, run a loop from start to the length of the string and increase the sum with that character(in this case it is numeric)
Below is the implementation:
C++
#include <iostream>
usingnamespacestd;
intgetProduct(string str)
{
intproduct = 1;
// Traversing through the string
for(inti = 0; i < str.length(); i++) {
// Since ascii value of
// numbers starts from 48
// so we subtract it from sum
product = product * (str[i] - 48);
}
returnproduct;
}
// Driver Code
intmain()
{
string st = "4513";
cout << getProduct(st);
return0;
}
Java
importjava.io.*;
classGFG {
staticintgetProduct(String str)
{
intproduct = 1;
// Traversing through the string
for(inti = 0; i < str.length(); i++)
{
// Since ascii value of
// numbers starts from 48
// so we subtract it from sum
product *= str.charAt(i) - '0';
}
returnproduct;
}
// Driver Code
publicstaticvoidmain(String[] args)
{
String st = "4513";
System.out.println(getProduct(st));
}
}
//this code is contributed by shivanisinghss2110
Python3
# Python3 program to compute
# product of digits in the number.
# Function to get product of digits
defgetProduct(n):
product =1
# Converting integer to string
num =str(n)
# Traversing the string
fori innum:
product =product *int(i)
returnproduct
# Driver Code
n =4513
print(getProduct(n))
# This code is contributed by vikkycirus
C#
usingSystem;
usingSystem.Collections;
classGFG
{
staticintgetProduct(String str)
{
intproduct = 1;
// Traversing through the string
for(inti = 0; i < str.Length; i++)
{
// Since ascii value of
// numbers starts from 48
// so we subtract it from sum
product = product * (str[i] - 48);
}
returnproduct;
}
// Driver Code
publicstaticvoidMain(String[] args)
{
String st = "4513";
Console.Write(getProduct(st));
}
}
//This code is contributed by shivanisinghss2110
Javascript
<script>
functiongetProduct(str)
{
let product = 1;
// Traversing through the string
for(let i = 0; i < str.length; i++)
{
// Since ascii value of
// numbers starts from 48
// so we subtract it from sum
product = product * (parseInt(str[i]));
}
returnproduct;
}
// Driver Code
let st = "4513";
document.write(getProduct(st));
// This code is contributed by unknown2108
</script>
Output
60
Time Complexity: O(N) Auxiliary Space: O(1)
Method #3: Recursion
Get the number
Get the remainder and pass the next remaining digits
Get the rightmost digit of the number with help of the remainder ‘%’ operator by dividing it by 10 and multiply it to the product.
Divide the number by 10 with help of ‘/’ operator to remove the rightmost digit
Check the base case with n = 0
Print or return the product
C++
//Recursive function to get product of the digits
#include <iostream>
usingnamespacestd;
intgetProduct(intn){
// Base Case
if(n == 0){
return1 ;
}
// get the last digit and multiply it with remaining digits
return(n%10) * getProduct(n/10) ;
}
intmain() {
// call the function
cout<<getProduct(125) ;
return0;
}
Java
// Recursive function to get product of the digits
importjava.util.*;
classGFG
{
staticintgetProduct(intn)
{
// Base Case
if(n == 0){
return1;
}
// get the last digit and multiply it with remaining digits
return(n%10) * getProduct(n/10) ;
}
publicstaticvoidmain(String[] args)
{
// call the function
System.out.println(getProduct(125));
}
}
// This code is contributed by phasing17
Python3
# Python3 program to implement the approach
# Recursive function to get product of the digits
defgetProduct(n):
# Base Case
if(n ==0):
return1
# get the last digit and multiply it with remaining digits
return(n%10) *getProduct(n//10) ;
# Driver Code
# call the function
print(getProduct(125));
# This code is contributed by phasing17
C#
// Recursive function to get product of the digits
usingSystem;
usingSystem.Collections.Generic;
classGFG
{
staticintgetProduct(intn)
{
// Base Case
if(n == 0){
return1 ;
}
// get the last digit and multiply it with remaining digits
return(n%10) * getProduct(n/10) ;
}
publicstaticvoidMain(string[] args)
{
// call the function
Console.WriteLine(getProduct(125));
}
}
// This code is contributed by phasing17
Javascript
// JS program to implement the approach
// Recursive function to get product of the digits
functiongetProduct(n){
// Base Case
if(n == 0){
return1 ;
}
// get the last digit and multiply it with remaining digits
return(n%10) * getProduct(Math.floor(n/10)) ;
}
// Driver Code
// call the function
console.log(getProduct(125));
// This code is contributed by phasing17
Output
10
Time Complexity: O(log10N) Auxiliary Space: O(1)
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, we can’t take that number as an integer since the range of long long int doesn’t satisfy the given number. So take input as a string, run a loop from start to the length of the string and increase the sum with that character(in this case it is numeric)
