Find maximum product of digits among numbers less than or equal to N

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Given an integer N > 0, the task is to find the maximum product of digits among numbers less than or equal to N.
Examples:

Input: N = 390
Output: 216
Maximum possible product is given by the number 389
3 * 8 * 9 = 216

Input: N = 432
Output: 243

Approach: This problem can also be solved using the method described in this article taking lower limit as 1 and upper limit as N. Another method to solve this problem is by using recursion. The conditions for recursion are as follows:

If N = 0 then return 1.
If N < 10 then return N.
Otherwise, return max(maxProd(N / 10) * (N % 10), maxProd((N / 10) – 1) * 9

At each step of recursion, either the last digit or 9 is taken to maximize the product of digit.
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function that returns the maximum product of
// digits among numbers less than or equal to N
int maxProd(int N)
{
    if (N == 0)
        return 1;
    if (N < 10)
        return N;
    return max(maxProd(N / 10) * (N % 10),
               maxProd(N / 10 - 1) * 9);
}
 
// Driver code
int main()
{
    int N = 390;
    cout << maxProd(N);
 
    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
 
class GFG
{
     
// Function that returns the maximum product of
// digits among numbers less than or equal to N
static int maxProd(int N)
{
    if (N == 0)
        return 1;
    if (N < 10)
        return N;
    return Math.max(maxProd(N / 10) * (N % 10),
            maxProd(N / 10 - 1) * 9);
}
 
// Driver code
public static void main (String[] args)
{
    int N = 390;
    System.out.println (maxProd(N));
}
}
 
// This code is contributed by ajit.

Python3

# Python3 implementation of the approach
 
# Function that returns the maximum product of
# digits among numbers less than or equal to N
def maxProd(N):
 
    if (N == 0):
        return 1
    if (N < 10):
        return N
    return max(maxProd(N // 10) * (N % 10),
               maxProd(N // 10 - 1) * 9)
 
# Driver code
N = 390
print(maxProd(N))
 
# This code is contributed by mohit kumar

C#

// C# implementation of the approach
using System;
 
class GFG
{
         
// Function that returns the maximum product of
// digits among numbers less than or equal to N
static int maxProd(int N)
{
    if (N == 0)
        return 1;
    if (N < 10)
        return N;
    return Math.Max(maxProd(N / 10) * (N % 10),
            maxProd(N / 10 - 1) * 9);
}
 
// Driver code
static public void Main ()
{
    int N = 390;
    Console.WriteLine(maxProd(N));
}
}
 
// This code is contributed by Tushil..

PHP

<?php
// PHP implementation of the approach
 
// Function that returns the maximum product of
// digits among numbers less than or equal to N
function maxProd($N)
{
    if ($N == 0)
        return 1;
    if ($N < 10)
        return $N;
    return max(maxProd((int)($N / 10)) * ($N % 10),
               maxProd((int)($N / 10) - 1) * 9);
}
 
// Driver code
$N = 390;
echo maxProd($N);
 
// This code is contributed by Akanksha Rai
?>

Javascript

<script>
 
// Javascript implementation of the approach
 
// Function that returns the maximum product of
// digits among numbers less than or equal to N
function maxProd(N)
{
    if (N == 0)
        return 1;
    if (N < 10)
        return N;
    return Math.max(maxProd(parseInt(N / 10)) * (N % 10),
            maxProd(parseInt(N / 10) - 1) * 9);
}
 
// Driver code
let N = 390;
document.write(maxProd(N));
 
// This code is contributed
// by bobby
 
</script>

Output:

Time complexity: O(logN)
Auxiliary space: O(logN) for recursive stack space.

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