696. Count Binary Substrings
1. Description
Give a string s, count the number of non-empty (contiguous) 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.
2. Example
Example 1:
Input: “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: “10101”
Output: 4
Explanation: There are 4 substrings: “10”, “01”, “10”, “01” that have equal number of consecutive 1’s and 0’s.
3. Note
- s.length will be between 1 and 50,000.
- s will only consist of “0” or “1” characters.
4. Solutions
My Accepted Solution
n = i_str.size()
Time complexity: O(n)
Space complexity: O(1)
// we can record the length of string with same-character
// and the valid result occurs during two successive strings of '0' or '1'
// the valid substring could be '01', '0011', '000111', ...
// if the substring is '00011111', the valid substring is also '000111', that is the min length of two strings
class Solution
{
public:
// int countBinarySubstrings(string s)
int countBinarySubstrings(string &i_str)
{
if(i_str.empty()) return 0;
int result = 0, previousCount = 0, currentCount = 1;
for(int i = 1; i < i_str.size(); i++)
{
if(i_str[i] != i_str[i - 1])
{
result += min(previousCount, currentCount);
previousCount = currentCount;
currentCount = 1;
}
else
{
currentCount++;
}
}
return result + min(previousCount, currentCount);
}
};