Count of N-digit numbers whose bitwise AND of adjacent digits equals 0

Last Updated : 22 Jul, 2021

Given a positive integer N, the task is to count the number of N-digit numbers such that the bitwise AND of adjacent digits equals 0.

Examples:

Input: N = 1
Output: 10
Explanation: All numbers from 0 to 9 satisfy the given condition as there is only one digit.

Input: N = 3
Output: 264

Naive Approach: The simplest approach to solve the given problem is to iterate over all possible N-digit numbers and count those numbers whose bitwise AND of adjacent digits is 0. After checking for all the numbers, print the value of count as the result.

Time Complexity: O(N × 10^N)
Auxiliary Space: O(1)

Efficient Approach: The above approach can also be optimized by using Dynamic Programming because the above problem has Overlapping subproblems and an Optimal substructure. The subproblems can be stored in dp[][] table using memoization where dp[digit][prev] stores the answer from the digit^th position till the end, when the previous digit selected is prev. Follow the steps below to solve the problem:

Define a recursive function, say countOfNumbers(digit, prev) by performing the following steps.
- If the value of digit is equal to N + 1 then return 1 as a valid N-digit number is formed.
- If the result of the state dp[digit][prev] is already computed, return this state dp[digit][prev].
- If the current digit is 1, then any digit from [1, 9] can be placed. If N = 1, then 0 can be placed as well.
- Otherwise, iterate through all the numbers from i = 0 to i = 9, and check if the condition ((i & prev) == 0) holds valid or not and accordingly place satisfying ‘i’ values in the current position.
- 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.
Print the value returned by the function countOfNumbers(1, 0, N) as the result.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
int dp[100][10];
 
// Function to calculate count of 'N' digit
// numbers such that bitwise AND of adjacent
// digits is 0.
int countOfNumbers(int digit, int prev, int n)
{
    // If digit = n + 1, a valid
    // n-digit number has been formed
    if (digit == n + 1) {
        return 1;
    }
 
    // If the state has
    // already been computed
    int& val = dp[digit][prev];
    if (val != -1) {
        return val;
    }
    val = 0;
 
    // If current position is 1,
    // then any digit from [1-9] can be placed.
    // If n = 1, 0 can be also placed.
    if (digit == 1) {
        for (int i = (n == 1 ? 0 : 1); i <= 9; ++i) {
            val += countOfNumbers(digit + 1, i, n);
        }
    }
 
    // For remaining positions,
    // any digit from [0-9] can be placed
    // after checking the conditions.
    else {
        for (int i = 0; i <= 9; ++i) {
 
            // Check if bitwise AND
            // of current digit and
            // previous digit is 0.
            if ((i & prev) == 0) {
                val += countOfNumbers(digit + 1, i, n);
            }
        }
    }
    // Return answer
    return val;
}
 
// Driver code
int main()
{
    // Initialize dp array with -1.
    memset(dp, -1, sizeof dp);
 
    // Given Input
    int N = 3;
 
    // Function call
    cout << countOfNumbers(1, 0, N) << endl;
}

Java

// Java program for the above approach
import java.util.*;
  
class GFG{
 
static int dp[][] = new int[100][10];
 
// Function to calculate count of 'N' digit
// numbers such that bitwise AND of adjacent
// digits is 0.
static int countOfNumbers(int digit, int prev, int n)
{
     
    // If digit = n + 1, a valid
    // n-digit number has been formed
    if (digit == n + 1) 
    {
        return 1;
    }
 
    // If the state has
    // already been computed
    int val = dp[digit][prev];
     
    if (val != -1) 
    {
        return val;
    }
    val = 0;
 
    // If current position is 1,
    // then any digit from [1-9] can be placed.
    // If n = 1, 0 can be also placed.
    if (digit == 1)
    {
        for(int i = (n == 1 ? 0 : 1); i <= 9; ++i)
        {
            val += countOfNumbers(digit + 1, i, n);
        }
    }
 
    // For remaining positions,
    // any digit from [0-9] can be placed
    // after checking the conditions.
    else
    {
        for(int i = 0; i <= 9; ++i) 
        {
             
            // Check if bitwise AND
            // of current digit and
            // previous digit is 0.
            if ((i & prev) == 0) 
            {
                val += countOfNumbers(digit + 1, i, n);
            }
        }
    }
     
    // Return answer
    return val;
}
 
// Driver code
public static void main(String[] args)
{
     
    // Initializing dp array with -1.
    for(int i = 0; i < 100; i++)
    {
        for(int j = 0; j < 10; j++)
        {
            dp[i][j] = -1;
        }
    }
 
    // Given Input
    int N = 3;
 
