Calculating N-digit numbers containing each possible digit at least once

Introduction

Often, we encounter problems where we need to calculate a number of specific types of numbers. One such problem is to calculate the number of N-digit numbers containing each possible digit at least once. This problem is important in various fields, including computer science, number theory, and combinatorics.

In this article, we will explore this problem and present different solutions and algorithms to solve it.

Problem Statement

Given an integer value N, we need to find the number of N-digit numbers that contain each possible digit at least once. For example, if N=3, then we need to find the number of 3-digit numbers such that each digit 0-9 appears at least once in the number.

Solution 1: Brute Force

One possible way to solve this problem is to use brute force. We can generate all possible N-digit numbers and check if they contain each possible digit at least once. However, this approach is not efficient and becomes increasingly slow as N increases.

def count_numbers_brute_force(N):
    count = 0
    for i in range(10**N):
        digits = set(str(i))
        if len(digits) == 10:
            count += 1
    return count

Solution 2: Inclusion-Exclusion Principle

Another approach to solve this problem is to use the inclusion-exclusion principle. We can first count the total number of N-digit numbers, which is simply 10^N (since we have 10 choices for each digit). Then, we can exclude the number of N-digit numbers that do not contain a specific digit. The total number of N-digit numbers not containing a specific digit is 9^N (since we have 9 choices for each digit if we exclude the specific digit). We can repeat this process for each digit and exclude the count of those numbers. However, this approach can lead to overcounting since we are excluding the numbers that do not contain a specific digit more than once. Thus, we need to add back the count of N-digit numbers that do not contain two specific digits and repeat the process for three, four, and all ten digits.

def count_numbers_inclusion_exclusion(N):
    count = 0
    for i in range(10):
        sign = (-1) ** i
        excluded = math.comb(10, i) * (9 ** (N-i))
        count += sign * excluded
    return 10 ** N + count

Solution 3: Dynamic Programming

Finally, we can solve this problem using dynamic programming. We can define a state dp[i][mask] as the number of i-digit numbers that contain all the digits in the binary representation of mask. The binary representation of mask indicates which digits are already present in a number.

def count_numbers_dynamic_programming(N):
    dp = [[0] * (1 << 10) for _ in range(N+1)]
    for i in range(10):
        dp[1][1<<i] = 1
    for i in range(2, N+1):
        for mask in range(1<<10):
            for j in range(10):
                if (mask & (1 << j)) == 0:
                    new_mask = mask | (1 << j)
                    dp[i][new_mask] += dp[i-1][mask]
    return dp[N][(1 << 10)-1]

Conclusion

In this article, we presented three different solutions to the problem of calculating the number of N-digit numbers containing each possible digit at least once. While brute force is not efficient, the inclusion-exclusion principle and dynamic programming provide more efficient algorithms to solve this problem.