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.
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
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]
Time Complexity : O(n2)
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