# Find the number of binary strings of length N with at least 3 consecutive 1s

Given an integer N. The task is to find the number of all possible distinct binary strings of length N which have at least 3 consecutive 1s.
Examples:

Input: N = 3
Output:
The only string of length 3 possible is “111”.
Input: N = 4
Output:
The 3 strings are “1110”, “0111” and “1111”.

Naive approach: Consider all possible strings.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;`   `// Function that returns true if s contains` `// three consecutive 1's` `bool` `check(string& s)` `{` `    ``int` `n = s.length();` `    ``for` `(``int` `i = 2; i < n; i++) {` `        ``if` `(s[i] == ``'1'` `&& s[i - 1] == ``'1'` `&& s[i - 2] == ``'1'``)` `            ``return` `1;` `    ``}` `    ``return` `0;` `}`   `// Function to return the count` `// of required strings` `int` `countStr(``int` `i, string& s)` `{` `    ``if` `(i < 0) {` `        ``if` `(check(s))` `            ``return` `1;` `        ``return` `0;` `    ``}` `    ``s[i] = ``'0'``;` `    ``int` `ans = countStr(i - 1, s);` `    ``s[i] = ``'1'``;` `    ``ans += countStr(i - 1, s);` `    ``s[i] = ``'0'``;` `    ``return` `ans;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `N = 4;` `    ``string s(N, ``'0'``);` `    ``cout << countStr(N - 1, s);`   `    ``return` `0;` `}`

## Java

 `// Java implementation of the approach` `import` `java.util.*;`   `class` `GFG` `{`   `// Function that returns true if s contains` `// three consecutive 1's` `static` `boolean` `check(String s)` `{` `    ``int` `n = s.length();` `    ``for` `(``int` `i = ``2``; i < n; i++) ` `    ``{` `        ``if` `(s.charAt(i) == ``'1'` `&& ` `          ``s.charAt(i-``1``) == ``'1'` `&& ` `          ``s.charAt(i-``2``) == ``'1'``)` `            ``return` `true``;` `    ``}` `    ``return` `false``;` `}`   `// Function to return the count` `// of required strings` `static` `int` `countStr(``int` `i, String s)` `{` `    ``if` `(i < ``0``)` `    ``{` `        ``if` `(check(s))` `            ``return` `1``;` `        ``return` `0``;` `    ``}` `    ``char``[] myNameChars = s.toCharArray();` `    ``myNameChars[i] = ``'0'``;` `    ``s = String.valueOf(myNameChars);`   `    ``int` `ans = countStr(i - ``1``, s);` `    ``char``[] myChar = s.toCharArray();` `    ``myChar[i] = ``'1'``;` `    ``s = String.valueOf(myChar);`   `    ``ans += countStr(i - ``1``, s);` `    ``char``[]myChar1 = s.toCharArray();` `    ``myChar1[i] = ``'0'``;` `    ``s = String.valueOf(myChar1);`   `    ``return` `ans;` `}`   `// Driver code` `public` `static` `void` `main(String args[])` `{` `    ``int` `N = ``4``;` `    ``String s = ``"0000"``;` `    ``System.out.println(countStr(N - ``1``, s));` `}` `}`   `// This code is contributed by` `// Surendra_Gangwar`

## Python3

 `# Python3 implementation of the approach`   `# Function that returns true if s contains` `# three consecutive 1's` `def` `check(s) :` `    ``n ``=` `len``(s);` `    ``for` `i ``in` `range``(``2``, n) :` `        ``if` `(s[i] ``=``=` `'1'` `and` `s[i ``-` `1``] ``=``=` `'1'` `and` `                            ``s[i ``-` `2``] ``=``=` `'1'``) :` `            ``return` `1``;`   `# Function to return the count` `# of required strings` `def` `countStr(i, s) :` `    `  `    ``if` `(i < ``0``) :` `        ``if` `(check(s)) :` `            ``return` `1``;` `        ``return` `0``;` `    `  `    ``s[i] ``=` `'0'``;` `    ``ans ``=` `countStr(i ``-` `1``, s);` `    `  `    ``s[i] ``=` `'1'``;` `    ``ans ``+``=` `countStr(i ``-` `1``, s);` `    `  `    ``s[i] ``=` `'0'``;` `    ``return` `ans;`   `# Driver code` `if` `__name__ ``=``=` `"__main__"` `:` `    ``N ``=` `4``;` `    ``s ``=` `list``(``'0'` `*` `N);` `    `  `    ``print``(countStr(N ``-` `1``, s));`   `# This code is contributed by Ryuga`

