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 × 10N)
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 digitth 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++
#include <bits/stdc++.h>
using namespace std;
int dp[100][10];
int countOfNumbers( int digit, int prev, int n)
{
if (digit == n + 1) {
return 1;
}
int & val = dp[digit][prev];
if (val != -1) {
return val;
}
val = 0;
if (digit == 1) {
for ( int i = (n == 1 ? 0 : 1); i <= 9; ++i) {
val += countOfNumbers(digit + 1, i, n);
}
}
else {
for ( int i = 0; i <= 9; ++i) {
if ((i & prev) == 0) {
val += countOfNumbers(digit + 1, i, n);
}
}
}
return val;
}
int main()
{
memset (dp, -1, sizeof dp);
int N = 3;
cout << countOfNumbers(1, 0, N) << endl;
}
|
Java
import java.util.*;
class GFG{
static int dp[][] = new int [ 100 ][ 10 ];
static int countOfNumbers( int digit, int prev, int n)
{
if (digit == n + 1 )
{
return 1 ;
}
int val = dp[digit][prev];
if (val != - 1 )
{
return val;
}
val = 0 ;
if (digit == 1 )
{
for ( int i = (n == 1 ? 0 : 1 ); i <= 9 ; ++i)
{
val += countOfNumbers(digit + 1 , i, n);
}
}
else
{
for ( int i = 0 ; i <= 9 ; ++i)
{
if ((i & prev) == 0 )
{
val += countOfNumbers(digit + 1 , i, n);
}
}
}
return val;
}
public static void main(String[] args)
{
for ( int i = 0 ; i < 100 ; i++)
{
for ( int j = 0 ; j < 10 ; j++)
{
dp[i][j] = - 1 ;
}
}
int N = 3 ;
System.out.println(countOfNumbers( 1 , 0 , N));
}
}
|
Python3
dp = [[ - 1 for i in range ( 10 )]
for j in range ( 100 )]
val = 0
def countOfNumbers(digit, prev, n):
global val
global dp
if (digit = = n + 1 ):
return 1
val = dp[digit][prev]
if (val ! = - 1 ):
return val
val = 0
if (digit = = 1 ):
i = 0 if n = = 1 else 1
while (i < = 9 ):
val + = countOfNumbers(digit + 1 , i, n)
i + = 1
else :
for i in range ( 10 ):
if ((i & prev) = = 0 ):
val + = countOfNumbers(digit + 1 , i, n)
return val
if __name__ = = '__main__' :
N = 3
print (countOfNumbers( 1 , 0 , N))
|
C#
using System;
class GFG
{
static int [,] dp = new int [100, 10];
static int countOfNumbers( int digit, int prev, int n)
{
if (digit == n + 1)
{
return 1;
}
int val = dp[digit, prev];
if (val != -1)
{
return val;
}
val = 0;
if (digit == 1)
{
for ( int i = (n == 1 ? 0 : 1); i <= 9; ++i)
{
val += countOfNumbers(digit + 1, i, n);
}
}
else
{
for ( int i = 0; i <= 9; ++i)
{
if ((i & prev) == 0)
{
val += countOfNumbers(digit + 1, i, n);
}
}
}
return val;
}
public static void Main( string [] args)
{
for ( int i = 0; i < 100; i++)
{
for ( int j = 0; j < 10; j++)
{
dp[i, j] = -1;
}
}
int N = 3;
Console.WriteLine(countOfNumbers(1, 0, N));
}
}
|
Javascript
<script>
function countOfNumbers(digit, prev, n)
{
if (digit == n + 1) {
return 1;
}
let val = dp[digit][prev];
if (val != -1) {
return val;
}
val = 0;
if (digit == 1) {
for (let i = (n == 1 ? 0 : 1); i <= 9; ++i) {
val += countOfNumbers(digit + 1, i, n);
}
}
else {
for (let i = 0; i <= 9; ++i) {
if ((i & prev) == 0) {
val += countOfNumbers(digit + 1, i, n);
}
}
}
return val;
}
let dp = Array(100).fill().map(() =>
Array(10).fill(-1));
let N = 3;
document.write(countOfNumbers(1, 0, N) + "<br>" );
</script>
|
Time Complexity: O(N × 102)
Auxiliary Space: O(N × 10)
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