Count Binary Substrings

Problem Description

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.

Examples

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

• $1 <= s.length <= 10^5$
• s[i] is either '0' or '1'.

Solution for Count Binary Substrings

Approach 1: Group By Character

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 its 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.

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

Code in Different Languages

Java

Written by @Shreyash3087

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;
    }
}

Python

Written by @Shreyash3087

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

Complexity Analysis

Time Complexity: $O(N)$

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

Space Complexity: $O(N)$

Reason: the space used by groups.

Approach 2: Linear Scan

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.

Code in Different Languages

Java

Written by @Shreyash3087

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);
    }
}

Python

Written by @Shreyash3087

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)

Complexity Analysis

Time Complexity: $O(N)$

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

Space Complexity: $O(1)$

Reason: the space used by prev, cur, and ans.

References

• LeetCode Problem: Count Binary Substrings

• Solution Link: Count Binary Substrings

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