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696.count-binary-substrings.cpp
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53 lines (50 loc) · 1.46 KB
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// Tag: Two Pointers, String
// Time: O(N)
// Space: O(1)
// Ref: -
// Note: -
// 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'.
//
//
class Solution {
public:
int countBinarySubstrings(string s) {
int n = s.size();
int pre = 0;
int cur = 1;
int res = 0;
for (int i = 1; i < n; i++) {
if (s[i] == s[i - 1]) {
cur++;
} else {
pre = cur;
cur = 1;
}
if (pre >= cur) {
res += 1;
}
}
return res;
}
};