    // Function call
    System.out.println(countOfNumbers(1, 0, N));
}
}
 
// This code is contributed by sanjoy_62

Python3

# Python3 program for the above approach
dp = [[-1 for i in range(10)] 
          for j in range(100)]
 
val = 0
 
# Function to calculate count of 'N' digit
# numbers such that bitwise AND of adjacent
# digits is 0.
def countOfNumbers(digit, prev, n):
     
    global val
    global dp
     
    # If digit = n + 1, a valid
    # n-digit number has been formed
    if (digit == n + 1):
        return 1
 
    # If the state has
    # already been computed
    val = dp[digit][prev]
    if (val != -1):
        return val
         
    val = 0
 
    # If current position is 1,
    # then any digit from [1-9] can be placed.
    # If n = 1, 0 can be also placed.
    if (digit == 1):
        i = 0 if n == 1 else 1
         
        while (i <= 9):
            val += countOfNumbers(digit + 1, i, n)
            i += 1
 
    # For remaining positions,
    # any digit from [0-9] can be placed
    # after checking the conditions.
    else:
        for i in range(10):
             
            # Check if bitwise AND
            # of current digit and
            # previous digit is 0.
            if ((i & prev) == 0):
                val += countOfNumbers(digit + 1, i, n)
 
    # Return answer
    return val
 
# Driver code
if __name__ == '__main__':
     
    # Given Input
    N = 3
 
    # Function call
    print(countOfNumbers(1, 0, N))
 
# This code is contributed by SURENDRA_GANGWAR

C#

// C# program for the above approach
using System;
 
class GFG
{
 
  static int[,] dp = new int[100, 10];
 
  // Function to calculate count of 'N' digit
  // numbers such that bitwise AND of adjacent
  // digits is 0.
  static int countOfNumbers(int digit, int prev, int n)
  {
 
    // If digit = n + 1, a valid
    // n-digit number has been formed
    if (digit == n + 1) 
    {
      return 1;
    }
 
    // If the state has
    // already been computed
    int val = dp[digit, prev];
 
    if (val != -1) 
    {
      return val;
    }
    val = 0;
 
    // If current position is 1,
    // then any digit from [1-9] can be placed.
    // If n = 1, 0 can be also placed.
    if (digit == 1)
    {
      for(int i = (n == 1 ? 0 : 1); i <= 9; ++i)
      {
        val += countOfNumbers(digit + 1, i, n);
      }
    }
 
    // For remaining positions,
    // any digit from [0-9] can be placed
    // after checking the conditions.
    else
    {
      for(int i = 0; i <= 9; ++i) 
      {
 
        // Check if bitwise AND
        // of current digit and
        // previous digit is 0.
        if ((i & prev) == 0) 
        {
          val += countOfNumbers(digit + 1, i, n);
        }
      }
    }
 
    // Return answer
    return val;
  }
 
  // Driver code
  public static void Main(string[] args)
  {
 
    // Initializing dp array with -1.
    for(int i = 0; i < 100; i++)
    {
      for(int j = 0; j < 10; j++)
      {
        dp[i, j] = -1;
      }
    }
 
    // Given Input
    int N = 3;
 
    // Function call
    Console.WriteLine(countOfNumbers(1, 0, N));
  }
}
 
// This code is contributed by avijitmondal1998.

Javascript

<script>
        // JavaScript program for the above approach
 
        // Function to calculate count of 'N' digit
        // numbers such that bitwise AND of adjacent
        // digits is 0.
        function countOfNumbers(digit, prev, n)
        {
         
            // If digit = n + 1, a valid
            // n-digit number has been formed
            if (digit == n + 1) {
                return 1;
            }
 
            // If the state has
            // already been computed
            let val = dp[digit][prev];
            if (val != -1) {
                return val;
            }
            val = 0;
 
            // If current position is 1,
            // then any digit from [1-9] can be placed.
            // If n = 1, 0 can be also placed.
            if (digit == 1) {
                for (let i = (n == 1 ? 0 : 1); i <= 9; ++i) {
                    val += countOfNumbers(digit + 1, i, n);
                }
            }
 
            // For remaining positions,
            // any digit from [0-9] can be placed
            // after checking the conditions.
            else {
                for (let i = 0; i <= 9; ++i) {
 
                    // Check if bitwise AND
                    // of current digit and
                    // previous digit is 0.
                    if ((i & prev) == 0) {
                        val += countOfNumbers(digit + 1, i, n);
                    }
                }
            }
            // Return answer
            return val;
        }
 
        // Initialize dp array with -1.
        let dp = Array(100).fill().map(() =>
            Array(10).fill(-1));
             
        // Given Input
        let N = 3;
 
        // Function call
        document.write(countOfNumbers(1, 0, N) + "<br>");
 
    // This code is contributed by Potta Lokesh
    </script>