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 = 216Input: 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++ 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 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 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# 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 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 ?> |
<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> |
216
Time complexity: O(logN)
Auxiliary space: O(logN) for recursive stack space.