Given two integers n and k, count the number of binary strings of length n with k as number of times adjacent 1’s appear.

**Examples:**

Input : n = 5, k = 2 Output : 6 Explanation: Binary strings of length 5 in which k number of times two adjacent set bits appear. 001110111011100110111011111101 Input : n = 4, k = 1 Output : 3 Explanation: Binary strings of length 3 in which k number of times two adjacent set bits appear. 00111100 0110

Lets try writing the recursive function for the above problem statement:

1) n = 1, only two binary strings exist with length 1, not having any adjacent 1’s

String 1 : “0”

String 2 : “1”

2) For all n > 1 and all k, two cases arise

a) Strings ending with 0 : String of length n can be created by appending 0 to all **strings of length n-1 having k times two adjacent 1’s ending with both 0 and 1** (Having 0 at n’th position will not change the count of adjacent 1’s).

b) Strings ending with 1 : String of length n can be created by appending 1 to all strings of **length n-1 having k times adjacent 1’s and ending with 0** and to all strings of **length n-1 having k-1 adjacent 1’s and ending with 1**.

Example: let s = 011 i.e. a string ending with 1 having adjacent count as 1. Adding 1 to it, s = 0111 increase the count of adjacent 1.

Let there be an array dp[i][j][2] wheredp[i][j][0]denotes number of binary strings with length i having j number of two adjacent 1's and ending with 0. Similarlydp[i][j][1]denotes the same binary strings with length i and j adjacent 1's but ending with 1. Then: dp[1][0][0] = 1 and dp[1][0][1] = 1 For all other i and j, dp[i][j][0] = dp[i-1][j][0] + dp[i-1][j][1] dp[i][j][1] = dp[i-1][j][0] + dp[i-1][j-1][1] Then, output dp[n][k][0] + dp[n][k][1]

## C++

`// C++ program to count number of binary strings ` `// with k times appearing consecutive 1's. ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `int` `countStrings(` `int` `n, ` `int` `k) ` `{ ` ` ` `// dp[i][j][0] stores count of binary ` ` ` `// strings of length i with j consecutive ` ` ` `// 1's and ending at 0. ` ` ` `// dp[i][j][1] stores count of binary ` ` ` `// strings of length i with j consecutive ` ` ` `// 1's and ending at 1. ` ` ` `int` `dp[n + 1][k + 1][2]; ` ` ` `memset` `(dp, 0, ` `sizeof` `(dp)); ` ` ` ` ` `// If n = 1 and k = 0. ` ` ` `dp[1][0][0] = 1; ` ` ` `dp[1][0][1] = 1; ` ` ` ` ` `for` `(` `int` `i = 2; i <= n; i++) { ` ` ` ` ` `// number of adjacent 1's can not exceed i-1 ` ` ` `for` `(` `int` `j = 0; j <= k; j++) { ` ` ` `dp[i][j][0] = dp[i - 1][j][0] + dp[i - 1][j][1]; ` ` ` `dp[i][j][1] = dp[i - 1][j][0]; ` ` ` ` ` `if` `(j - 1 >= 0) ` ` ` `dp[i][j][1] += dp[i - 1][j - 1][1]; ` ` ` `} ` ` ` `} ` ` ` ` ` `return` `dp[n][k][0] + dp[n][k][1]; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `n = 5, k = 2; ` ` ` `cout << countStrings(n, k); ` ` ` `return` `0; ` `}` |

