Count N-length Binary Strings consisting of "11" as substring

Given a positive integer N, the task is to find the number of binary strings of length N which contains "11" as a substring.

Examples:

Input: N = 2
Output: 1
Explanation: The only string of length 2 that has "11" as a substring is "11".

Input: N = 12
Output: 3719

Approach: The idea is to derive the number of possibilities of having "11" as a substring for binary representations starting with 0 or 1 based on the following observations:

• If the first bit is 0, then the starting bit does not contribute to the string having "11" as a substring. Therefore, the remaining (N - 1) bits have to form a string having "11" as a substring.
• If the first bit is 1 and the following bit is also 1, then there exists 2^{(N - 2)} strings having "11" as a substring.
• If the first bit is 1 but the following bit is 0, then a string having "11" as a substring can be _formed with remaining (N - 2) bits.
• Therefore, the recurrence relation to generate all the binary strings of length N is:

dp[i] = dp[i - 1] + dp[i - 2] + 2^{(i - 2)}
where, 
dp[i] is the string of length i having "11" as a substring.
and dp[0] = dp[1] = 0.

Recursive Solution (Naive Solution):

here we have to find the number of possibilities having "11" as a substring for binary representation starting with 0 or 1 based on the observations. so above mentioned approach is for that.  but in a recursive solution if we want to find the solution for i then that should be calculated as follows,
binaryStrings(i) = binaryStrings(i-1) + binaryStrings(i-2) + 2^(i-2)

which can be solved recursively as follow.

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

// Function to count binary strings of length N having substring "11"
int binaryStrings(int N) {
    // Base Cases
    if (N == 0 || N == 1) {
        return 0;
    }

    // Recursively calculate answer for current state
    return binaryStrings(N - 1) + binaryStrings(N - 2) + (1 << (N - 2)); // 1<<(i-2) means power of 2^(i-2)
}

// Driver Code
int main()
{
    int N = 12;
    cout << binaryStrings(N);

    return 0;
}

import java.util.*;

public class Main 
{

  // Function to count binary strings of length N having
  // substring "11"
  static int BinaryStrings(int N)
  {
    // Base Cases
    if (N == 0 || N == 1) {
      return 0;
    }

    // Recursively calculate answer for current state
    return BinaryStrings(N - 1) + BinaryStrings(N - 2)
      + (1
         << (N
             - 2)); // 1<<(i-2) means power of 2^(i-2)
  }

  // Driver Code
  public static void main(String[] args)
  {
    int N = 12;
    System.out.println(BinaryStrings(N));
  }
}

# Function to count binary strings of length N having substring "11"
def binaryStrings(N: int) -> int:
    # Base Cases
    if N == 0 or N == 1:
        return 0

    # Recursively calculate answer for current state
    return binaryStrings(N - 1) + binaryStrings(N - 2) + (1 << (N - 2))  # 1<<(i-2) means power of 2^(i-2)


# Driver Code
if __name__ == "__main__":
    N = 12
    print(binaryStrings(N))

using System;

public class Program {
    // Function to count binary strings of length N having substring "11"
    static int BinaryStrings(int N) {
        // Base Cases
        if (N == 0 || N == 1) {
            return 0;
        }

        // Recursively calculate answer for current state
        return BinaryStrings(N - 1) + BinaryStrings(N - 2) + (1 << (N - 2)); // 1<<(i-2) means power of 2^(i-2)
    }

    // Driver Code
    public static void Main() {
        int N = 12;
        Console.WriteLine(BinaryStrings(N));
    }
}

// Function to count binary strings of length N having substring "11"
function binaryStrings(N) 
{
    // Base Cases
    if (N == 0 || N == 1)
    {
        return 0;
    } 
    
    // Recursively calculate answer for current state
    return binaryStrings(N - 1) + binaryStrings(N - 2) + (1 << (N - 2)); // 1<<(i-2) means power of 2^(i-2)
}

// Driver Code
let N = 12;
console.log(binaryStrings(N));

3719

Optimized Solution:

Follow the steps below to solve the problem:

• Initialize an array, say dp[], of size (N + 1) and assign dp[0] as 0 and dp[1] as 0.
• Precompute the first N powers of 2 and store it in an array, say power[].
• Iterate over the range [2, N] and update dp[i] as (dp[i - 1] + dp[i - 2] + power[i - 2]).
• After completing the above steps, print the value of dp[N] as the result.

