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

The optimal solution has a time complexity of O(n).

What data structures are used to solve Maximum Count of Positive Integer and Negative Integer?

This problem uses Array, Binary Search, Counting. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Maximum Count of Positive Integer and Negative Integer?

Insight 1: Since the array is sorted, we don't need binary search or any sophisticated algorithm. A simple linear scan is sufficient. Insight 2: We can use two counters to keep track of positive and negative numbers simultaneously during a single pass through the array.

Which companies ask LeetCode 2529. Maximum Count of Positive Integer and Negative Integer in interviews?

This problem is frequently asked at ShareChat.

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Maximum Count of Positive Integer and Negative Integer - LeetCode 2529 Solution

Maximum Count of Positive Integer and Negative Integer - Complete Solution Guide

Maximum Count of Positive Integer and Negative Integer is LeetCode problem 2529, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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.

Detailed Explanation

The problem asks you to find the maximum between the number of positive integers and the number of negative integers in a sorted array. The input is a sorted array of integers (`nums`). The output is a single integer representing the maximum of the count of positive numbers and the count of negative numbers. Zero is considered neither positive nor negative. The array is guaranteed to be sorted in non-decreasing order, and the elements are within a specified range.

Solution Approach

The provided solutions all employ a linear scan approach. They iterate through the array, incrementing a counter for positive numbers (`pos_count` or `pos`) if a number is greater than 0, and incrementing a counter for negative numbers (`neg_count` or `neg`) if a number is less than 0. Finally, the maximum of the two counters is returned.

Step-by-Step Algorithm

Step 1: Initialize two counters, `pos_count` (or `pos`) and `neg_count` (or `neg`), to 0.
Step 2: Iterate through the input array `nums`.
Step 3: For each number `num` in the array:
Step 4: If `num > 0`, increment `pos_count` (or `pos`).
Step 5: If `num < 0`, increment `neg_count` (or `neg`).
Step 6: After iterating through the entire array, return `max(pos_count, neg_count)` (or `max(pos, neg)`).

Key Insights

Insight 1: Since the array is sorted, we don't need binary search or any sophisticated algorithm. A simple linear scan is sufficient.
Insight 2: We can use two counters to keep track of positive and negative numbers simultaneously during a single pass through the array.
Insight 3: The problem's constraint of a sorted array is not strictly necessary for this solution; the solution would work for unsorted arrays as well.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Binary Search, Counting.

Companies

Asked at: ShareChat.

Maximum Count of Positive Integer and Negative Integer | LeetCode 2529

Easy

Array Binary Search Counting

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int maximumCount(int[] nums) {
        int pos = 0;
        int neg = 0;
        for (int num : nums) {
            if (num > 0) {
                pos++;
            } else if (num < 0) {
                neg++;
            }
        }
        return Math.max(pos, neg);
    }
}

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Problem Info

Description

Problem 2529: Maximum Count of Positive Integer and Negative Integer - Detailed explanation and solution approaches.

Topics

Array Binary Search Counting

Companies

ShareChat

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:2529

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Initialize two counters, `pos_count` (or `pos`) and `neg_count` (or `neg`), to 0.
2
Step 2: Iterate through the input array `nums`.
3
Step 3: For each number `num` in the array:
4
Step 4: If `num > 0`, increment `pos_count` (or `pos`).
5
Step 5: If `num < 0`, increment `neg_count` (or `neg`).
6
Step 6: After iterating through the entire array, return `max(pos_count, neg_count)` (or `max(pos, neg)`).

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The algorithm iterates through the array once, performing a constant amount of work for each element. Therefore, the time complexity is linear, directly proportional to the size of the input array. The space complexity is constant because only a fixed number of variables (two counters) are used regardless of the input size.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Using `count_if` (C++) or similar functions

Utilize standard library functions like `std::count_if` in C++ to count elements satisfying a condition (positive or negative).

Time: O(n)Space: O(1)

Trade-offs: Might be slightly less readable for beginners, but offers a more concise solution in some languages. Performance is essentially the same.

Binary Search (Follow-up)

Since the array is sorted, we can find the index of the first positive number and the index of the last negative number using binary search. The difference between these indices will give us the count of positives and negatives. This approach is more efficient if a majority of the numbers are positive or negative.

Time: O(log n)Space: O(1)

Trade-offs: More complex to implement, but offers logarithmic time complexity which is significantly faster for very large arrays. Not as efficient for smaller arrays due to the overhead of binary search.

✨

Key Insights

1Insight 1: Since the array is sorted, we don't need binary search or any sophisticated algorithm. A simple linear scan is sufficient.
2Insight 2: We can use two counters to keep track of positive and negative numbers simultaneously during a single pass through the array.
3Insight 3: The problem's constraint of a sorted array is not strictly necessary for this solution; the solution would work for unsorted arrays as well.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Edge Cases

Edge case 1: Empty array. The solution correctly handles this, returning 0 as the maximum of 0 positive and 0 negative numbers.
Edge case 2: Array with only positive or only negative numbers. The solution accurately counts all positive or negative numbers and returns the correct count.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly handling zero. Beginners might mistakenly count zero as either positive or negative.

✗Common mistake 2: Forgetting to consider the case of an empty array.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Analyzing sentiment in text data. Positive and negative words can be counted efficiently using this approach.

🏦Real-world scenario 2: Analyzing financial data where you need to quickly determine the number of positive and negative returns (profits and losses).

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 10, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Can binary search be applied to non-sorted arrays?

Binary search requires a monotonic property, not necessarily sorted data. It can be applied to any search space where you can determine if the answer is in the left or right half. Examples include searching rotated arrays or finding minimum in optimization problems.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Maximum Count of Positive Integer and Negative Integer"?

For this easy-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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