Count N-digit numbers that contains every possible digit atleast once

Given a positive integer N, the task is to count the number of N-digit numbers such that every digit from [0-9] occurs at least once.  

 Examples : 

Input: N = 10
Output : 3265920

Input: N = 5
Output: 0
Explanation: Since the number of digits is less than the minimum number of digits required [0-9], the answer is 0.

Naive Approach: The simplest approach to solve the problem is to generate over all possible N-digit numbers and for each such number, check if all its digits satisfy the required condition or not.
Time Complexity: O(10^N*N)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is to use Dynamic Programming as it has overlapping subproblems and optimal substructure. The subproblems can be stored in dp[][] table using memoization, where dp[digit][mask] stores the answer from the digit^th position till the end, when the included digits are represented using a mask.

Follow the steps below to solve this problem:

• Define a recursive function, say countOfNumbers(digit, mask), and perform the following steps:
  • Base Case: If the value of digit is equal to N+1, then check if the value of the mask is equal to (1 << 10 – 1). If found to be true, return 1 as a valid N-digit number is formed.
  • If the result of the state dp[digit][mask] is already computed, return this state dp[digit][mask].
  • If the current position is 1, then any digit from [1-9] can be placed. If N is equal to 1, then 0 can be placed as well.
  • For any other position, any digit from [0-9] can be placed.
  • If a particular digit ‘i’ is included, then update mask as mask = mask | (1 << i ).
  • After making a valid placement, recursively call the countOfNumbers function for index digit+1.
  • Return the sum of all possible valid placements of digits as the answer.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Stores the dp-states
long long dp[100][1 << 10];

// Function to calculate the count of
// N-digit numbers that contains all
// digits from [0-9] atleast once
long long countOfNumbers(int digit,
                         int mask, int N)
{
    // If all digits are traversed
    if (digit == N + 1) {

        // Check if all digits are
        // included in the mask
        if (mask == (1 << 10) - 1)
            return 1;
        return 0;
    }

    long long& val = dp[digit][mask];

    // If the state has
    // already been computed
    if (val != -1)
        return val;

    val = 0;

    // If the current digit is 1, any
    // digit from [1-9] can be placed.
    // If N==1, 0 can also be placed
    if (digit == 1) {
        for (int i = (N == 1 ? 0 : 1); i <= 9; ++i) {

            val += countOfNumbers(digit + 1,
                                  mask | (1 << i), N);
        }
    }

    // For other positions, any digit
    // from [0-9] can be placed
    else {
        for (int i = 0; i <= 9; ++i) {

            val += countOfNumbers(digit + 1,
                                  mask | (1 << i), N);
        }
    }

    // Return the answer
    return val;
}

// Driver Code
int main()
{
    // Initializing dp array with -1.
    memset(dp, -1, sizeof dp);

    // Given Input
    int N = 10;

    // Function Call
    cout << countOfNumbers(1, 0, N);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

  // Stores the dp-states
  static int dp[][] = new int[100][1 << 10];

  // Function to calculate the count of
  // N-digit numbers that contains all
  // digits from [0-9] atleast once
  static long countOfNumbers(int digit,
                             int mask, int N)
  {
    // If all digits are traversed
    if (digit == N + 1) {

      // Check if all digits are
      // included in the mask
      if (mask == (1 << 10) - 1)
        return 1;

      return 0;
    }


    long val = dp[digit][mask];

    // If the state has
    // already been computed
    if (val != -1)
      return val;

    val = 0;

    // If the current digit is 1, any
    // digit from [1-9] can be placed.
    // If N==1, 0 can also be placed
    if (digit == 1) {
      for (int i = (N == 1 ? 0 : 1); i <= 9; ++i) {

        val += countOfNumbers(digit + 1,
                              mask | (1 << i), N);
      }
    }

    // For other positions, any digit
    // from [0-9] can be placed
    else {
      for (int i = 0; i <= 9; ++i) {

        val += countOfNumbers(digit + 1,
                              mask | (1 << i), N);
      }
    }

    // Return the answer
    return val;
  }

  // Driver Code
  public static void main(String[] args)
  {

    // Initializing dp array with -1.
    for(int i =0;i<dp.length;i++)
      Arrays.fill(dp[i], -1);

    // Given Input
    int N = 10;

    // Function Call
    System.out.print(countOfNumbers(1, 0, N));