Below is the implementation of the above approach:

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

// Function to count binary strings
// of length N having substring "11"
void binaryStrings(int N)
{
    // Initialize dp[] of size N + 1
    int dp[N + 1];

    // Base Cases
    dp[0] = 0;
    dp[1] = 0;



    // Iterate over the range [2, N]
    for (int i = 2; i <= N; i++) {
        dp[i] = dp[i - 1]
                + dp[i - 2]        
                + (1<<(i-2));   // 1<<(i-2) means power of 2^(i-2)
    }

    // Print total count of substrings
    cout << dp[N];
}

// Driver Code
int main()
{
    int N = 12;
    binaryStrings(N);

    return 0;
}

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

class GFG{

// Function to count binary strings
// of length N having substring "11"
static void binaryStrings(int N)
{
    
    // Initialize dp[] of size N + 1
    int[] dp = new int[N + 1];

    // Base Cases
    dp[0] = 0;
    dp[1] = 0;

    // Stores the first N powers of 2
    int[] power = new int[N + 1];
    power[0] = 1;

    // Generate
    for(int i = 1; i <= N; i++) 
    {
        power[i] = 2 * power[i - 1];
    }

    // Iterate over the range [2, N]
    for(int i = 2; i <= N; i++) 
    {
        dp[i] = dp[i - 1] + dp[i - 2] + power[i - 2];
    }
    
    // Print total count of substrings
    System.out.println(dp[N]);
}

// Driver Code
public static void main(String[] args)
{
    int N = 12;
    
    binaryStrings(N);
}
}

// This code is contributed by ukasp

# Python3 program for the above approach

# Function to count binary strings
# of length N having substring "11"
def binaryStrings(N):
    
    # Initialize dp[] of size N + 1
    dp = [0]*(N + 1)

    # Base Cases
    dp[0] = 0
    dp[1] = 0

    # Stores the first N powers of 2
    power = [0]*(N + 1)
    power[0] = 1

    # Generate
    for i in range(1, N + 1):
        power[i] = 2 * power[i - 1]

    # Iterate over the range [2, N]
    for i in range(2, N + 1):
        dp[i] = dp[i - 1] + dp[i - 2] + power[i - 2]

    # Prtotal count of substrings
    print (dp[N])

# Driver Code
if __name__ == '__main__':
    N = 12
    binaryStrings(N)

    # This code is contributed by mohit kumar 29.

// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{

  // Function to count binary strings
  // of length N having substring "11"
  static void binaryStrings(int N)
  {

    // Initialize dp[] of size N + 1
    int []dp = new int[N + 1];

    // Base Cases
    dp[0] = 0;
    dp[1] = 0;

    // Stores the first N powers of 2
    int []power = new int[N + 1];
    power[0] = 1;

    // Generate 
    for (int i = 1; i <= N; i++) {
      power[i] = 2 * power[i - 1];
    }

    // Iterate over the range [2, N]
    for (int i = 2; i <= N; i++) {
      dp[i] = dp[i - 1]
        + dp[i - 2]
        + power[i - 2];
    }

    // Print total count of substrings
    Console.WriteLine(dp[N]);
  }

  // Driver Code
  public static void Main()
  {
    int N = 12;
    binaryStrings(N);
  }
}

// This code is contributed by bgangwar59.

<script>

// JavaScript program for the above approach

    // Function to count binary strings
    // of length N having substring "11"
    function binaryStrings(N) {

        // Initialize dp of size N + 1
        var dp = Array(N + 1).fill(0);

        // Base Cases
        dp[0] = 0;
        dp[1] = 0;

        // Stores the first N powers of 2
        var power = Array(N+1).fill(0);
        power[0] = 1;

        // Generate
        for (i = 1; i <= N; i++) {
            power[i] = 2 * power[i - 1];
        }

        // Iterate over the range [2, N]
        for (i = 2; i <= N; i++) {
            dp[i] = dp[i - 1] + dp[i - 2] + power[i - 2];
        }

        // Print total count of substrings
        document.write(dp[N]);
    }

    // Driver Code
    
        var N = 12;

        binaryStrings(N);

// This code contributed by aashish1995

</script>

3719

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