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Count N-length Binary Strings consisting of “11” as substring

  • Last Updated : 21 May, 2021

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:

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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.

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++




// 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




// 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




# 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#




// 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.

Javascript




<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>
Output
3719

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




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