Problem Description

You are given an array nums that is sorted in non-decreasing order (smallest to largest). Your task is to count how many positive integers and how many negative integers are in the array, then return whichever count is larger.

Key points to understand:

• The array is already sorted from smallest to largest
• You need to count positive integers (numbers greater than 0)
• You need to count negative integers (numbers less than 0)
• The number 0 is neither positive nor negative, so don't count it
• Return the maximum between these two counts

For example, if an array has 3 negative numbers and 5 positive numbers, you would return 5 since that's the larger count.

The solution approach is straightforward - traverse through the array once, counting elements that are greater than 0 for positive integers and elements that are less than 0 for negative integers. Then return the maximum of these two counts using max(a, b).

Intuition

Since we need to find the maximum between the count of positive and negative integers, the most direct approach is to simply count both types of numbers and compare them.

The key insight is that we don't need to do anything complex - we just need two counters. We can traverse the array once and for each element, determine if it's positive (x > 0) or negative (x < 0). Zero values are automatically ignored since they don't satisfy either condition.

Python's built-in sum() function combined with generator expressions makes this particularly elegant. The expression sum(x > 0 for x in nums) counts how many elements satisfy the condition x > 0 by creating a series of True/False values (which Python treats as 1/0) and summing them up. This gives us the count of positive numbers. Similarly, sum(x < 0 for x in nums) counts negative numbers.

While the array being sorted might suggest we could use binary search to find the boundaries between negative, zero, and positive regions more efficiently, for most practical cases, a simple linear scan is sufficient and the code remains much cleaner and easier to understand. The time complexity is still O(n) which is optimal since we need to count elements anyway.

Learn more about Binary Search patterns.

Solution Approach

The solution uses a simple traversal approach to count positive and negative integers separately, then returns the maximum of the two counts.

Implementation Steps:

1. Count positive integers: We use sum(x > 0 for x in nums) to count all positive numbers. This generator expression iterates through each element x in nums, evaluates the condition x > 0 (which returns True or False), and sum() adds up these boolean values (treating True as 1 and False as 0). The result is stored in variable a.

2. Count negative integers: Similarly, we use sum(x < 0 for x in nums) to count all negative numbers. The condition x < 0 identifies negative integers, and the sum gives us the total count, stored in variable b.

3. Return the maximum: We use max(a, b) to return the larger of the two counts.

Why this works:

• The conditions x > 0 and x < 0 naturally exclude zeros, since 0 is neither positive nor negative
• Each element is checked exactly once, making this an O(n) time complexity solution
• The space complexity is O(1) as we only use two variables to store the counts

Example walkthrough: If nums = [-3, -2, -1, 0, 0, 1, 2]:

• Positive count: sum(x > 0 for x in nums) evaluates to 2 (for elements 1 and 2)
• Negative count: sum(x < 0 for x in nums) evaluates to 3 (for elements -3, -2, -1)
• Return max(2, 3) = 3

Example Walkthrough

Let's walk through the solution with the array nums = [-4, -1, 0, 3, 5, 6]:

Step 1: Count positive integers

• We iterate through each element and check if it's greater than 0
• -4 > 0? False (0)
• -1 > 0? False (0)
• 0 > 0? False (0)
• 3 > 0? True (1)
• 5 > 0? True (1)
• 6 > 0? True (1)
• Sum: 0 + 0 + 0 + 1 + 1 + 1 = 3 positive integers

Step 2: Count negative integers

• We iterate through each element and check if it's less than 0
• -4 < 0? True (1)
• -1 < 0? True (1)
• 0 < 0? False (0)
• 3 < 0? False (0)
• 5 < 0? False (0)
• 6 < 0? False (0)
• Sum: 1 + 1 + 0 + 0 + 0 + 0 = 2 negative integers

Step 3: Return the maximum

• We have 3 positive integers and 2 negative integers
• max(3, 2) = 3
• Return 3

Notice how the zero in the array is automatically excluded from both counts since it's neither positive nor negative. The solution correctly identifies and counts only the relevant integers, then returns the larger count.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the array nums. This is because the code iterates through the entire array twice - once to count positive numbers using sum(x > 0 for x in nums) and once to count negative numbers using sum(x < 0 for x in nums). Each iteration takes O(n) time, and O(n) + O(n) = O(n).

The space complexity is O(1). The algorithm only uses two additional variables a and b to store the counts of positive and negative numbers respectively, which requires constant extra space regardless of the input size. The generator expressions used in the sum() functions don't create intermediate lists but instead process elements one at a time, maintaining constant space usage.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

Pitfall 1: Inefficient O(n) traversal for sorted arrays

The Problem: While the current solution correctly counts positive and negative integers with O(n) time complexity, it doesn't leverage the fact that the array is already sorted. In a sorted array, we can find the boundaries between negative, zero, and positive numbers using binary search, achieving O(log n) time complexity instead.

Why it matters: For large arrays, the difference between O(n) and O(log n) can be significant. If you have an array with millions of elements, binary search would be exponentially faster.

The Solution: Use binary search to find:

1. The index of the first non-negative number (end of negative numbers)
2. The index of the first positive number (start of positive numbers)

class Solution:
    def maximumCount(self, nums: List[int]) -> int:
        # Binary search for the first non-negative number (>= 0)
        left, right = 0, len(nums)
        while left < right:
            mid = (left + right) // 2
            if nums[mid] < 0:
                left = mid + 1
            else:
                right = mid
        negative_count = left
      
        # Binary search for the first positive number (> 0)
        left, right = 0, len(nums)
        while left < right:
            mid = (left + right) // 2
            if nums[mid] <= 0:
                left = mid + 1
            else:
                right = mid
        positive_count = len(nums) - left
      
        return max(negative_count, positive_count)

Pitfall 2: Counting zeros accidentally

The Problem: Using incorrect comparison operators like >= or <= instead of > and < would incorrectly include zeros in the count.

Example of incorrect code:

# WRONG - this would count zeros as positive
positive_count = sum(1 for num in nums if num >= 0)

The Solution: Always use strict inequalities (> for positive, < for negative) to exclude zeros from both counts.