Problem
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.
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.
Solution
public class Solution {
/**
* @param s: a string
* @return: the number of substrings
*/
public int countBinarySubstrings(String s) {
// Write your code here
int countZero = 0, countOne = 0, result = 0;
char[] str = s.toCharArray();
for (int i = 0; i < str.length; i++) {
if (i == 0) {
if (str[i] == '0') {
countZero++;
} else countOne++;
} else {
if (str[i] == '0') {
if (str[i-1] == '0') countZero++;
else countZero = 1;
if (countOne >= countZero) result++;
} else {
if (str[i-1] == '1') countOne++;
else countOne = 1;
if (countZero >= countOne) result++;
}
}
}
return result;
}
}
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