Skip to content
Related Articles
Number of binary strings such that there is no substring of length ≥ 3
• Difficulty Level : Medium
• Last Updated : 25 May, 2021

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:

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] = dp[i – 1] + dp[i – 1] + dp[i – 1] as in i position, 0 will be put.
dp[i] = dp[i – 1] as there is no 1 at the (i – 1)th position so we take that value.
dp[i] = dp[i – 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, dp = 1, dp = 0. So, find all the value dp[N] + dp[N] + dp[N] 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];` `    ``// Fill 0's in the dp array``    ``memset``(dp, 0, ``sizeof``(dp));` `    ``// Base cases``    ``dp = 1;``    ``dp = 1;``    ``dp = 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] = (dp[i - 1] + dp[i - 1]``                    ``+ dp[i - 1])``                   ``% MOD;` `        ``// Taking previously calculated value``        ``dp[i] = dp[i - 1] % MOD;``        ``dp[i] = dp[i - 1] % MOD;``    ``}` `    ``// Taking all the possible cases that``    ``// can appear at the Nth position``    ``long` `ans = (dp[N] + dp[N] + dp[N]) % 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

 ``
Output:

`149`

Time Complexity: O(N)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer Geeks Classes Live

My Personal Notes arrow_drop_up