  }
}

// This code is contributed by Rajput-Ji

Python3

# Python program for the above approach

# Stores the dp-states
dp = [[-1 for j in range(1 << 10)]for i in range(100)]

# Function to calculate the count of
# N-digit numbers that contains all
# digits from [0-9] atleast once
def countOfNumbers(digit, mask, N):

    # If all digits are traversed
    if (digit == N + 1):

        # Check if all digits are
        # included in the mask
        if (mask == (1 << 10) - 1):
            return 1
        return 0

    val = dp[digit][mask]

    # If the state has
    # already been computed
    if (val != -1):
        return val

    val = 0

    # If the current digit is 1, any
    # digit from [1-9] can be placed.
    # If N==1, 0 can also be placed
    if (digit == 1):
        for i in range((0 if (N == 1) else 1),10):
            val += countOfNumbers(digit + 1, mask | (1 << i), N)


    # For other positions, any digit
    # from [0-9] can be placed
    else:
        for i in range(10):
            val += countOfNumbers(digit + 1, mask | (1 << i), N)

    # Return the answer
    return val

# Driver Code

# Given Input
N = 10

# Function Call
print(countOfNumbers(1, 0, N))

# This code is contributed by shinjanpatra

C#

// C# program to implement above approach
using System;
using System.Collections.Generic;

class GFG
{

    // Stores the dp-states
    static long[][] dp = new long[100][];

    // Function to calculate the count of
    // N-digit numbers that contains all
    // digits from [0-9] atleast once
    static long countOfNumbers(int digit, int mask, int N)
    {
        // If all digits are traversed
        if (digit == N + 1) {

            // Check if all digits are
            // included in the mask
            if (mask == (1 << 10) - 1){
                return 1;
            }

            return 0;
        }


        long val = dp[digit][mask];

        // If the state has
        // already been computed
        if (val != -1){
            return val;
        }

        val = 0;

        // If the current digit is 1, any
        // digit from [1-9] can be placed.
        // If N==1, 0 can also be placed
        if (digit == 1) {
            for (int i = (N == 1 ? 0 : 1) ; i <= 9 ; ++i) {

                val += countOfNumbers(digit + 1, mask | (1 << i), N);
            }
        }
        // For other positions, any digit
        // from [0-9] can be placed
        else {
            for (int i = 0; i <= 9; ++i) {

                val += countOfNumbers(digit + 1, mask | (1 << i), N);
            }
        }

        dp[digit][mask] = val;
        // Return the answer
        return val;
    }


    // Driver Code
    public static void Main(string[] args){

        // Initializing dp array with -1.
        for(int i = 0 ; i < dp.Length ; i++){
            dp[i] = new long[1 << 10];
            for(int j = 0 ; j < (1 << 10) ; j++){
                dp[i][j] = -1;
            }
        }

        // Given Input
        int N = 10;

        // Function Call
        Console.Write(countOfNumbers(1, 0, N));

    }
}

// This code is contributed by subhamgoyal2014.

Javascript

<script>

// JavaScript program for the above approach

// Stores the dp-states
let dp = new Array(100);
for (let i = 0; i < 100; i++) {
    dp[i] = new Array(1 << 10).fill(-1);
}

// Function to calculate the count of
// N-digit numbers that contains all
// digits from [0-9] atleast once
function countOfNumbers(digit, mask, N)
{
    // If all digits are traversed
    if (digit == N + 1) {

        // Check if all digits are
        // included in the mask
        if (mask == (1 << 10) - 1)
            return 1;
        return 0;
    }

    let val = dp[digit][mask];

    // If the state has
    // already been computed
    if (val != -1)
        return val;

    val = 0;

    // If the current digit is 1, any
    // digit from [1-9] can be placed.
    // If N==1, 0 can also be placed
    if (digit == 1) {
        for (let i = (N == 1 ? 0 : 1); i <= 9; ++i) {

            val += countOfNumbers(digit + 1,
                                  mask | (1 << i), N);
        }
    }

    // For other positions, any digit
    // from [0-9] can be placed
    else {
        for (let i = 0; i <= 9; ++i) {

            val += countOfNumbers(digit + 1,
                                  mask | (1 << i), N);
        }
    }

    // Return the answer
    return val;
}

// Driver Code

    // Given Input
    let N = 10;

    // Function Call
    document.write(countOfNumbers(1, 0, N));

</script>

Output

3265920

Time Complexity: O(N²*2¹⁰)
Auxiliary Space: O(N*2¹⁰)

Efficient approach : Using DP Tabulation method ( Iterative approach )

The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memoization(top-down) because memoization method needs extra stack space of recursion calls.

