# Number of binary strings such that there is no substring of length ≥ 3

Given an integer N, the task is to count the number of binary strings possible such that there is no substring of length ≥ 3 of all 1’s. This count can become very large so print the answer modulo 109 + 7.

Examples:

Input: N = 4
Output: 13
All possible valid strings are 0000, 0001, 0010, 0100,
1000, 0101, 0011, 1010, 1001, 0110, 1100, 1101 and 1011.

Input: N = 2
Output: 4

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: For every value from 1 to N, the only required strings are in which the number of substrings in which ‘1’ appears consecutively for just two times, one time or zero times. This can be calculated from 2 to N recursively. Dynamic programming can be used for memoization where dp[i][j] will store the number of possible strings such that 1 just appeared consecutively j times upto the ith index and j will be 0, 1, 2, …, i (may vary from 1 to N).
dp[i][0] = dp[i – 1][0] + dp[i – 1][1] + dp[i – 1][2] as in i position, 0 will be put.
dp[i][1] = dp[i – 1][0] as there is no 1 at the (i – 1)th position so we take that value.
dp[i][2] = dp[i – 1][1] as first 1 appeared at (i – 1)th position (consecutively) so we take that value directly.
The base cases are for length 1 string i.e. dp[1][0] = 1, dp[1][1] = 1, dp[1][2] = 0. So, find all the value dp[N][0] + dp[N][1] + dp[N][2] ans sum of all possible cases at the Nth position.

Below is the implementation of the above approach:

## CPP

 // C++ implementation of the approach #include using namespace std;    const long MOD = 1000000007;    // Function to return the count of // all possible binary strings long countStr(long N) {        long dp[N + 1][3];        // Fill 0's in the dp array     memset(dp, 0, sizeof(dp));        // Base cases     dp[1][0] = 1;     dp[1][1] = 1;     dp[1][2] = 0;        for (int i = 2; i <= N; i++) {            // dp[i][j] is the number of possible         // strings such that '1' just appeared         // consecutively j times upto ith index         dp[i][0] = (dp[i - 1][0] + dp[i - 1][1]                     + dp[i - 1][2])                    % MOD;            // Taking previously calculated value         dp[i][1] = dp[i - 1][0] % MOD;         dp[i][2] = dp[i - 1][1] % MOD;     }        // Taking all the possible cases that     // can appear at the Nth position     long ans = (dp[N][0] + dp[N][1] + dp[N][2]) % MOD;        return ans; }    // Driver code int main() {     long N = 8;        cout << countStr(N);        return 0; }

## Java

 // Java implementation of the approach  class GFG  {            final static long MOD = 1000000007;             // Function to return the count of      // all possible binary strings      static long countStr(int N)      {          long dp[][] = new long[N + 1][3];                 // Fill 0's in the dp array          //memset(dp, 0, sizeof(dp));                 // Base cases          dp[1][0] = 1;          dp[1][1] = 1;          dp[1][2] = 0;                 for (int i = 2; i <= N; i++)          {                     // dp[i][j] is the number of possible              // strings such that '1' just appeared              // consecutively j times upto ith index              dp[i][0] = (dp[i - 1][0] + dp[i - 1][1]                          + dp[i - 1][2]) % MOD;                     // Taking previously calculated value              dp[i][1] = dp[i - 1][0] % MOD;              dp[i][2] = dp[i - 1][1] % MOD;          }                 // Taking all the possible cases that          // can appear at the Nth position          long ans = (dp[N][0] + dp[N][1] + dp[N][2]) % MOD;                 return ans;      }             // Driver code      public static void main (String[] args)     {          int N = 8;                 System.out.println(countStr(N));      }  }    // This code is contributed by AnkitRai01

## Python

 # Python3 implementation of the approach MOD = 1000000007    # Function to return the count of # all possible binary strings def countStr(N):        dp = [[0 for i in range(3)] for i in range(N + 1)]        # Base cases     dp[1][0] = 1     dp[1][1] = 1     dp[1][2] = 0        for i in range(2, N + 1):            # dp[i][j] is the number of possible         # strings such that '1' just appeared         # consecutively j times upto ith index         dp[i][0] = (dp[i - 1][0] + dp[i - 1][1] +                     dp[i - 1][2]) % MOD            # Taking previously calculated value         dp[i][1] = dp[i - 1][0] % MOD         dp[i][2] = dp[i - 1][1] % MOD        # Taking all the possible cases that     # can appear at the Nth position     ans = (dp[N][0] + dp[N][1] + dp[N][2]) % MOD        return ans    # Driver code if __name__ == '__main__':     N = 8        print(countStr(N))    # This code is contributed by mohit kumar 29

## C#

 // C# implementation of the approach  using System;    class GFG  {            static long MOD = 1000000007;             // Function to return the count of      // all possible binary strings      static long countStr(int N)      {          long [,]dp = new long[N + 1, 3];                // Base cases          dp[1, 0] = 1;          dp[1, 1] = 1;          dp[1, 2] = 0;                 for (int i = 2; i <= N; i++)          {                     // dp[i,j] is the number of possible              // strings such that '1' just appeared              // consecutively j times upto ith index              dp[i, 0] = (dp[i - 1, 0] + dp[i - 1, 1]                          + dp[i - 1, 2]) % MOD;                     // Taking previously calculated value              dp[i, 1] = dp[i - 1, 0] % MOD;              dp[i, 2] = dp[i - 1, 1] % MOD;          }                 // Taking all the possible cases that          // can appear at the Nth position          long ans = (dp[N, 0] + dp[N, 1] + dp[N, 2]) % MOD;                 return ans;      }             // Driver code      public static void Main ()     {          int N = 8;                 Console.WriteLine(countStr(N));      }  }    // This code is contributed by AnkitRai01

Output:

149

Time Complexity: O(N)

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Improved By : mohit kumar 29, AnkitRai01