Problem Description

Given a binary string consisting of only '0's and '1's, you need to count how many substrings satisfy two conditions:

1. The substring has an equal number of 0's and 1's
2. All 0's are grouped together consecutively, and all 1's are grouped together consecutively

For example, in the string "00110011":

• "0011" is valid (2 zeros followed by 2 ones)
• "01" is valid (1 zero followed by 1 one)
• "1100" is valid (2 ones followed by 2 zeros)
• "0110" is NOT valid (the digits alternate instead of being grouped)

The solution works by first grouping consecutive identical characters. For the string "00110011", this creates groups: [2, 2, 2, 2] representing 2 zeros, 2 ones, 2 zeros, 2 ones.

Then, for each pair of adjacent groups, the number of valid substrings that can be formed between them equals the minimum of the two group sizes. This is because:

• If we have 3 consecutive 0's followed by 2 consecutive 1's ("00011"), we can form:
  • "01" (using the last 0 and first 1)
  • "0011" (using the last 2 zeros and both 1's)
  • Total: min(3, 2) = 2 valid substrings

The algorithm counts all such valid substrings by summing up the minimum values between each pair of adjacent groups.

Intuition

The key insight is recognizing that valid substrings must have a specific pattern: consecutive 0's followed by consecutive 1's (or vice versa). This means we can't have alternating digits like "0101" as a valid substring.

When we look at any valid substring, it must span exactly two groups of consecutive identical characters. For instance, if we see "000111", the valid substrings are formed at the boundary between the 0's and 1's.

The critical observation is that for any two adjacent groups, the number of valid substrings we can form is limited by the smaller group. Why? Because to maintain equal counts of 0's and 1's, we need to match characters from both groups:

• With groups of size 3 and 2, we can only match at most 2 characters from each group
• We can take 1 from each group ("01"), or 2 from each group ("0011")

This leads us to transform the problem: instead of examining every possible substring, we can:

1. Compress the string into counts of consecutive characters
2. For each adjacent pair of counts, the number of valid substrings equals min(count1, count2)

For example, "00110011" becomes [2,2,2,2], and between each adjacent pair we get min(2,2) = 2 valid substrings. The total is 2+2+2 = 6.

This approach is efficient because it reduces the problem from checking all possible substrings to simply counting consecutive groups and comparing adjacent pairs.

Learn more about Two Pointers patterns.

Solution Approach

The implementation follows a two-phase approach:

Phase 1: Group Consecutive Characters

We iterate through the string and count consecutive identical characters:

i, n = 0, len(s)
t = []
while i < n:
    cnt = 1
    while i + 1 < n and s[i + 1] == s[i]:
        cnt += 1
        i += 1
    t.append(cnt)
    i += 1

This loop:

• Starts at each new character group
• Counts how many consecutive identical characters follow
• Stores each count in list t
• Moves to the next different character

For example, "00110011" produces t = [2, 2, 2, 2].

Phase 2: Calculate Valid Substrings

We then iterate through adjacent pairs in the groups list:

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

For each adjacent pair (t[i-1], t[i]):

• These represent two consecutive groups of different characters (0's and 1's)
• The number of valid substrings between them is min(t[i-1], t[i])
• We accumulate this count in ans

Time Complexity: O(n) where n is the length of the string - we traverse the string once to create groups and then traverse the groups once.

Space Complexity: O(n) in the worst case where characters alternate (creating n groups), but typically much less as consecutive characters are compressed into single counts.

Example Walkthrough

Let's walk through the solution with the string s = "00110011".

Step 1: Group Consecutive Characters

Starting from index 0:

• Position 0-1: We see "00" (2 consecutive 0's) → Add 2 to groups
• Position 2-3: We see "11" (2 consecutive 1's) → Add 2 to groups
• Position 4-5: We see "00" (2 consecutive 0's) → Add 2 to groups
• Position 6-7: We see "11" (2 consecutive 1's) → Add 2 to groups

Groups array: [2, 2, 2, 2]

Step 2: Count Valid Substrings Between Adjacent Groups

Now we examine each pair of adjacent groups:

• Groups[0] and Groups[1]: [2, 2] representing "00" and "11"