Steps to solve this problem :

• Create a DP to store the solution of the subproblems and initialize it with 0.\
• Initialize the DP  with base cases dp[0][0] = 1.
• Now Iterate over subproblems to get the value of current problem form previous computation of subproblems stored in DP
• Create a variable ans to store final result and iterate over DP and update ans.
• Return the final solution stored in ans.

Implementation :

C++

#include <bits/stdc++.h>
using namespace std;

// Function to calculate the count of
// N-digit numbers that contains all
// digits from [0-9] atleast once
long long countOfNumbers(int N) {
    // Stores the dp-states
    long long dp[N + 1][1 << 10];

    // Initializing dp array with 0
    memset(dp, 0, sizeof(dp));

    // Setting initial state
    dp[0][0] = 1;

    // Computing DP table
    for (int digit = 1; digit <= N; ++digit) {
        // If the current digit is 1, any
        // digit from [1-9] can be placed.
        // If N==1, 0 can also be placed
        if (digit == 1) {
            for (int i = (N == 1 ? 0 : 1); i <= 9; ++i) {
                for (int mask = 0; mask < (1 << 10); ++mask) {
                    dp[digit][mask | (1 << i)] += dp[digit - 1][mask];
                }
            }
        }
        // For other positions, any digit
        // from [0-9] can be placed
        else {
            for (int i = 0; i <= 9; ++i) {
                for (int mask = 0; mask < (1 << 10); ++mask) {
                    dp[digit][mask | (1 << i)] += dp[digit - 1][mask];
                }
            }
        }
    }

    // Check if all digits are
    // included in the mask
    long long ans = 0;
    for (int mask = 0; mask < (1 << 10); ++mask) {
        if (mask == (1 << 10) - 1)
            ans += dp[N][mask];
    }

    // Return the answer
    return ans;
}

// Driver Code
int main()
{
    // Given Input
    int N = 10;

    // Function Call
    cout << countOfNumbers(N);

    return 0;
}
// --- by bhardwajji

Java

import java.util.Arrays;

public class GFG {

    // Function to calculate the count of N-digit numbers that contain all digits from [0-9] at least once
    static long countOfNumbers(int N) {
        // Stores the dp-states
        long[][] dp = new long[N + 1][1 << 10];

        // Initializing dp array with 0
        for (long[] row : dp) {
            Arrays.fill(row, 0);
        }

        // Setting initial state
        dp[0][0] = 1;

        // Computing DP table
        for (int digit = 1; digit <= N; ++digit) {
            // If the current digit is 1, any digit from [1-9] can be placed.
            // If N==1, 0 can also be placed
            if (digit == 1) {
                for (int i = (N == 1 ? 0 : 1); i <= 9; ++i) {
                    for (int mask = 0; mask < (1 << 10); ++mask) {
                        dp[digit][mask | (1 << i)] += dp[digit - 1][mask];
                    }
                }
            }
            // For other positions, any digit from [0-9] can be placed
            else {
                for (int i = 0; i <= 9; ++i) {
                    for (int mask = 0; mask < (1 << 10); ++mask) {
                        dp[digit][mask | (1 << i)] += dp[digit - 1][mask];
                    }
                }
            }
        }

        // Check if all digits are included in the mask
        long ans = 0;
        for (int mask = 0; mask < (1 << 10); ++mask) {
            if (mask == (1 << 10) - 1) {
                ans += dp[N][mask];
            }
        }

        // Return the answer
        return ans;
    }

    // Driver Code
    public static void main(String[] args) {
        // Given Input
        int N = 10;

        // Function Call
        System.out.println(countOfNumbers(N));

        // This Code is Contributed by Shivam Tiwari
    }
}

Python

# Function to calculate the count of
# N-digit numbers that contains all
# digits from [0-9] atleast once
def countOfNumbers(N):
    # Stores the dp-states
    dp = [[0]*(1 << 10) for i in range(N + 1)]

