Given an integer N, the task is to find the number of binary strings possible of length N having same frequency of 0s and 1s. If such string is possible of length N, print -1.
Note: Since the count can be very large, return the answer modulo 109+7.
Input: N = 2
All possible binary strings of length 2 are “00”, “01”, “10” and “11”.
Among them, “10” and “01” have the same frequency of 0s and 1s.
Hence, the answer is 2.
Strings “0011”, “0101”, “0110”, “1100”, “1010” and “1001” have same frequency of 0s and 1s.
Hence, the answer is 6.
The simplest approach is to generate all possible permutations of a string of length N having equal number of ‘0’ and ‘1’. For every permutation generated, increase the count. Print the total count of permutatiosn generated.
Time Complexity: O(N*N!)
Auxiliary Space: O(N)
The above approach can be optimized by using concepts of Permutation and Combination. Follow the steps below to solve the problem:
- Since N positions need to be filled with equal number of 0‘s and 1‘s, select N/2 positions from the N positions in C(N, N/2) % mod( where mod = 109 + 7) ways to fill with only 1’s.
- Fill the remaining positions in C(N/2, N/2) % mod (i.e 1) way with only 0’s.
Below is the implementation of the above approach:
Time Complexity: O(N2)
Space 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.
- 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
- Check whether two strings can be made equal by reversing substring of equal length from both strings
- Count number of binary strings of length N having only 0's and 1's
- Count of Binary strings of length N having atmost M consecutive 1s or 0s alternatively exactly K times
- Count number of binary strings such that there is no substring of length greater than or equal to 3 with all 1's
- Count the number of subsequences of length k having equal LCM and HCF
- Count ways to split a Binary String into three substrings having equal count of zeros
- Count of binary strings of length N with even set bit count and at most K consecutive 1s
- Count of Substrings having Sum equal to their Length
- Count of subarrays having sum equal to its length
- Split an Array A into Subsets having equal Sum and sizes equal to elements of Array B
- Count of same length Strings that exists lexicographically in between two given Strings
- Count of strings that become equal to one of the two strings after one removal
- Print all Strings from array A having all strings from array B as subsequence
- Count ways to generate pairs having Bitwise XOR and Bitwise AND equal to X and Y respectively
- Check if all substrings of length K of a Binary String has equal count of 0s and 1s
- Number of ways to select equal sized subarrays from two arrays having atleast K equal pairs of elements
- Longest palindrome formed by concatenating and reordering strings of equal length
- Count of binary strings of given length consisting of at least one 1
- Count of Binary Strings of length N such that frequency of 1's exceeds frequency of 0's
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to email@example.com. 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.