  • min(2, 2) = 2 valid substrings
  • These are: "01" (taking 1 from each) and "0011" (taking 2 from each)

• Groups[1] and Groups[2]: [2, 2] representing "11" and "00"

  • min(2, 2) = 2 valid substrings
  • These are: "10" (taking 1 from each) and "1100" (taking 2 from each)

• Groups[2] and Groups[3]: [2, 2] representing "00" and "11"

  • min(2, 2) = 2 valid substrings
  • These are: "01" (taking 1 from each) and "0011" (taking 2 from each)

Total count: 2 + 2 + 2 = 6 valid substrings

The complete list of valid substrings in "00110011":

1. "01" at position [1:3]
2. "0011" at position [0:4]
3. "10" at position [2:4]
4. "1100" at position [2:6]
5. "01" at position [5:7]
6. "0011" at position [4:8]

Each substring has equal 0's and 1's with all identical digits grouped consecutively.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n) where n is the length of the input string s.

The algorithm consists of two main phases:

1. The first while loop traverses the entire string once to count consecutive groups of 0s and 1s. Each character is visited exactly once, resulting in O(n) time.
2. The second for loop iterates through the array t which has at most n elements (in the worst case when characters alternate), taking O(n) time.

Therefore, the overall time complexity is O(n) + O(n) = O(n).

Space Complexity: O(n) in the worst case.

The additional space is used by the array t which stores the counts of consecutive groups. In the worst case scenario where the string alternates between 0 and 1 (e.g., "010101..."), we would have n groups, each of size 1. This would result in an array t of size n, giving us O(n) space complexity. In the best case where the string consists of all same characters, t would only have 1 element, but we consider worst-case for complexity analysis.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Attempting to Check All Substrings Directly

A common mistake is trying to validate every possible substring by checking if it has equal 0's and 1's and if they're grouped consecutively:

# Inefficient approach - DON'T DO THIS
def countBinarySubstrings(s):
    count = 0
    n = len(s)
    for i in range(n):
        for j in range(i+1, n+1):
            substring = s[i:j]
            if isValid(substring):  # Check equal 0s/1s and consecutive grouping
                count += 1
    return count

Why it's problematic: This has O(n³) time complexity - O(n²) for generating substrings and O(n) for validation.

Solution: Use the grouping approach that recognizes valid substrings only form between adjacent groups of different characters.

2. Misunderstanding Valid Substring Formation

Developers often think valid substrings can span multiple groups:

# Incorrect thinking: "001100" might seem valid (equal 0s and 1s)
# But it's actually formed from 3 groups: [2 zeros, 2 ones, 2 zeros]

Why it's wrong: Valid substrings must have exactly one transition point between 0's and 1's. They can only form between two adjacent groups.

Solution: Remember that valid substrings have the pattern "000...111..." or "111...000..." with exactly one transition.

3. Off-by-One Error in Group Counting

A subtle bug can occur when counting consecutive characters:

# Buggy version
while i < n:
    cnt = 1
    while i < n - 1 and s[i] == s[i + 1]:  # Wrong comparison!
        cnt += 1
        i += 1
    t.append(cnt)
    i += 1

Why it's problematic: Comparing s[i] with s[i+1] instead of s[i+1] with s[i] can lead to incorrect indexing.

Solution: Use the correct comparison s[i+1] == s[i] and ensure i+1 < n for bounds checking.

4. Space Optimization Oversight

While the provided solution is correct, it can be optimized to use O(1) space:

def countBinarySubstrings(self, s: str) -> int:
    prev_group = 0
    curr_group = 1
    result = 0
  
    for i in range(1, len(s)):
        if s[i] == s[i-1]:
            curr_group += 1
        else:
            result += min(prev_group, curr_group)
            prev_group = curr_group
            curr_group = 1
  
    result += min(prev_group, curr_group)  # Don't forget the last pair!
    return result

Why it's better: Instead of storing all groups, we only track the current and previous group sizes, reducing space complexity from O(n) to O(1).

5. Forgetting Edge Cases

Common edge cases that are missed:

• Empty string or single character string
• String with all same characters (e.g., "0000" or "1111")
• Alternating characters (e.g., "010101")

Solution: The grouping approach naturally handles these cases, but ensure your implementation doesn't assume at least two different groups exist.