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

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  • Difficulty Level : Medium
  • Last Updated : 15 Feb, 2022

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
Javascript



<script>
 
// Javascript implementation of the approach
 
var MOD = 1000000007;
 
// Function to return the count of
// all possible binary strings
function countStr(N)
{
 
    var dp = Array.from(Array(N+1), ()=> Array(3).fill(0));
 
    // Base cases
    dp[1][0] = 1;
    dp[1][1] = 1;
    dp[1][2] = 0;
 
    for (var 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
    var ans = (dp[N][0] + dp[N][1] + dp[N][2]) % MOD;
 
    return ans;
}
 
// Driver code
var N = 8;
document.write( countStr(N));
 
// This code is contributed by itsok.
</script>
Output: 
149
 
Time Complexity: O(N)
Auxiliary Space: O(N)
 

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souradeep
@souradeep
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