How do you solve Longer Contiguous Segments of Ones than Zeros on LeetCode?

Longer Contiguous Segments of Ones than Zeros (LeetCode #1869) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The longest contiguous segment of 1s or 0s can be found by counting as we traverse the string.

What is the brute force solution for Longer Contiguous Segments of Ones than Zeros?

The brute force approach for Longer Contiguous Segments of Ones than Zeros is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check every possible segment of the string to find the longest contiguous segments of 1s and 0s. This approach is straightforward but inefficient as it involves checking all possible substrings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Longer Contiguous Segments of Ones than Zeros?

Longer Contiguous Segments of Ones than Zeros uses the following concepts: String. The recommended patterns to study are: Two Pointers, Sliding Window, Counting Segments.

What are the interview tips for Longer Contiguous Segments of Ones than Zeros?

Always consider edge cases, such as strings with only 1s or only 0s. Explain your thought process clearly while coding to show your understanding. Practice writing clean and efficient code, as readability is important.

What are common mistakes when solving Longer Contiguous Segments of Ones than Zeros?

Not resetting the count variable correctly when switching between 1s and 0s. Failing to check the last segment after exiting the loop.

#1869

Longer Contiguous Segments of Ones than Zeros

Easy

StringTwo PointersSliding WindowCounting Segments

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

We can traverse the string once while counting the lengths of contiguous segments of 1s and 0s. This approach is efficient because it only requires a single pass through the string.

⚙️

Algorithm

5 steps

1Step 1: Initialize variables to track the maximum lengths of contiguous 1s and 0s.
2Step 2: Use a single loop to iterate through the string, counting the length of the current segment.
3Step 3: When the character changes, update the maximum length for the previous character.
4Step 4: At the end of the loop, ensure to check the last segment.
5Step 5: Compare the maximum lengths of 1s and 0s and return the result.

solution.py17 lines

1def checkZeroOnes(s):
2    max_ones = max_zeros = 0
3    current_count = 1
4    for i in range(1, len(s)):
5        if s[i] == s[i - 1]:
6            current_count += 1
7        else:
8            if s[i - 1] == '1':
9                max_ones = max(max_ones, current_count)
10            else:
11                max_zeros = max(max_zeros, current_count)
12            current_count = 1
13    if s[-1] == '1':
14        max_ones = max(max_ones, current_count)
15    else:
16        max_zeros = max(max_zeros, current_count)
17    return max_ones > max_zeros

ℹ

Complexity note: The time complexity is O(n) because we only traverse the string once, counting segments as we go. The space complexity is O(1) since we are using a constant amount of space for our variables.

1The longest contiguous segment of 1s or 0s can be found by counting as we traverse the string.
2A single pass through the string is sufficient to determine the maximum lengths.

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