## C#

 `// C# implementation of the approach` `// value x` `using` `System; `   `class` `GFG` `{`   `// Function that returns true if s contains` `// three consecutive 1's` `static` `bool` `check(String s)` `{` `    ``int` `n = s.Length;` `    ``for` `(``int` `i = 2; i < n; i++)` `    ``{` `        ``if` `(s[i] == ``'1'` `&& s[i - 1] == ``'1'` `&& s[i - 2] == ``'1'``)` `            ``return` `true``;` `    ``}` `    ``return` `false``;` `}`   `// Function to return the count` `// of required strings` `static` `int` `countStr(``int` `i, String s)` `{` `    ``if` `(i < 0)` `    ``{` `        ``if` `(check(s))` `            ``return` `1;` `        ``return` `0;` `    ``}` `    ``char``[] myNameChars = s.ToCharArray();` `    ``myNameChars[i] = ``'0'``;` `    ``s = String.Join(``""``, myNameChars);`   `    ``int` `ans = countStr(i - 1, s);` `    ``char``[] myChar = s.ToCharArray();` `    ``myChar[i] = ``'1'``;` `    ``s = String.Join(``""``, myChar);`   `    ``ans += countStr(i - 1, s);` `    ``char``[]myChar1 = s.ToCharArray();` `    ``myChar1[i] = ``'0'``;` `    ``s = String.Join(``""``, myChar1);`   `    ``return` `ans;` `}`   `// Driver code` `public` `static` `void` `Main(String []args)` `{` `    ``int` `N = 4;` `    ``String s = ``"0000"``;` `    ``Console.WriteLine(countStr(N - 1, s));` `}` `}`   `/* This code contributed by PrinciRaj1992 */`

## Javascript

 ``

Output:

`3`

Time Complexity: O(2N
Space Complexity: O(N) because of recurrence stack space.
Efficient approach: We use dynamic programming for computing the number of strings.
State of dp: dp(i, x) : denotes number of strings of length i with x consecutive 1s in position i + 1 to i + x.
Recurrence: dp(i, x) = dp(i – 1, 0) + dp(i – 1, x + 1)
The recurrence is based on the fact that either the string can have a ‘0’ at position i or a ‘1’.

1. If it has a ‘0’ at position i then for (i-1)th position value of x = 0.
2. If it has a ‘1’ at position i the for (i-1)th position value of x = value of x at position i + 1.

Base Condition: dp(i, 3) = 2i. Because once you have 3 consecutive ‘1’s you don’t care what characters are there at position 1, 2…i of string as all the strings are valid.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;`   `int` `n;`   `// Function to return the count` `// of required strings` `int` `solve(``int` `i, ``int` `x, ``int` `dp[][4])` `{` `    ``if` `(i < 0)` `        ``return` `x == 3;` `    ``if` `(dp[i][x] != -1)` `        ``return` `dp[i][x];`   `    ``// '0' at ith position` `    ``dp[i][x] = solve(i - 1, 0, dp);`   `    ``// '1' at ith position` `    ``dp[i][x] += solve(i - 1, x + 1, dp);` `    ``return` `dp[i][x];` `}`   `// Driver code` `int` `main()` `{` `    ``n = 4;` `    ``int` `dp[n][4];`   `    ``for` `(``int` `i = 0; i < n; i++)` `        ``for` `(``int` `j = 0; j < 4; j++)` `            ``dp[i][j] = -1;`   `    ``for` `(``int` `i = 0; i < n; i++) {`   `        ``// Base condition:` `        ``// 2^(i+1) because of 0 indexing` `        ``dp[i][3] = (1 << (i + 1));` `    ``}` `    ``cout << solve(n - 1, 0, dp);`   `    ``return` `0;` `}`

