Count ways to generate Binary String not containing “0100” Substring

Given the number N, count the number of ways to create a binary string (the string that contains characters as zero or one) of size N such that it does not contain “0100” as a substring.  

A substring is a contiguous sequence of characters within a string.

Examples: 

Input: N = 4
Output: 15
Explanation: The answer will contain all possible substrings of size 4 except 0100 itself.

Input: N = 5
Output: 28

Naive approach: The article can be solved based on the following idea:

There are N positions to fill of binary string with either 1 or 0 while avoiding “0100”  substring, for any position i recursive function will be called by setting that position with either 1 or 0 except when last three characters are 010 only recursive function is called for 1 (to form 0101) not for 0 since if 0 is called then 0100 will be present as a substring in given count of strings. Now to keep track of last three characters make use of bitmask. 

Follow the steps below to solve the problem:

• Create a recursive function that takes two parameters one is the position that needs to fill and the other is the last three characters in form of a bitmask.
• Check the base cases. If the value of i is equal to N return 1.
• If the last three characters are not 010 Call the function recursively for 1 and  0, and sum up the values that are returned.
• else if the last three characters are 010 then call the function only for 1.
• return the value sum.

Below is the implementation of the above approach.

C++

// C++ code to implement the approach

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

// Returns number of strings that does
// not contain 0100 in its substring
int recur(int i, int lastThreeDigits, int N)
{
    // Base Case
    if (i == N)
        return 1;

    int ans = 0LL;

    if (i >= 3 and lastThreeDigits == 2) {

        // If last three substring is 010
        // then only one option we have to
        // choose for next character which
        // is 1 so lastThreeDigits will be
        // 5 which is 101
        ans += recur(i + 1, 5, N);
    }
    else {

        // Choosing 1 as a next character by
        // left shifting lastThreeDigits by
        // 1 then taking its bitwise AND with
        // 7 so that only first three bits
        // remain active and others become zero
        ans += recur(i + 1, lastThreeDigits << 1 & 7 | 1,
                     N);

        // Choosing 0 as a next character by
        // left shifting lastThreeDigits by
        // 1 then taking its bitwise AND with
        // 7 so that only first three bits
        // remain active and others become zero
        ans += recur(i + 1, lastThreeDigits << 1 & 7, N);
    }
    return ans;
}

// Function to count number of binary
// strings that does not contain 0100
// as substring
void countBinStrings(int N)
{
    cout << recur(0, 0, N) << endl;
}

// Driver Code
int main()
{
    int N = 4;

    // Function Call
    countBinStrings(N);

    int N1 = 5;

    // Function Call
    countBinStrings(N1);
    return 0;
}

Java

// Java code to implement the approach
import java.io.*;
import java.util.*;

class GFG {

  static int recur(int i, int lastThreeDigits, int N)
  {

    // Base Case
    if (i == N)
      return 1;

    int ans = 0;

    if (i >= 3 && lastThreeDigits == 2)
    {

      // If last three substring is 010
      // then only one option we have to
      // choose for next character which
      // is 1 so lastThreeDigits will be
      // 5 which is 101
      ans += recur(i + 1, 5, N);
    }
    else {
      // Choosing 1 as a next character by
      // left shifting lastThreeDigits by
      // 1 then taking its bitwise AND with
      // 7 so that only first three bits
      // remain active and others become zero
      ans += recur(i + 1,
                   lastThreeDigits << 1 & 7 | 1, N);

      // Choosing 0 as a next character by
      // left shifting lastThreeDigits by
      // 1 then taking its bitwise AND with
      // 7 so that only first three bits
      // remain active and others become zero
      ans += recur(i + 1, lastThreeDigits << 1 & 7,
                   N);
    }
    return ans;
  }

  // Function to count number of binary
  // strings that does not contain 0100
  // as substring
  static void countBinStrings(int N)
  {
    System.out.println(recur(0, 0, N));
  }

  // Driver Code
  public static void main(String[] args)
  {
    int N = 4;

    // Function Call
    countBinStrings(N);

    int N1 = 5;

    // Function Call
    countBinStrings(N1);
  }
}

// This code is contributed by lokesh.

