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++
// 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 codeint 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 approachimport 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 codepublic 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 approachdp = [[-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 codeif __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 approachusing 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> |
Output
264
Time Complexity: O(N × 102)
Auxiliary Space: O(N × 10)
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