What is the time complexity of Count Binary Substrings?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Count Binary Substrings?

This problem uses Two Pointers, String. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Count Binary Substrings?

Insight 1: The core idea is to identify consecutive groups of identical characters (either 0s or 1s). For example, in "0011100", the groups are [2, 3, 2] corresponding to "00", "111", and "00". Insight 2: Once the lengths of these consecutive groups are determined, the number of valid substrings formed at the boundary of two adjacent groups is the minimum of the lengths of the two groups. For example, if we have groups [2, 3], we can form `min(2, 3) = 2` valid substrings: "0011" and "01".

Which companies ask LeetCode 696. Count Binary Substrings in interviews?

This problem is frequently asked at ByteDance, IBM, J.P. Morgan, Morgan Stanley, Salesforce.

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Count Binary Substrings - LeetCode 696 Solution

Count Binary Substrings - Complete Solution Guide

Count Binary Substrings is LeetCode problem 696, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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. 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 ar

Detailed Explanation

The problem requires us to find the number of non-empty substrings in a binary string `s` that have an equal number of consecutive 0s and 1s. A crucial constraint is that the 0s and 1s must be grouped consecutively. For instance, "0011" and "10" are valid, but "0101" is not because the 0s and 1s are not grouped together. The output should be the total count of such valid substrings, counting duplicates as many times as they appear.

Solution Approach

The solution involves two main steps. First, we iterate through the string and identify the consecutive groups of identical characters (0s or 1s). We record the lengths of these groups. Second, we iterate through the list of group lengths and calculate the number of valid substrings at the boundaries of adjacent groups by taking the minimum of their lengths. We accumulate these counts to arrive at the final answer.

Step-by-Step Algorithm

Step 1: Initialize an empty list or array `groups` (or two variables `prev` and `curr`) to store the lengths of consecutive groups of 0s or 1s.
Step 2: Initialize a counter `count` to 1 to track the length of the current group.
Step 3: Iterate through the string `s` from the second character (index 1) to the end.
Step 4: For each character, compare it to the previous character. If they are the same, increment the `count` by 1 (the current group is extending).
Step 5: If the current character is different from the previous character, it indicates the end of the previous group. Append the current `count` to the `groups` list (or update `prev` with the current `curr` and set `curr` to 1), and reset `count` to 1 to start counting the new group.
Step 6: After the loop finishes, append the final `count` to the `groups` list (or handle the edge case for the last group if using the `prev` and `curr` variables).
Step 7: Initialize a variable `ans` to 0 to store the total number of valid substrings.
Step 8: Iterate through the `groups` list from the second element (index 1) to the end.
Step 9: For each element, calculate the minimum of the current group length and the previous group length (groups[i-1] and groups[i]). Add this minimum value to `ans`.
Step 10: Return the final value of `ans`.

Key Insights

Insight 1: The core idea is to identify consecutive groups of identical characters (either 0s or 1s). For example, in "0011100", the groups are [2, 3, 2] corresponding to "00", "111", and "00".
Insight 2: Once the lengths of these consecutive groups are determined, the number of valid substrings formed at the boundary of two adjacent groups is the minimum of the lengths of the two groups. For example, if we have groups [2, 3], we can form `min(2, 3) = 2` valid substrings: "0011" and "01".
Insight 3: We can avoid using extra space for a `groups` array by keeping track of the previous and current group size only.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Two Pointers, String.

Companies

Asked at: ByteDance, IBM, J.P. Morgan, Morgan Stanley, Salesforce.

Count Binary Substrings | LeetCode 696

Easy

Two Pointers String

Time: O(n)Space: O(n)

Solution Code

Available in:

class Solution {
    public int countBinarySubstrings(String s) {
        int[] groups = new int[s.length()];
        int i = 0, j = 0;
        while (i < s.length()) {
            char c = s.charAt(i);
            int count = 0;
            while (i < s.length() && s.charAt(i) == c) {
                i++;
                count++;
            }
            groups[j++] = count;
        }
        int ans = 0;
        for (int k = 1; k < j; k++) {
            ans += Math.min(groups[k - 1], groups[k]);
        }
        return ans;
    }
}

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Problem Info

Description

Problem 696: Count Binary Substrings - Detailed explanation and solution approaches.

