Approach #1: Group By Character [Accepted]

Intuition

We can convert the string s into an array groups that represents the length of same-character contiguous blocks within the string. For example, if s = "110001111000000", then groups = [2, 3, 4, 6].

For every binary string of the form '0' * k + '1' * k or '1' * k + '0' * k, the middle of this string must occur between two groups.

Let's try to count the number of valid binary strings between groups[i] and groups[i+1]. If we have groups[i] = 2, groups[i+1] = 3, then it represents either "00111" or "11000". We clearly can make min(groups[i], groups[i+1]) valid binary strings within this string. Because the binary digits to the left or right of this string must change at the boundary, our answer can never be larger.

Algorithm

Let's create groups as defined above. The first element of s belongs in it's own group. From then on, each element either doesn't match the previous element, so that it starts a new group of size 1; or it does match, so that the size of the most recent group increases by 1.

Afterwards, we will take the sum of min(groups[i-1], groups[i]).

Python

class Solution(object):
    def countBinarySubstrings(self, s):
        groups = [1]
        for i in xrange(1, len(s)):
            if s[i-1] != s[i]:
                groups.append(1)
            else:
                groups[-1] += 1

        ans = 0
        for i in xrange(1, len(groups)):
            ans += min(groups[i-1], groups[i])
        return ans

Alternate Implentation

class Solution(object):
    def countBinarySubstrings(self, s):
        groups = [len(list(v)) for _, v in itertools.groupby(s)]
        return sum(min(a, b) for a, b in zip(groups, groups[1:]))

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;
    }
}

Complexity Analysis

• Time Complexity: $$O(N)$$ , where $$N$$ is the length of s. Every loop is through $$O(N)$$ items with $$O(1)$$ work inside the for-block.

• Space Complexity: $$O(N)$$ , the space used by groups.

Approach #2: Linear Scan [Accepted]

Intuition and Algorithm

We can amend our Approach #1 to calculate the answer on the fly. Instead of storing groups, we will remember only prev = groups[-2] and cur = groups[-1]. Then, the answer is the sum of min(prev, cur) over each different final (prev, cur) we see.

Python

class Solution(object):
    def countBinarySubstrings(self, s):
        ans, prev, cur = 0, 0, 1
        for i in xrange(1, len(s)):
            if s[i-1] != s[i]:
                ans += min(prev, cur)
                prev, cur = cur, 1
            else:
                cur += 1

        return ans + min(prev, cur)

Java

class Solution {
    public int countBinarySubstrings(String s) {
        int ans = 0, prev = 0, cur = 1;
        for (int i = 1; i < s.length(); i++) {
            if (s.charAt(i-1) != s.charAt(i)) {
                ans += Math.min(prev, cur);
                prev = cur;
                cur = 1;
            } else {
                cur++;
            }
        }
        return ans + Math.min(prev, cur);
    }
}

Complexity Analysis

• Time Complexity: $$O(N)$$ , where $$N$$ is the length of s. Every loop is through $$O(N)$$ items with $$O(1)$$ work inside the for-block.

• Space Complexity: $$O(1)$$ , the space used by prev, cur, and ans.

Analysis written by: @awice.