Python3

#Python code for the above approach
def recur(i, lastThreeDigits, N):
    # Base Case
    if i == N:
        return 1

    ans = 0

    if i >= 3 and lastThreeDigits == 2:
        # If last three substring is 010
        # then only one option we have to
        # choose for next character which
        # is 1 so lastThreeDigits will be
        # 5 which is 101
        ans += recur(i + 1, 5, N)
    else:
        # Choosing 1 as a next character by
        # left shifting lastThreeDigits by
        # 1 then taking its bitwise AND with
        # 7 so that only first three bits
        # remain active and others become zero
        ans += recur(i + 1, (lastThreeDigits << 1) & 7 | 1, N)

        # Choosing 0 as a next character by
        # left shifting lastThreeDigits by
        # 1 then taking its bitwise AND with
        # 7 so that only first three bits
        # remain active and others become zero
        ans += recur(i + 1, (lastThreeDigits << 1) & 7, N)

    return ans

# Function to count number of binary
# strings that does not contain 0100
# as substring
def countBinStrings(N):
    print(recur(0, 0, N))

# Driver code
N = 4

# Function Call
countBinStrings(N)

N1 = 5

# Function Call
countBinStrings(N1)
#This code is contributed by Potta Lokesh

C#

// C# code to implement the approach

using System;
using System.Collections.Generic;

public class Gfg {

    // Returns number of strings that does
    // not contain 0100 in its substring
    static int recur(int i, int lastThreeDigits, int N)
    {
        // Base Case
        if (i == N)
            return 1;

        int ans = 0;

        if (i >= 3 && lastThreeDigits == 2) {

            // If last three substring is 010
            // then only one option we have to
            // choose for next character which
            // is 1 so lastThreeDigits will be
            // 5 which is 101
            ans += recur(i + 1, 5, N);
        }
        else {

            // Choosing 1 as a next character by
            // left shifting lastThreeDigits by
            // 1 then taking its bitwise AND with
            // 7 so that only first three bits
            // remain active and others become zero
            ans += recur(i + 1, lastThreeDigits << 1 & 7 | 1,
                         N);

            // Choosing 0 as a next character by
            // left shifting lastThreeDigits by
            // 1 then taking its bitwise AND with
            // 7 so that only first three bits
            // remain active and others become zero
            ans += recur(i + 1, lastThreeDigits << 1 & 7, N);
        }
        return ans;
    }

    // Function to count number of binary
    // strings that does not contain 0100
    // as substring
    static void countBinStrings(int N)
    {
        Console.WriteLine(recur(0, 0, N));
    }

    // Driver Code
    public static void Main(string[] args)
    {
        int N = 4;

        // Function Call
        countBinStrings(N);

        int N1 = 5;

        // Function Call
        countBinStrings(N1);
    }
}

Javascript

// Javascript code for the above approach

// Returns number of strings that does
// not contain 0100 in its substring
function recur(i, lastThreeDigits, N)
{

    // Base Case
    if (i == N)
        return 1;

    let ans = 0;

    if (i >= 3 && lastThreeDigits == 2) {

        // If last three substring is 010
        // then only one option we have to
        // choose for next character which
        // is 1 so lastThreeDigits will be
        // 5 which is 101
        ans += recur(i + 1, 5, N);
    }
    else {

        // Choosing 1 as a next character by
        // left shifting lastThreeDigits by
        // 1 then taking its bitwise AND with
        // 7 so that only first three bits
        // remain active and others become zero
        ans += recur(i + 1, lastThreeDigits << 1 & 7 | 1,
                     N);

        // Choosing 0 as a next character by
        // left shifting lastThreeDigits by
        // 1 then taking its bitwise AND with
        // 7 so that only first three bits
        // remain active and others become zero
        ans += recur(i + 1, lastThreeDigits << 1 & 7, N);
    }
    return ans;
}

// Function to count number of binary
// strings that does not contain 0100
// as substring
function countBinStrings(N)
{
    console.log(recur(0, 0, N));
}

// Driver Code
let N = 4;

    // Function Call
    countBinStrings(N);

    let N1 = 5;

    // Function Call
    countBinStrings(N1);

  // This code is contributed by poojaagarwal2.

Output

15
28

Time Complexity: O(2^N)
Auxiliary Space: O(1)

Efficient Approach:  The above approach can be optimized based on the following idea:

The idea is similar, but it can be observed that there are N * 8 states but the recursive function is called several times. That means that some states are called repeatedly. So the idea is to store the value of states. This can be done using recursive structure intact and just store the value in a HashMap and whenever the function is called, return the value store without computing .

Follow the steps below to solve the problem:

• Create a 2d array of dp[N + 1][8] initially filled with -1.
• If the answer for a particular state is computed then save it in dp[i][lastThreeDigits].
• If the answer for a particular state is already computed then just return dp[i][lastThreeDigits].

Below is the implementation of the above approach.

