How do you solve Count the Number of Substrings With Dominant Ones on LeetCode?

Count the Number of Substrings With Dominant Ones (LeetCode #3234) 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: Understanding the condition for dominant ones is crucial for substring evaluation.

What is the time complexity of Count the Number of Substrings With Dominant Ones?

The optimal solution for Count the Number of Substrings With Dominant Ones runs in O(n) time and uses O(1) space. The time complexity is O(n) because we traverse the string with two pointers, ensuring each character is processed at most twice.

What is the brute force solution for Count the Number of Substrings With Dominant Ones?

The brute force approach for Count the Number of Substrings With Dominant Ones is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks all possible substrings of the binary string. For each substring, we count the number of zeros and ones to determine if the substring has dominant ones. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count the Number of Substrings With Dominant Ones?

Count the Number of Substrings With Dominant Ones uses the following concepts: String, Enumeration. The recommended patterns to study are: Sliding Window, Two Pointers.

What are the interview tips for Count the Number of Substrings With Dominant Ones?

Always consider edge cases, such as strings with all zeros or all ones. Explain your thought process clearly while coding, as it helps the interviewer understand your approach. Practice writing code with time complexity in mind to improve efficiency.

What are common mistakes when solving Count the Number of Substrings With Dominant Ones?

Not correctly counting zeros and ones in substrings. Failing to optimize the substring evaluation process, leading to unnecessary computations.

#3234

Count the Number of Substrings With Dominant Ones

Medium

StringEnumerationSliding WindowTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

The optimal approach uses a sliding window technique to efficiently count dominant substrings by maintaining a count of zeros and ones as we expand the window. This avoids the need to re-count for every substring.

⚙️

Algorithm

5 steps

1Step 1: Initialize a count for dominant substrings and a hashmap to store the count of zeros.
2Step 2: Use two pointers to represent the current window of the substring.
3Step 3: Expand the right pointer and update the counts of zeros and ones.
4Step 4: If the substring is not dominant, move the left pointer to shrink the window until it becomes dominant again.
5Step 5: Count all valid substrings ending at the right pointer.

solution.py24 lines

1# Full working Python code
2
3def countDominantSubstrings(s):
4    count = 0
5    left = 0
6    ones = 0
7    zeros = 0
8    n = len(s)
9    for right in range(n):
10        if s[right] == '1':
11            ones += 1
12        else:
13            zeros += 1
14        while ones < zeros * zeros:
15            if s[left] == '1':
16                ones -= 1
17            else:
18                zeros -= 1
19            left += 1
20        count += right - left + 1
21    return count
22
23# Example usage
24print(countDominantSubstrings('00011'))  # Output: 5

ℹ

Complexity note: The time complexity is O(n) because we traverse the string with two pointers, ensuring each character is processed at most twice.

1Understanding the condition for dominant ones is crucial for substring evaluation.
2Using a sliding window approach can significantly reduce the time complexity.

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