Topics

Two Pointers String

Companies

ByteDanceIBMJ.P. MorganMorgan StanleySalesforce

Stats

Difficulty:Easy

Languages:4

Companies:5

Problem #:696

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

10 steps

1
Step 1: Initialize an empty list or array `groups` (or two variables `prev` and `curr`) to store the lengths of consecutive groups of 0s or 1s.
2
Step 2: Initialize a counter `count` to 1 to track the length of the current group.
3
Step 3: Iterate through the string `s` from the second character (index 1) to the end.
4
Step 4: For each character, compare it to the previous character. If they are the same, increment the `count` by 1 (the current group is extending).
5
Step 5: If the current character is different from the previous character, it indicates the end of the previous group. Append the current `count` to the `groups` list (or update `prev` with the current `curr` and set `curr` to 1), and reset `count` to 1 to start counting the new group.
6
Step 6: After the loop finishes, append the final `count` to the `groups` list (or handle the edge case for the last group if using the `prev` and `curr` variables).
7
Step 7: Initialize a variable `ans` to 0 to store the total number of valid substrings.
8
Step 8: Iterate through the `groups` list from the second element (index 1) to the end.
9
Step 9: For each element, calculate the minimum of the current group length and the previous group length (groups[i-1] and groups[i]). Add this minimum value to `ans`.
10
Step 10: Return the final value of `ans`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is directly proportional to the length of the string because we iterate through it a fixed number of times. The space complexity depends on whether we store the lengths of all the groups. If we do, the size of the `groups` list can grow up to `n/2` in the worst case (e.g., "010101..."). If we just store the two previous group lengths, the space used is constant.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Optimized One-Pass Solution

This approach optimizes the space complexity to O(1) by maintaining only the previous and current group lengths instead of storing all group lengths in an array/list. This involves tracking the previous and current counts and updating the answer as we iterate through the string.

Time: O(n)Space: O(1)

Trade-offs: This is the best approach if space is a concern, as it has the same time complexity as the initial approach but significantly lower space complexity. It requires a bit more careful handling of the last group.

✨

Key Insights

1Insight 1: The core idea is to identify consecutive groups of identical characters (either 0s or 1s). For example, in "0011100", the groups are [2, 3, 2] corresponding to "00", "111", and "00".
2Insight 2: Once the lengths of these consecutive groups are determined, the number of valid substrings formed at the boundary of two adjacent groups is the minimum of the lengths of the two groups. For example, if we have groups [2, 3], we can form `min(2, 3) = 2` valid substrings: "0011" and "01".
3Insight 3: We can avoid using extra space for a `groups` array by keeping track of the previous and current group size only.

📚

Key Terms Explained

For Students

Two Pointers

Using two indices to traverse data from different positions simultaneously.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Empty string: The problem states `1 <= s.length <= 10^5`, so an empty string is not a valid input, and this case doesn't need to be handled explicitly. However, if it were, one could add a check at the beginning of the function to return 0 if the string is empty.
Edge case 2: String with only one character type (e.g., "00000"): In this case, the `groups` list will contain only one element, and the loop that calculates the substring count will not execute, resulting in a correct answer of 0. This is handled correctly by the provided solutions.
Edge case 3: The last group of characters must be accounted for correctly. If using the `prev` and `curr` optimization, you must process the last group after the loop finishes.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the consecutive nature of 0s and 1s: Beginners may try to count all substrings with an equal number of 0s and 1s, regardless of whether they are consecutive, leading to an incorrect answer.

✗Common mistake 2: Incorrectly calculating the number of valid substrings at the boundary of two groups: The number of valid substrings is the minimum of the two group lengths, not their sum or product. Using the max or sum will lead to an overcount of substrings.

✗Common mistake 3: Forgetting to process the last group's count if using the optimized O(1) space version.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Data compression: The problem demonstrates a pattern recognition approach. Finding consecutive sequences can be applied in data compression algorithms to reduce redundancy.

🏦Real-world scenario 2: Signal processing: Analyzing signal patterns often involves identifying consecutive sequences of similar values. This can be useful in audio or image processing to detect features or reduce noise.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

The two-pointer technique is powerful for sorted arrays. The key insight is recognizing when elements can be processed from both ends simultaneously.

🎯 Interview Tip

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

The most common mistake is not handling the case when pointers meet. Always consider the termination condition.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 4, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

❓

Frequently Asked Questions

When is the two-pointer technique most effective?

Two pointers work best on sorted arrays or when searching for pairs/triplets that satisfy a condition. Common patterns include: moving pointers inward from both ends (like in two-sum variants), fast/slow pointers for cycle detection, or left/right pointers for partitioning.

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Count Binary Substrings"?

For this easy-level Two Pointers problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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