## Java

 `// Java implementation of the above approach ` `import` `java.util.*;`   `class` `GFG ` `{`   `    ``static` `int` `n;`   `    ``// Function to return the count` `    ``// of required strings` `    ``static` `int` `solve(``int` `i, ``int` `x, ``int` `dp[][]) ` `    ``{` `        ``if` `(i < ``0``) ` `        ``{` `            ``return` `x == ``3` `? ``1` `: ``0``;` `        ``}` `        ``if` `(dp[i][x] != -``1``) ` `        ``{` `            ``return` `dp[i][x];` `        ``}`   `        ``// '0' at ith position` `        ``dp[i][x] = solve(i - ``1``, ``0``, dp);`   `        ``// '1' at ith position` `        ``dp[i][x] += solve(i - ``1``, x + ``1``, dp);` `        ``return` `dp[i][x];` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String[] args) ` `    ``{` `        ``n = ``4``;` `        ``int` `dp[][] = ``new` `int``[n][``4``];`   `        ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``{` `            ``for` `(``int` `j = ``0``; j < ``4``; j++) ` `            ``{` `                ``dp[i][j] = -``1``;` `            ``}` `        ``}`   `        ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``{`   `            ``// Base condition:` `            ``// 2^(i+1) because of 0 indexing` `            ``dp[i][``3``] = (``1` `<< (i + ``1``));` `        ``}` `        ``System.out.print(solve(n - ``1``, ``0``, dp));` `    ``}` `}`   `// This code has been contributed by 29AjayKumar`

## Python 3

 `# Python 3 implementation of the approach`   `# Function to return the count` `# of required strings` `def` `solve(i, x, dp):` `    ``if` `(i < ``0``):` `        ``return` `x ``=``=` `3` `    ``if` `(dp[i][x] !``=` `-``1``):` `        ``return` `dp[i][x]`   `    ``# '0' at ith position` `    ``dp[i][x] ``=` `solve(i ``-` `1``, ``0``, dp)`   `    ``# '1' at ith position` `    ``dp[i][x] ``+``=` `solve(i ``-` `1``, x ``+` `1``, dp)` `    ``return` `dp[i][x]`     `# Driver code` `if` `__name__ ``=``=` `"__main__"``:`   `    ``n ``=` `4``;` `    ``dp ``=` `[[``0` `for` `i ``in` `range``(n)] ``for` `j ``in` `range``(``4``)]`   `    ``for` `i ``in` `range``(n):` `        ``for` `j ``in` `range``(``4``):` `            ``dp[i][j] ``=` `-``1`   `    ``for` `i ``in` `range``(n) :`   `        ``# Base condition:` `        ``# 2^(i+1) because of 0 indexing` `        ``dp[i][``3``] ``=` `(``1` `<< (i ``+` `1``))` `    `  `    ``print``(solve(n ``-` `1``, ``0``, dp))`   `# This code is contributed by ChitraNayal`

## C#

 `// C# implementation of the above approach ` `using` `System; `   `class` `GFG ` `{`   `    ``static` `int` `n;`   `    ``// Function to return the count` `    ``// of required strings` `    ``static` `int` `solve(``int` `i, ``int` `x, ``int` `[,]dp) ` `    ``{` `        ``if` `(i < 0) ` `        ``{` `            ``return` `x == 3 ? 1 : 0;` `        ``}` `        ``if` `(dp[i,x] != -1) ` `        ``{` `            ``return` `dp[i,x];` `        ``}`   `        ``// '0' at ith position` `        ``dp[i,x] = solve(i - 1, 0, dp);`   `        ``// '1' at ith position` `        ``dp[i,x] += solve(i - 1, x + 1, dp);` `        ``return` `dp[i,x];` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `Main(String[] args) ` `    ``{` `        ``n = 4;` `        ``int` `[,]dp = ``new` `int``[n, 4];`   `        ``for` `(``int` `i = 0; i < n; i++) ` `        ``{` `            ``for` `(``int` `j = 0; j < 4; j++) ` `            ``{` `                ``dp[i, j] = -1;` `            ``}` `        ``}`   `        ``for` `(``int` `i = 0; i < n; i++) ` `        ``{`   `            ``// Base condition:` `            ``// 2^(i+1) because of 0 indexing` `            ``dp[i,3] = (1 << (i + 1));` `        ``}` `        ``Console.Write(solve(n - 1, 0, dp));` `    ``}` `}`   `// This code contributed by Rajput-Ji`

## Javascript

 ``

Output:

`3`

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

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