696. Count Binary Substrings
Description
Given a binary string s
, return the number of non-empty substrings that have the same number of 0
's and 1
's, and all the 0
's and all the 1
's in these substrings are grouped consecutively.
Substrings that occur multiple times are counted the number of times they occur.
Example 1:
Input: s = "00110011" Output: 6 Explanation: There are 6 substrings that have equal number of consecutive 1's and 0's: "0011", "01", "1100", "10", "0011", and "01". Notice that some of these substrings repeat and are counted the number of times they occur. Also, "00110011" is not a valid substring because all the 0's (and 1's) are not grouped together.
Example 2:
Input: s = "10101" Output: 4 Explanation: There are 4 substrings: "10", "01", "10", "01" that have equal number of consecutive 1's and 0's.
Constraints:
1 <= s.length <= 105
s[i]
is either'0'
or'1'
.
Solution
count-binary-substrings.py
class Solution:
def countBinarySubstrings(self, s: str) -> int:
chunks, consecutive, res, n = [], 1, 0, len(s)
for i in range(1, n):
if s[i] == s[i - 1]:
consecutive += 1
else:
chunks.append(consecutive)
consecutive = 1
chunks.append(consecutive)
for i in range(1, len(chunks)):
res += min(chunks[i], chunks[i - 1])
return res