Given a binary string S of length N and an even integer K, the task is to check if all substrings of length K contains an equal number of 0s and 1s. If found to be true, print “Yes”. Otherwise, print “No”.
Input: S = “101010”, K = 2
Since all the substrings of length 2 has equal number of 0s and 1s, the answer is Yes.
Input: S = “101011”, K = 4
Since substring “1011” has unequal count of 0s and 1s, the answer is No..
Naive Approach: The simplest approach to solve the problem is to generate all substrings of length K and check it if it contains an equal count of 1s and 0s or not.
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: The main observation for optimizing the above approach is, for the string S to have an equal count of 0 and 1 in substrings of length K, S[i] must be equal to S[i + k]. Follow the steps below to solve the problem:
- Traverse the string and for every ith character, check if S[i] = S[j] where (i == (j % k))
- If found to be false at any instant, then print “No”.
- Otherwise, check the first substring of length K and if it contains an equal count of 1s and 0s or not. If found to be true, then print “Yes”. Otherwise, print “No”.
Below is the implementation of the above approach:
Time Complexity: O(N)
Auxiliary Space: O(1)
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