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++ 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<i; 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; ` `}` |

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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.

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