C++

// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;

int dp[100001][8];

// Returns number of strings that does
// not contain 0100 in its substring
int recur(int i, int lastThreeDigits, int N)
{

    // Base Case
    if (i == N)
        return 1;

    // If answer for current state is
    // already calculated then just
    // return the answer.
    if (dp[i][lastThreeDigits] != -1)
        return dp[i][lastThreeDigits];

    int ans = 0LL;

    if (i >= 3 and lastThreeDigits == 2) {

        // If last three substring is 010
        // then only one option we have to
        // choose for next character which
        // is 1 so lastThreeDigits will be
        // 5 which is 101
        ans += recur(i + 1, 5, N);
    }
    else {

        // Choosing 1 as a next character
        // by left shifting lastThreeDigits
        // by 1 then taking its bitwise
        // AND with 7 so that only first
        // three bits remain active and
        // others become zero
        ans += recur(i + 1, lastThreeDigits << 1 & 7 | 1,
                     N);

        // Choosing 0 as a next character
        // by left shifting lastThreeDigits
        // by 1 then taking its bitwise AND
        // with 7 so that only first three
        // bits remain active and others
        // become zero
        ans += recur(i + 1, lastThreeDigits << 1 & 7, N);
    }

    // Saving and returning sum
    return dp[i][lastThreeDigits] = ans;
}

// Function to count number of binary
// strings that does not contain 0100
// as substring
void countBinStrings(int N)
{

    // Initializing value of dp with -1
    memset(dp, -1, sizeof(dp));

    cout << recur(0, 0, N) << endl;
}

// Driver Code
int main()
{
    int N = 4;

    // Function Call
    countBinStrings(N);

    int N1 = 5;

    // Function Call
    countBinStrings(N1);
    return 0;
}

Java

// Java code to implement the approach
import java.io.*;

class GFG {

static int[][] dp=new int[100001][8];

// Returns number of strings that does
// not contain 0100 in its substring
static int recur(int i, int lastThreeDigits, int N)
{

    // Base Case
    if (i == N)
        return 1;

    // If answer for current state is
    // already calculated then just
    // return the answer.
    if (dp[i][lastThreeDigits] != -1)
        return dp[i][lastThreeDigits];

    int ans = 0;

    if (i >= 3 && lastThreeDigits == 2) {

        // If last three substring is 010
        // then only one option we have to
        // choose for next character which
        // is 1 so lastThreeDigits will be
        // 5 which is 101
        ans += recur(i + 1, 5, N);
    }
    else {

        // Choosing 1 as a next character
        // by left shifting lastThreeDigits
        // by 1 then taking its bitwise
        // AND with 7 so that only first
        // three bits remain active and
        // others become zero
        ans += recur(i + 1, lastThreeDigits << 1 & 7 | 1,
                    N);

        // Choosing 0 as a next character
        // by left shifting lastThreeDigits
        // by 1 then taking its bitwise AND
        // with 7 so that only first three
        // bits remain active and others
        // become zero
        ans += recur(i + 1, lastThreeDigits << 1 & 7, N);
    }

    // Saving and returning sum
    return dp[i][lastThreeDigits] = ans;
}

// Function to count number of binary
// strings that does not contain 0100
// as substring
static void countBinStrings(int N)
{

    // Initializing value of dp with -1
    for(int i=0;i<dp.length;i++)
    {
        for(int j=0;j<dp[0].length;j++)
        {
            dp[i][j]=-1;
        }
    }

    System.out.println(recur(0, 0, N));
}

// Driver Code
public static void main (String[] args)
{
    int N = 4;

    // Function Call
    countBinStrings(N);

    int N1 = 5;

    // Function Call
    countBinStrings(N1);
}

}

// This code is contributed by Aman Kumar

Python3

# Python code to implement the approach

# Returns number of strings that does
# not contain 0100 in its substring
def recur(i, lastThreeDigits, N, dp):

    # Base Case
    if i == N:
        return 1

    # If answer for current state is already
    # calculated then just return the answer.
    if dp[i][lastThreeDigits] != -1:
        return dp[i][lastThreeDigits]

    ans = 0

    if i >= 3 and lastThreeDigits == 2:

        # If last three substring is 010 then only one option
        # we have to choose for next character which is 1 so
        # lastThreeDigits will be 5 which is 101
        ans += recur(i + 1, 5, N, dp)
    else:

        # Choosing 1 as a next character by left shifting
        # lastThreeDigits by 1 then taking its bitwise AND
        # with 7 so that only first three bits remain active
        # and others become zero
        ans += recur(i + 1, (lastThreeDigits << 1) & 7 | 1,
                     N, dp)

        # Choosing 0 as a next character by left shifting
        # lastThreeDigits by 1 then taking its bitwise AND
        # with 7 so that only first three bits remain active
        # and others become zero
        ans += recur(i + 1, (lastThreeDigits << 1) & 7,
                     N, dp)

