Given an integer N, the task is to count the number of binary strings possible such that there is no substring of length ≥ 3 of all 1’s. This count can become very large so print the answer modulo 109 + 7.
Examples:
Input: N = 4
Output: 13
All possible valid strings are 0000, 0001, 0010, 0100,
1000, 0101, 0011, 1010, 1001, 0110, 1100, 1101 and 1011.Input: N = 2
Output: 4
Approach: For every value from 1 to N, the only required strings are in which the number of substrings in which ‘1’ appears consecutively for just two times, one time or zero times. This can be calculated from 2 to N recursively. Dynamic programming can be used for memoization where dp[i][j] will store the number of possible strings such that 1 just appeared consecutively j times upto the ith index and j will be 0, 1, 2, …, i (may vary from 1 to N).
dp[i][0] = dp[i – 1][0] + dp[i – 1][1] + dp[i – 1][2] as in i position, 0 will be put.
dp[i][1] = dp[i – 1][0] as there is no 1 at the (i – 1)th position so we take that value.
dp[i][2] = dp[i – 1][1] as first 1 appeared at (i – 1)th position (consecutively) so we take that value directly.
The base cases are for length 1 string i.e. dp[1][0] = 1, dp[1][1] = 1, dp[1][2] = 0. So, find all the value dp[N][0] + dp[N][1] + dp[N][2] ans sum of all possible cases at the Nth position.
Below is the implementation of the above approach:
CPP
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; const long MOD = 1000000007; // Function to return the count of // all possible binary strings long countStr(long N) { long dp[N + 1][3]; // Fill 0's in the dp array memset(dp, 0, sizeof(dp)); // Base cases dp[1][0] = 1; dp[1][1] = 1; dp[1][2] = 0; for (int i = 2; i <= N; i++) { // dp[i][j] is the number of possible // strings such that '1' just appeared // consecutively j times upto ith index dp[i][0] = (dp[i - 1][0] + dp[i - 1][1] + dp[i - 1][2]) % MOD; // Taking previously calculated value dp[i][1] = dp[i - 1][0] % MOD; dp[i][2] = dp[i - 1][1] % MOD; } // Taking all the possible cases that // can appear at the Nth position long ans = (dp[N][0] + dp[N][1] + dp[N][2]) % MOD; return ans; } // Driver code int main() { long N = 8; cout << countStr(N); return 0; } |
Java
// Java implementation of the approach class GFG { final static long MOD = 1000000007; // Function to return the count of // all possible binary strings static long countStr(int N) { long dp[][] = new long[N + 1][3]; // Fill 0's in the dp array //memset(dp, 0, sizeof(dp)); // Base cases dp[1][0] = 1; dp[1][1] = 1; dp[1][2] = 0; for (int i = 2; i <= N; i++) { // dp[i][j] is the number of possible // strings such that '1' just appeared // consecutively j times upto ith index dp[i][0] = (dp[i - 1][0] + dp[i - 1][1] + dp[i - 1][2]) % MOD; // Taking previously calculated value dp[i][1] = dp[i - 1][0] % MOD; dp[i][2] = dp[i - 1][1] % MOD; } // Taking all the possible cases that // can appear at the Nth position long ans = (dp[N][0] + dp[N][1] + dp[N][2]) % MOD; return ans; } // Driver code public static void main (String[] args) { int N = 8; System.out.println(countStr(N)); } } // This code is contributed by AnkitRai01 |
Python
# Python3 implementation of the approach MOD = 1000000007 # Function to return the count of # all possible binary strings def countStr(N): dp = [[0 for i in range(3)] for i in range(N + 1)] # Base cases dp[1][0] = 1 dp[1][1] = 1 dp[1][2] = 0 for i in range(2, N + 1): # dp[i][j] is the number of possible # strings such that '1' just appeared # consecutively j times upto ith index dp[i][0] = (dp[i - 1][0] + dp[i - 1][1] + dp[i - 1][2]) % MOD # Taking previously calculated value dp[i][1] = dp[i - 1][0] % MOD dp[i][2] = dp[i - 1][1] % MOD # Taking all the possible cases that # can appear at the Nth position ans = (dp[N][0] + dp[N][1] + dp[N][2]) % MOD return ans # Driver code if __name__ == '__main__': N = 8 print(countStr(N)) # This code is contributed by mohit kumar 29 |
C#
// C# implementation of the approach using System; class GFG { static long MOD = 1000000007; // Function to return the count of // all possible binary strings static long countStr(int N) { long [,]dp = new long[N + 1, 3]; // Base cases dp[1, 0] = 1; dp[1, 1] = 1; dp[1, 2] = 0; for (int i = 2; i <= N; i++) { // dp[i,j] is the number of possible // strings such that '1' just appeared // consecutively j times upto ith index dp[i, 0] = (dp[i - 1, 0] + dp[i - 1, 1] + dp[i - 1, 2]) % MOD; // Taking previously calculated value dp[i, 1] = dp[i - 1, 0] % MOD; dp[i, 2] = dp[i - 1, 1] % MOD; } // Taking all the possible cases that // can appear at the Nth position long ans = (dp[N, 0] + dp[N, 1] + dp[N, 2]) % MOD; return ans; } // Driver code public static void Main () { int N = 8; Console.WriteLine(countStr(N)); } } // This code is contributed by AnkitRai01 |
149
Time Complexity: O(N)
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.
Recommended Posts:
- Count number of binary strings such that there is no substring of length greater than or equal to 3 with all 1's
- Check whether two strings can be made equal by reversing substring of equal length from both strings
- Construct a string of length L such that each substring of length X has exactly Y distinct letters
- Count of binary strings of length N having equal count of 0's and 1's and count of 1's ≥ count of 0's in each prefix substring
- Convert the undirected graph into directed graph such that there is no path of length greater than 1
- Length of the largest substring which have character with frequency greater than or equal to half of the substring
- Minimum length of substring whose rotation generates a palindromic substring
- Partition given string in such manner that i'th substring is sum of (i-1)'th and (i-2)'th substring
- Check if there exists a permutation of given string which doesn't contain any monotonous substring
- Generate all binary permutations such that there are more or equal 1's than 0's before every point in all permutations
- Count of Binary Strings of length N such that frequency of 1's exceeds frequency of 0's
- Maximum splits in binary string such that each substring is divisible by given odd number
- Longest Even Length Substring such that Sum of First and Second Half is same
- Minimum K such that every substring of length atleast K contains a character c
- Maximum length palindromic substring such that it starts and ends with given char
- Minimize count of flips required such that no substring of 0s have length exceeding K
- Python | Check if there are K consecutive 1's in a binary number
- Check if there is any common character in two given strings
- Smallest index such that there are no 0 or 1 to its right
- Count different numbers that can be generated such that there digits sum is equal to 'n'
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 Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.