*chevron_right*

*filter_none*

## Java

`// Java program to count number of binary strings ` `// with k times appearing consecutive 1's. ` `class` `GFG { ` ` ` ` ` `static` `int` `countStrings(` `int` `n, ` `int` `k) ` ` ` `{ ` ` ` `// dp[i][j][0] stores count of binary ` ` ` `// strings of length i with j consecutive ` ` ` `// 1's and ending at 0. ` ` ` `// dp[i][j][1] stores count of binary ` ` ` `// strings of length i with j consecutive ` ` ` `// 1's and ending at 1. ` ` ` `int` `dp[][][] = ` `new` `int` `[n + ` `1` `][k + ` `1` `][` `2` `]; ` ` ` ` ` `// If n = 1 and k = 0. ` ` ` `dp[` `1` `][` `0` `][` `0` `] = ` `1` `; ` ` ` `dp[` `1` `][` `0` `][` `1` `] = ` `1` `; ` ` ` ` ` `for` `(` `int` `i = ` `2` `; i <= n; i++) { ` ` ` ` ` `// number of adjacent 1's can not exceed i-1 ` ` ` `for` `(` `int` `j = ` `0` `; j < i && j < k + ` `1` `; j++) { ` ` ` `dp[i][j][` `0` `] = dp[i - ` `1` `][j][` `0` `] + dp[i - ` `1` `][j][` `1` `]; ` ` ` `dp[i][j][` `1` `] = dp[i - ` `1` `][j][` `0` `]; ` ` ` ` ` `if` `(j - ` `1` `>= ` `0` `) { ` ` ` `dp[i][j][` `1` `] += dp[i - ` `1` `][j - ` `1` `][` `1` `]; ` ` ` `} ` ` ` `} ` ` ` `} ` ` ` ` ` `return` `dp[n][k][` `0` `] + dp[n][k][` `1` `]; ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `main(String[] args) ` ` ` `{ ` ` ` `int` `n = ` `5` `, k = ` `2` `; ` ` ` `System.out.println(countStrings(n, k)); ` ` ` `} ` `} ` ` ` `// This code has been contributed by 29AjayKumar ` |

*chevron_right*

*filter_none*

## Python3

`# Python3 program to count number of ` `# binary strings with k times appearing ` `# consecutive 1's. ` `def` `countStrings(n, k): ` ` ` ` ` `# dp[i][j][0] stores count of binary ` ` ` `# strings of length i with j consecutive ` ` ` `# 1's and ending at 0. ` ` ` `# dp[i][j][1] stores count of binary ` ` ` `# strings of length i with j consecutive ` ` ` `# 1's and ending at 1. ` ` ` `dp ` `=` `[[[` `0` `, ` `0` `] ` `for` `__ ` `in` `range` `(k ` `+` `1` `)] ` ` ` `for` `_ ` `in` `range` `(n ` `+` `1` `)] ` ` ` ` ` `# If n = 1 and k = 0. ` ` ` `dp[` `1` `][` `0` `][` `0` `] ` `=` `1` ` ` `dp[` `1` `][` `0` `][` `1` `] ` `=` `1` ` ` ` ` `for` `i ` `in` `range` `(` `2` `, n ` `+` `1` `): ` ` ` ` ` `# number of adjacent 1's can not exceed i-1 ` ` ` `for` `j ` `in` `range` `(k ` `+` `1` `): ` ` ` `dp[i][j][` `0` `] ` `=` `(dp[i ` `-` `1` `][j][` `0` `] ` `+` ` ` `dp[i ` `-` `1` `][j][` `1` `]) ` ` ` `dp[i][j][` `1` `] ` `=` `dp[i ` `-` `1` `][j][` `0` `] ` ` ` `if` `j >` `=` `1` `: ` ` ` `dp[i][j][` `1` `] ` `+` `=` `dp[i ` `-` `1` `][j ` `-` `1` `][` `1` `] ` ` ` ` ` `return` `dp[n][k][` `0` `] ` `+` `dp[n][k][` `1` `] ` ` ` `# Driver Code ` `if` `__name__ ` `=` `=` `'__main__'` `: ` ` ` `n ` `=` `5` ` ` `k ` `=` `2` ` ` `print` `(countStrings(n, k)) ` ` ` `# This code is contributed by vibhu4agarwal ` |

