Given an integer N > 0, the task is to find the maximum product of digits among numbers less than or equal to N.
Input: N = 390
Maximum possible product is given by the number 389
3 * 8 * 9 = 216
Input: N = 432
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:
- Find the maximum sum of digits of the product of two numbers
- Find the number in a range having maximum product of the digits
- Find four factors of N with maximum product and sum equal to N
- Find four factors of N with maximum product and sum equal to N | Set-2
- Find four factors of N with maximum product and sum equal to N | Set 3
- Smallest number k such that the product of digits of k is equal to n
- Numbers with sum of digits equal to the sum of digits of its all prime factor
- Check if product of digits of a number at even and odd places is equal
- Sort the numbers according to their product of digits
- Number of digits in the product of two numbers
- Maximum of sum and product of digits until number is reduced to a single digit
- Count different numbers that can be generated such that there digits sum is equal to 'n'
- Find the first N integers such that the sum of their digits is equal to 10
- Find a Number X whose sum with its digits is equal to N
- Find a number x such that sum of x and its digits is equal to given n.
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