How do you solve Count Binary Substrings on LeetCode?

Count Binary Substrings (LeetCode #696) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Grouping consecutive characters helps in efficiently counting valid substrings.

What is the time complexity of Count Binary Substrings?

The optimal solution for Count Binary Substrings runs in O(n) time and uses O(n) space. This complexity is due to the need to store the lengths of groups of '0's and '1's, but we only traverse the string a couple of times.

What is the brute force solution for Count Binary Substrings?

The brute force approach for Count Binary Substrings is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible substring of the binary string to see if it has equal numbers of consecutive '0's and '1's. This method is straightforward but inefficient for larger strings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Binary Substrings?

Count Binary Substrings uses the following concepts: Two Pointers, String. The recommended patterns to study are: Two Pointers, Sliding Window, Array.

What are the interview tips for Count Binary Substrings?

Always think about how to reduce the problem size — grouping is key here. Be prepared to explain your thought process clearly, especially when transitioning from brute force to optimal. Practice identifying patterns in problems; many string problems can be solved using similar techniques.

What are common mistakes when solving Count Binary Substrings?

Not considering the grouping of characters, leading to incorrect substring checks. Overlooking the requirement for consecutive '0's and '1's.

#696

Count Binary Substrings

Easy

Two PointersStringTwo PointersSliding WindowArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution leverages the observation that valid substrings can be counted by grouping consecutive '0's and '1's. By counting the lengths of these groups, we can determine how many valid substrings can be formed.

⚙️

Algorithm

3 steps

1Step 1: Count the lengths of consecutive '0's and '1's and store them in an array.
2Step 2: Iterate through the array of counts and for each pair of consecutive counts, the number of valid substrings is the minimum of the two counts.
3Step 3: Sum these minimum counts to get the total number of valid substrings.

solution.py18 lines

1# Full working Python code
2
3def countBinarySubstrings(s):
4    groups = []
5    count = 1
6    for i in range(1, len(s)):
7        if s[i] == s[i - 1]:
8            count += 1
9        else:
10            groups.append(count)
11            count = 1
12    groups.append(count)  # Append the last group
13
14    result = 0
15    for i in range(1, len(groups)):
16        result += min(groups[i - 1], groups[i])
17    return result
18

ℹ

Complexity note: This complexity is due to the need to store the lengths of groups of '0's and '1's, but we only traverse the string a couple of times.

1Grouping consecutive characters helps in efficiently counting valid substrings.
2The minimum of consecutive counts determines the number of valid substrings that can be formed.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.