Count positives and negatives
Easy
Java
Variables
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Problem Statement
Task: return [posCount, negCount, zeroCount] for the array.
Input Format
An integer n
Then n integers
Then n integers
Output Format
An array [posCount, negCount, zeroCount]
Example
6 3 -1 0 5 -2 0
[2, 2, 2]
Constraints
• n ≥ 0
Concept Explanation
We count how many numbers are:
• Greater than 0 → positive
• Less than 0 → negative
• Equal to 0 → zero
• Greater than 0 → positive
• Less than 0 → negative
• Equal to 0 → zero
Step-by-Step Explanation
1.Read n and array.
2.Initialize pos = 0, neg = 0, zero = 0.
3.Loop through array:
• If value > 0 → increment pos.
• Else if value < 0 → increment neg.
• Else → increment zero.
4.Return [pos, neg, zero].
2.Initialize pos = 0, neg = 0, zero = 0.
3.Loop through array:
• If value > 0 → increment pos.
• Else if value < 0 → increment neg.
• Else → increment zero.
4.Return [pos, neg, zero].
Concept Explanation
We count how many numbers are:
• Greater than 0 → positive
• Less than 0 → negative
• Equal to 0 → zero
• Greater than 0 → positive
• Less than 0 → negative
• Equal to 0 → zero
Step-by-Step Explanation
1.Read n and array.
2.Initialize pos = 0, neg = 0, zero = 0.
3.Loop through array:
• If value > 0 → increment pos.
• Else if value < 0 → increment neg.
• Else → increment zero.
4.Return [pos, neg, zero].
2.Initialize pos = 0, neg = 0, zero = 0.
3.Loop through array:
• If value > 0 → increment pos.
• Else if value < 0 → increment neg.
• Else → increment zero.
4.Return [pos, neg, zero].
Input / Output Format
Input Format
An integer n
Then n integers
Then n integers
Output Format
An array [posCount, negCount, zeroCount]
Constraints
• n ≥ 0
Examples
Input:
6
3 -1 0 5 -2 0
Output:
[2, 2, 2]
Example Solution (Public)
static int[] countSigns(int[] a){int p=0,n=0,z=0;for(int x:a){if(x>0) p++; else if(x<0) n++; else z++;}return new int[]{p,n,z};}
Official Solution Code
static int[] countSigns(int[] a){int p=0,n=0,z=0;for(int x:a){if(x>0) p++; else if(x<0) n++; else z++;}return new int[]{p,n,z};}
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