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Suppose, we have one integer N > 0. The task is to find the maximum product of digits among numbers less than or equal to N. If the N is 390, then the result is 216, as the number 389 is making maximum product 3 * 8 * 9 = 216.

To solve this problem, we will use the recursive approach. So if N = 0, then return 1, if the number N < 10, then return N, otherwise return max(max_product(N/10) * (N mod 10), max_product((N/10) - 1)*9)

#include<iostream> using namespace std; int max_product(int N) { if (N == 0) return 1; if (N < 10) return N; return max(max_product(N / 10) * (N % 10), max_product(N / 10 - 1) * 9); } int main() { int N = 432; cout << "Maximum product is: " << max_product(N); }

Maximum product is: 243

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