C# Program to Count number of binary strings without consecutive 1âs
Write a C# program for a given positive integer N, the task is to count all possible distinct binary strings of length N such that there are no consecutive 1s.
Examples:
Input: N = 2
Output: 3
Explanation: The 3 strings are 00, 01, 10Input: N = 3
Output: 5
Explanation: The 5 strings are 000, 001, 010, 100, 101
C# Program to Count a number of binary strings without consecutive 1âs using Dynamic Programming:
Let a[i] be the number of binary strings of length i that do not contain any two consecutive 1s and which end in 0. Similarly, let b[i] be the number of such strings which end in 1. We can append either 0 or 1 to a string ending in 0, but we can only append 0 to a string ending in 1. This yields the recurrence relation:
a[i] = a[i â 1] + b[i â 1]
b[i] = a[i â 1]The base cases of above recurrence are a[1] = b[1] = 1. The total number of strings of length i is just a[i] + b[i].
Below is the implementation of the above approach:
C#
// C# program to count all distinct binary // strings without two consecutive 1's using System; class Subset_sum { static int countStrings( int n) { int []a = new int [n]; int []b = new int [n]; a[0] = b[0] = 1; for ( int i = 1; i < n; i++) { a[i] = a[i-1] + b[i-1]; b[i] = a[i-1]; } return (a[n-1] + b[n-1])%1000000007; } // Driver Code public static void Main () { Console.Write(countStrings(3)); } } // This code is contributed by nitin mittal |
5
Time Complexity: O(N)
Auxiliary Space: O(N)
Another method:
From the above method it is clear that we only want just previous value in the for loop which we can also do by replacing the array with the variable.
Below is the implementation of the above approach:
C#
// Include namespace system using System; public class Subset_sum { public static int countStrings( int n) { var a = 1; var b = 1; for ( int i = 1; i < n; i++) { // Here we have used the temp variable because // we want to assign the older value of a to b var temp = a + b; b = a; a = temp; } return (a + b)%1000000007; } // Driver program to test above function public static void Main(String[] args) { Console.WriteLine(Subset_sum.countStrings(3)); } } // This code is contributed by aadityaburujwale. |
5
Time Complexity: O(N)
Auxiliary Space: O(1)
Fibonacci in Log (n) without using Matrix Exponentiation:
As for calculating the n we can express it as the product of n/2 and n/2 ((n/2+1) for n is odd) as they are independent ,now as there are the cases when we combining these two half values to get full length there may be occurrence of consecutive 1âs at the joining. So we have to subtract those cases to get the final result.
f(n)=f(n/2)*f(n/2) â (those values which has consecutive 1âs at the joining part)
Below is the implementation of the above approach:
C#
using System; using System.Collections.Generic; class Solution { // Dictionary to store memoized values static Dictionary< long , long > mp = new Dictionary< long , long >(); // Recursive function to count strings public static long countStrings( long N) { // Base cases: if (N == 0) return 1; if (N == 1) return 2; if (N == 2) return 3; long mod = ( long )1e9 + 7; // Check if the value for N is already memoized long a = mp.ContainsKey(N / 2 - 1) ? mp[N / 2 - 1] : countStrings(N / 2 - 1); long b = mp.ContainsKey(N / 2) ? mp[N / 2] : countStrings(N / 2); long c = mp.ContainsKey(N / 2 + 1) ? mp[N / 2 + 1] : countStrings(N / 2 + 1); // Calculations based on whether N is even or odd if (N % 2 == 1) { // Memoize and return result for odd N return mp[N] = ((b * c % mod) - ((c - b) * (b - a) % mod) + mod) % mod; } else { // Memoize and return result for even N return mp[N] = ((b * b % mod) - ((b - a) * (b - a) % mod) + mod) % mod; } } } class Program { static void Main( string [] args) { long N = 10; Console.WriteLine(Solution.countStrings(N)); } } |
144
Time Complexity: O(log N)
Auxiliary Space: O(log N)
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