LeetCode #2529 — EASY

Maximum Count of Positive Integer and Negative Integer

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

Given an array nums sorted in non-decreasing order, return the maximum between the number of positive integers and the number of negative integers.

In other words, if the number of positive integers in nums is pos and the number of negative integers is neg, then return the maximum of pos and neg.

Note that 0 is neither positive nor negative.

Example 1:

Input: nums = [-2,-1,-1,1,2,3]
Output: 3
Explanation: There are 3 positive integers and 3 negative integers. The maximum count among them is 3.

Example 2:

Input: nums = [-3,-2,-1,0,0,1,2]
Output: 3
Explanation: There are 2 positive integers and 3 negative integers. The maximum count among them is 3.

Example 3:

Input: nums = [5,20,66,1314]
Output: 4
Explanation: There are 4 positive integers and 0 negative integers. The maximum count among them is 4.

Constraints:

1 <= nums.length <= 2000
-2000 <= nums[i] <= 2000
nums is sorted in a non-decreasing order.

Follow up: Can you solve the problem in O(log(n)) time complexity?

Step 01

Brute Force Baseline

Problem summary: Given an array nums sorted in non-decreasing order, return the maximum between the number of positive integers and the number of negative integers. In other words, if the number of positive integers in nums is pos and the number of negative integers is neg, then return the maximum of pos and neg. Note that 0 is neither positive nor negative.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search

Example 1

[-2,-1,-1,1,2,3]

Example 2

[-3,-2,-1,0,0,1,2]

Example 3

[5,20,66,1314]

Core Insight

What unlocks the optimal approach

Count how many positive integers and negative integers are in the array.
Since the array is sorted, can we use the binary search?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2529: Maximum Count of Positive Integer and Negative Integer
class Solution {
    public int maximumCount(int[] nums) {
        int a = 0, b = 0;
        for (int x : nums) {
            if (x > 0) {
                ++a;
            } else if (x < 0) {
                ++b;
            }
        }
        return Math.max(a, b);
    }
}

// Accepted solution for LeetCode #2529: Maximum Count of Positive Integer and Negative Integer
func maximumCount(nums []int) int {
	var a, b int
	for _, x := range nums {
		if x > 0 {
			a++
		} else if x < 0 {
			b++
		}
	}
	return max(a, b)
}

# Accepted solution for LeetCode #2529: Maximum Count of Positive Integer and Negative Integer
class Solution:
    def maximumCount(self, nums: List[int]) -> int:
        a = sum(x > 0 for x in nums)
        b = sum(x < 0 for x in nums)
        return max(a, b)

// Accepted solution for LeetCode #2529: Maximum Count of Positive Integer and Negative Integer
impl Solution {
    pub fn maximum_count(nums: Vec<i32>) -> i32 {
        let mut a = 0;
        let mut b = 0;

        for x in nums {
            if x > 0 {
                a += 1;
            } else if x < 0 {
                b += 1;
            }
        }

        std::cmp::max(a, b)
    }
}

// Accepted solution for LeetCode #2529: Maximum Count of Positive Integer and Negative Integer
function maximumCount(nums: number[]): number {
    let [a, b] = [0, 0];
    for (const x of nums) {
        if (x > 0) {
            ++a;
        } else if (x < 0) {
            ++b;
        }
    }
    return Math.max(a, b);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.