Count binary strings with k times appearing adjacent two set bits

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.
00111
01110
11100
11011
10111
11101

Input  : n = 4, k = 1
Output : 3
Explanation:
Binary strings of length 3 in which k number of times
two adjacent set bits appear.
0011
1100
0110

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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] where dp[i][j]
denotes number of binary strings with length i having
j number of two adjacent 1's and ending with 0.
Similarly dp[i][j] denotes the same binary strings
with length i and j adjacent 1's but ending with 1.
Then:
dp = 1 and dp = 1
For all other i and j,
dp[i][j] = dp[i-1][j] + dp[i-1][j]
dp[i][j] = dp[i-1][j] + dp[i-1][j-1]

Then, output dp[n][k] + dp[n][k]

C++

 // C++ program to count number of binary strings // with k times appearing consecutive 1's. #include using namespace std;    int countStrings(int n, int k) {     // dp[i][j] stores count of binary     // strings of length i with j consecutive     // 1's and ending at 0.     // dp[i][j] stores count of binary     // strings of length i with j consecutive     // 1's and ending at 1.     int dp[n + 1][k + 1];     memset(dp, 0, sizeof(dp));        // If n = 1 and k = 0.     dp = 1;     dp = 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] = dp[i - 1][j] + dp[i - 1][j];             dp[i][j] = dp[i - 1][j];                if (j - 1 >= 0)                 dp[i][j] += dp[i - 1][j - 1];         }     }        return dp[n][k] + dp[n][k]; }    // Driver code int main() {     int n = 5, k = 2;     cout << countStrings(n, k);     return 0; }

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] stores count of binary         // strings of length i with j consecutive         // 1's and ending at 0.         // dp[i][j] 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];            // If n = 1 and k = 0.         dp = 1;         dp = 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] = dp[i - 1][j] + dp[i - 1][j];                 dp[i][j] = dp[i - 1][j];                    if (j - 1 >= 0) {                     dp[i][j] += dp[i - 1][j - 1];                 }             }         }            return dp[n][k] + dp[n][k];     }        // 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

Python3

 # Python3 program to count number of  # binary strings with k times appearing  # consecutive 1's. def countStrings(n, k):        # dp[i][j] stores count of binary     # strings of length i with j consecutive     # 1's and ending at 0.     # dp[i][j] 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     dp = 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] = (dp[i - 1][j] +                             dp[i - 1][j])             dp[i][j] = dp[i - 1][j]             if j >= 1:                 dp[i][j] += dp[i - 1][j - 1]        return dp[n][k] + dp[n][k]    # Driver Code if __name__ == '__main__':     n = 5     k = 2     print(countStrings(n, k))    # This code is contributed by vibhu4agarwal

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] stores count of binary         // strings of length i with j consecutive         // 1's and ending at 0.         // dp[i][j] 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

PHP

 = 0 && isset(\$dp[\$i][\$j]))                 \$dp[\$i][\$j] += \$dp[\$i - 1][\$j - 1];         }     }        return \$dp[\$n][\$k] + \$dp[\$n][\$k]; }    // Driver code \$n=5; \$k=2; echo countStrings(\$n, \$k);        // This code is contributed by mits ?>

Output:

6

Time Complexity : O(n2)

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