What is the time complexity of Maximum Count of Positive Integer and Negative Integer?

The optimal solution for Maximum Count of Positive Integer and Negative Integer runs in O(log n) time and uses O(1) space. This complexity is due to the binary search, which efficiently narrows down the search space logarithmically.

What is the brute force solution for Maximum Count of Positive Integer and Negative Integer?

The brute force approach for Maximum Count of Positive Integer and Negative Integer is Brute Force, with O(n) time complexity and O(1) space complexity. We can simply iterate through the array and count the positive and negative integers. This straightforward approach gives us a clear understanding of the problem. The optimal Optimal Solution approach improves this to O(log n).

What algorithm or data structure is used to solve Maximum Count of Positive Integer and Negative Integer?

Maximum Count of Positive Integer and Negative Integer uses the following concepts: Array, Binary Search, Counting. The recommended patterns to study are: Binary Search, Array.

What are the interview tips for Maximum Count of Positive Integer and Negative Integer?

Always consider the properties of the input data (like sorted arrays). Explain your thought process clearly to the interviewer. Practice implementing both brute force and optimized solutions.

What are common mistakes when solving Maximum Count of Positive Integer and Negative Integer?

Not accounting for zeros in the count. Using a linear search when a binary search is more efficient.

#2529

Maximum Count of Positive Integer and Negative Integer

Easy

ArrayBinary SearchCountingBinary SearchArray

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Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(log n)Space O(1)

Since the array is sorted, we can use binary search to find the point where positive integers start. This allows us to count positives and negatives efficiently.

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Algorithm

4 steps

1Step 1: Use binary search to find the index of the first positive integer.
2Step 2: The count of positive integers is the length of the array minus this index.
3Step 3: The count of negative integers is simply this index.
4Step 4: Return the maximum of the counts from steps 2 and 3.

solution.py13 lines

1# Full working Python code
2
3def maxCount(nums):
4    left, right = 0, len(nums) - 1
5    while left <= right:
6        mid = (left + right) // 2
7        if nums[mid] < 0:
8            left = mid + 1
9        else:
10            right = mid - 1
11    neg_count = left
12    pos_count = len(nums) - neg_count
13    return max(pos_count, neg_count)

ℹ

Complexity note: This complexity is due to the binary search, which efficiently narrows down the search space logarithmically.

1The array is sorted, allowing for efficient counting using binary search.
2Zero does not affect the count of positive or negative integers.

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