*chevron_right*

*filter_none*

## C#

`// C# program to count number of binary strings ` `// with k times appearing consecutive 1's. ` `using` `System; ` ` ` `class` `GFG { ` ` ` ` ` `static` `int` `countStrings(` `int` `n, ` `int` `k) ` ` ` `{ ` ` ` `// dp[i][j][0] stores count of binary ` ` ` `// strings of length i with j consecutive ` ` ` `// 1's and ending at 0. ` ` ` `// dp[i][j][1] stores count of binary ` ` ` `// strings of length i with j consecutive ` ` ` `// 1's and ending at 1. ` ` ` `int` `[,, ] dp = ` `new` `int` `[n + 1, k + 1, 2]; ` ` ` ` ` `// If n = 1 and k = 0. ` ` ` `dp[1, 0, 0] = 1; ` ` ` `dp[1, 0, 1] = 1; ` ` ` ` ` `for` `(` `int` `i = 2; i <= n; i++) { ` ` ` ` ` `// number of adjacent 1's can not exceed i-1 ` ` ` `for` `(` `int` `j = 0; j < i && j < k + 1; j++) { ` ` ` `dp[i, j, 0] = dp[i - 1, j, 0] + dp[i - 1, j, 1]; ` ` ` `dp[i, j, 1] = dp[i - 1, j, 0]; ` ` ` ` ` `if` `(j - 1 >= 0) { ` ` ` `dp[i, j, 1] += dp[i - 1, j - 1, 1]; ` ` ` `} ` ` ` `} ` ` ` `} ` ` ` ` ` `return` `dp[n, k, 0] + dp[n, k, 1]; ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `Main(String[] args) ` ` ` `{ ` ` ` `int` `n = 5, k = 2; ` ` ` `Console.WriteLine(countStrings(n, k)); ` ` ` `} ` `} ` ` ` `// This code contributed by Rajput-Ji ` |

*chevron_right*

*filter_none*

## PHP

`<?php ` `// PHP program to count number of binary strings ` `// with k times appearing consecutive 1's. ` ` ` `function` `countStrings(` `$n` `, ` `$k` `) ` `{ ` ` ` `// dp[i][j][0] stores count of binary ` ` ` `// strings of length i with j consecutive ` ` ` `// 1's and ending at 0. ` ` ` `// dp[i][j][1] stores count of binary ` ` ` `// strings of length i with j consecutive ` ` ` `// 1's and ending at 1. ` ` ` `$dp` `=` `array_fill` `(0, ` `$n` `+ 1, ` `array_fill` `(0, ` `$k` `+ 1, ` `array_fill` `(0, 2, 0))); ` ` ` ` ` `// If n = 1 and k = 0. ` ` ` `$dp` `[1][0][0] = 1; ` ` ` `$dp` `[1][0][1] = 1; ` ` ` ` ` `for` `(` `$i` `= 2; ` `$i` `<= ` `$n` `; ` `$i` `++) ` ` ` `{ ` ` ` `// number of adjacent 1's can not exceed i-1 ` ` ` `for` `(` `$j` `= 0; ` `$j` `< ` `$i` `; ` `$j` `++) ` ` ` `{ ` ` ` `if` `(isset(` `$dp` `[` `$i` `][` `$j` `][0])||isset(` `$dp` `[` `$i` `][` `$j` `][1])){ ` ` ` `$dp` `[` `$i` `][` `$j` `][0] = ` `$dp` `[` `$i` `- 1][` `$j` `][0] + ` `$dp` `[` `$i` `- 1][` `$j` `][1]; ` ` ` `$dp` `[` `$i` `][` `$j` `][1] = ` `$dp` `[` `$i` `- 1][` `$j` `][0]; ` ` ` `} ` ` ` `if` `(` `$j` `- 1 >= 0 && isset(` `$dp` `[` `$i` `][` `$j` `][1])) ` ` ` `$dp` `[` `$i` `][` `$j` `][1] += ` `$dp` `[` `$i` `- 1][` `$j` `- 1][1]; ` ` ` `} ` ` ` `} ` ` ` ` ` `return` `$dp` `[` `$n` `][` `$k` `][0] + ` `$dp` `[` `$n` `][` `$k` `][1]; ` `} ` ` ` `// Driver code ` `$n` `=5; ` `$k` `=2; ` `echo` `countStrings(` `$n` `, ` `$k` `); ` ` ` `// This code is contributed by mits ` `?> ` |

*chevron_right*

*filter_none*

**Output:**

6

**Time Complexity :** O(n^{2})

This article is contributed by **Ekta Goel**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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.