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# 01/30/2022 696

Use an array group to record each consecutive length of 0 or 1. Afterwards, we will take the sum of `min(groups[i-1], groups[i])`.

```java
class Solution {
    public int countBinarySubstrings(String s) {
        int[] groups = new int[s.length()];
        int t = 0;
        groups[0] = 1;
        for (int i = 1; i < s.length(); i++) {
            if (s.charAt(i-1) != s.charAt(i)) {
                groups[++t] = 1;
            } else {
                groups[t]++;
            }
        }

        int ans = 0;
        for (int i = 1; i <= t; i++) {
            ans += Math.min(groups[i-1], groups[i]);
        }
        return ans;
    }
}
```

Maintain the current character run length and previous character run length. If prevRunLength >= curRunLength, we have found a valid string.

Time O(N)

Space O(1)

```
public int countBinarySubstrings(String s) {
    int prevRunLength = 0, curRunLength = 1, res = 0;
    for (int i=1;i<s.length();i++) {
        if (s.charAt(i) == s.charAt(i-1)) curRunLength++;
        else {
            prevRunLength = curRunLength;
            curRunLength = 1;
        }
        if (prevRunLength >= curRunLength) res++;
    }
    return res;
}
```