C++ Program to Count Positive and Negative Numbers with Explanation
C++
Easy
Arrays
31 views
2 min read
211 words
This problem helps you practice core C++ fundamentals in a practical way. It builds intuition around positive, negative, count. Let’s break it down step by step so you can implement it confidently.
Problem Statement
Given an array with positive and negative numbers, count how many are positive and how many are negative. This builds conditional logic skills.
Logic: Check each number's sign and increment respective counter
Input Format
An integer n (size of array)
Then n integers (can be positive or negative).
Output Format
Two integers:
• count of positive numbers
• count of negative numbers
Constraints
n ≥ 1
Zero is neither positive nor negative
Concept Explanation
Each element of the array is checked.
If the value is greater than 0, it is positive.
If the value is less than 0, it is negative
Step-by-Step Logic
1.Read array size n.
2.Initialize two counters:
• posCount = 0
• negCount = 0
3.Traverse the array from index 0 to n-1.
4.For each element:
• If value > 0 → increment posCount.
• Else if value < 0 → increment negCount.
5.Ignore value 0.
6.Print posCount and negCount.
Code Solution
This explanation is written for learning purposes and to help beginners understand the concept clearly.
void question4_count_pos_neg() {
int arr[] = {5, -3, 8, -1, 0, -7, 12};
int size = 7;
int positive = 0, negative = 0;
for(int i = 0; i < size; i++) {
if(arr[i] > 0) {
positive++;
} else if(arr[i] < 0) {
negative++;
}
}
cout << "Positive: " << positive << ", Negative: " << negative << endl;
}
Output Example
Output:
Positive = 3
Negative = 2
Common Mistakes
- Misreading input/output format.
- Not handling constraints and edge cases.
- Off-by-one errors in loops.
- Forgetting to reset variables between test cases (if any).
Solution Guide
Problem
Given an array with positive and negative numbers, count how many are positive and how many are negative. This builds conditional logic skills.
Logic: Check each number's sign and increment respective counter
Input / Output
Input
An integer n (size of array)
Then n integers (can be positive or negative).
Output
Two integers:
• count of positive numbers
• count of negative numbers
Constraints
n ≥ 1
Zero is neither positive nor negative
Examples
Output:
Positive = 3
Negative = 2
Explanation
Concept Explanation
Each element of the array is checked.
If the value is greater than 0, it is positive.
If the value is less than 0, it is negative
Step-by-Step Explanation
1.Read array size n.
2.Initialize two counters:
• posCount = 0
• negCount = 0
3.Traverse the array from index 0 to n-1.
4.For each element:
• If value > 0 → increment posCount.
• Else if value < 0 → increment negCount.
5.Ignore value 0.
6.Print posCount and negCount.
Details
Common Mistakes
- Misreading input/output format.
- Not handling constraints and edge cases.
- Off-by-one errors in loops.
- Forgetting to reset variables between test cases (if any).
Official Solution
void question4_count_pos_neg() {
int arr[] = {5, -3, 8, -1, 0, -7, 12};
int size = 7;
int positive = 0, negative = 0;
for(int i = 0; i < size; i++) {
if(arr[i] > 0) {
positive++;
} else if(arr[i] < 0) {
negative++;
}
}
cout << "Positive: " << positive << ", Negative: " << negative << endl;
}
Solutions (0)
No solutions submitted yet. Be the first!