    # Initializing dp array with 0
    for i in range(N+1):
        for j in range(1 << 10):
            dp[i][j] = 0

    # Setting initial state
    dp[0][0] = 1

    # Computing DP table
    for digit in range(1, N+1):
        # If the current digit is 1, any
        # digit from [1-9] can be placed.
        # If N==1, 0 can also be placed
        if digit == 1:
            for i in range(0 if N==1 else 1, 10):
                for mask in range(1 << 10):
                    dp[digit][mask | (1 << i)] += dp[digit - 1][mask]
        # For other positions, any digit
        # from [0-9] can be placed
        else:
            for i in range(10):
                for mask in range(1 << 10):
                    dp[digit][mask | (1 << i)] += dp[digit - 1][mask]

    # Check if all digits are
    # included in the mask
    ans = 0
    for mask in range(1 << 10):
        if mask == (1 << 10) - 1:
            ans += dp[N][mask]

    # Return the answer
    return ans

# Driver Code
if __name__ == '__main__':
    # Given Input
    N = 10

    # Function Call
    print(countOfNumbers(N))

C#

using System;

public class GFG
{
    // Function to calculate the count of N-digit numbers
    // that contain all digits from [0-9] at least once
    public static long CountOfNumbers(int N)
    {
        // Stores the dp-states
        long[,] dp = new long[N + 1, 1 << 10];

        // Initializing dp array with 0
        for (int i = 0; i <= N; i++)
        {
            for (int j = 0; j < (1 << 10); j++)
            {
                dp[i, j] = 0;
            }
        }

        // Setting initial state
        dp[0, 0] = 1;

        // Computing DP table
        for (int digit = 1; digit <= N; digit++)
        {
            // If the current digit is 1, any
            // digit from [1-9] can be placed.
            // If N==1, 0 can also be placed
            if (digit == 1)
            {
                for (int i = (N == 1 ? 0 : 1); i <= 9; i++)
                {
                    for (int mask = 0; mask < (1 << 10); mask++)
                    {
                        dp[digit, mask | (1 << i)] += dp[digit - 1, mask];
                    }
                }
            }
            // For other positions, any digit
            // from [0-9] can be placed
            else
            {
                for (int i = 0; i <= 9; i++)
                {
                    for (int mask = 0; mask < (1 << 10); mask++)
                    {
                        dp[digit, mask | (1 << i)] += dp[digit - 1, mask];
                    }
                }
            }
        }

        // Check if all digits are
        // included in the mask
        long ans = 0;
        for (int mask = 0; mask < (1 << 10); mask++)
        {
            if (mask == (1 << 10) - 1)
                ans += dp[N, mask];
        }

        // Return the answer
        return ans;
    }

    // Driver Code
    public static void Main()
    {
        // Given Input
        int N = 10;

        // Function Call
        Console.WriteLine(CountOfNumbers(N));

        // This code is contributed by Shivam Tiwari
    }
}

Javascript

// Function to calculate the count of N-digit
// numbers that contain all digits from [0-9] at least once
function countOfNumbers(N) {
    // Stores the dp-states
    let dp = new Array(N + 1);
    for (let i = 0; i <= N; i++) {
        dp[i] = new Array(1 << 10).fill(0);
    }

    // Setting initial state
    dp[0][0] = 1;

    // Computing DP table
    for (let digit = 1; digit <= N; ++digit) {
        // If the current digit is 1, any digit from [1-9] can be placed.
        // If N==1, 0 can also be placed
        if (digit === 1) {
            for (let i = (N === 1 ? 0 : 1); i <= 9; ++i) {
                for (let mask = 0; mask < (1 << 10); ++mask) {
                    dp[digit][mask | (1 << i)] += dp[digit - 1][mask];
                }
            }
        }
        // For other positions, any digit from [0-9] can be placed
        else {
            for (let i = 0; i <= 9; ++i) {
                for (let mask = 0; mask < (1 << 10); ++mask) {
                    dp[digit][mask | (1 << i)] += dp[digit - 1][mask];
                }
            }
        }
    }

    // Check if all digits are included in the mask
    let ans = 0;
    for (let mask = 0; mask < (1 << 10); ++mask) {
        if (mask === (1 << 10) - 1) {
            ans += dp[N][mask];
        }
    }

    // Return the answer
    return ans;
}

// Driver Code

// Given Input
let N = 10;

// Function Call
console.log(countOfNumbers(N));

// This Code is Contributed by Shivam Tiwari

Output

3265920

Time Complexity: O(N * 2^10)
Auxiliary Space: O(N * 2^10)