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Number of binary strings such that there is no substring of length ≥ 3
  • Difficulty Level : Medium
  • Last Updated : 24 Jan, 2020

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

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 <bits/stdc++.h>
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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