    # Saving and returning sum
    dp[i][lastThreeDigits] = ans
    return ans

# Function to count number of binary strings that does
# not contain 0100 as substring
def countBinStrings(N):
    # Initializing value of dp with -1
    dp = [[-1 for j in range(8)] for i in range(100001)]

    print(recur(0, 0, N, dp))


# Driver Code
if __name__ == "__main__":
    N = 4

    # Function Call
    countBinStrings(N)

    N1 = 5

    # Function Call
    countBinStrings(N1)

# This code is contributed by sankar.

C#

// C# code implementation for the above approach

using System;

public class GFG {

    static int[, ] dp = new int[100001, 8];

    // Returns number of strings that does not contain 0100
    // in its substring
    static int Recur(int i, int lastThreeDigits, int N)
    {

        // Base Case
        if (i == N)
            return 1;

        // If answer for current state is already calculated
        // then just return the answer.
        if (dp[i, lastThreeDigits] != -1)
            return dp[i, lastThreeDigits];

        int ans = 0;

        if (i >= 3 && lastThreeDigits == 2) {

            // If last three substring is 010 then only one
            // option we have to choose for next character
            // which is 1 so lastThreeDigits will be 5 which
            // is 101
            ans += Recur(i + 1, 5, N);
        }
        else {

            // Choosing 1 as a next character by left
            // shifting lastThreeDigits by 1 then taking its
            // bitwise AND with 7 so that only first three
            // bits remain active and others become zero
            ans += Recur(i + 1,
                         (lastThreeDigits << 1) & 7 | 1, N);

            // Choosing 0 as a next character by left
            // shifting lastThreeDigits by 1 then taking its
            // bitwise AND with 7 so that only first three
            // bits remain active and others become zero
            ans += Recur(i + 1, (lastThreeDigits << 1) & 7,
                         N);
        }

        // Saving and returning sum
        return dp[i, lastThreeDigits] = ans;
    }

    // Function to count number of binary strings that does
    // not contain 0100 as substring
    static void CountBinStrings(int N)
    {

        // Initializing value of dp with -1
        for (int i = 0; i < dp.GetLength(0); i++) {
            for (int j = 0; j < dp.GetLength(1); j++) {
                dp[i, j] = -1;
            }
        }

        Console.WriteLine(Recur(0, 0, N));
    }

    static public void Main()
    {

        // Code
        int N = 4;

        // Function Call
        CountBinStrings(N);

        int N1 = 5;

        // Function Call
        CountBinStrings(N1);
    }
}

// This code is contributed by karthik.

Javascript

// JS code to implement the approach

// dp table initialized with - 1
let dp=new Array(100001);
for(let i=0; i<100001; i++)
    dp[i]=new Array(8);

// Returns number of strings that does
// not contain 0100 in its substring
function recur( i,  lastThreeDigits,  N)
{

    // Base Case
    if (i == N)
        return 1;

    // If answer for current state is
    // already calculated then just
    // return the answer.
    if (dp[i][lastThreeDigits] != -1)
        return dp[i][lastThreeDigits];

    let ans = 0;

    if (i >= 3 && lastThreeDigits == 2) {

        // If last three substring is 010
        // then only one option we have to
        // choose for next character which
        // is 1 so lastThreeDigits will be
        // 5 which is 101
        ans += recur(i + 1, 5, N);
    }
    else {

        // Choosing 1 as a next character
        // by left shifting lastThreeDigits
        // by 1 then taking its bitwise
        // AND with 7 so that only first
        // three bits remain active and
        // others become zero
        ans += recur(i + 1, lastThreeDigits << 1 & 7 | 1,
                     N);

        // Choosing 0 as a next character
        // by left shifting lastThreeDigits
        // by 1 then taking its bitwise AND
        // with 7 so that only first three
        // bits remain active and others
        // become zero
        ans += recur(i + 1, lastThreeDigits << 1 & 7, N);
    }

    // Saving and returning sum
    return dp[i][lastThreeDigits] = ans;
}

// Function to count number of binary
// strings that does not contain 0100
// as substring
function countBinStrings( N)
{

    // Initializing value of dp with -1
    for(let i=0; i<100001; i++)
        for(let j=0; j<8; j++)
            dp[i][j]=-1;

    console.log(recur(0, 0, N)+"<br>");
}

// Driver Code
let N = 4;

// Function Call
countBinStrings(N);

let N1 = 5;

// Function Call
countBinStrings(N1);

Output

15
28

Time Complexity: O(N)
Auxiliary Space: O(N)