C program to count Positive and Negative numbers in an Array

Given an array arr of integers of size N, the task is to find the count of positive numbers and negative numbers in the array Examples:

Input: arr[] = {2, -1, 5, 6, 0, -3} Output: Positive elements = 3 Negative elements = 2 There are 3 positive, 2 negative, and 1 zero. Input: arr[] = {4, 0, -2, -9, -7, 1} Output: Positive elements = 2 Negative elements = 3 There are 2 positive, 3 negative, and 1 zero.

Approach:

1. Traverse the elements in the array one by one.
2. For each element, check if the element is less than 0. If it is, then increment the count of negative elements.
3. For each element, check if the element is greater than 0. If it is, then increment the count of positive elements.
4. Print the count of negative and positive elements.

Below is the implementation of the above approach: 

C

// C program to find the count of positive
// and negative integers in an array

#include <stdio.h>

// Function to find the count of
// positive integers in an array
int countPositiveNumbers(int* arr, int n)
{
 int pos_count = 0;
 int i;
 for (i = 0; i < n; i++) {
  if (arr[i] > 0)
   pos_count++;
 }
 return pos_count;
}

// Function to find the count of
// negative integers in an array
int countNegativeNumbers(int* arr, int n)
{
 int neg_count = 0;
 int i;
 for (i = 0; i < n; i++) {
  if (arr[i] < 0)
   neg_count++;
 }
 return neg_count;
}

// Function to print the array
void printArray(int* arr, int n)
{
 int i;

 printf("Array: ");
 for (i = 0; i < n; i++) {
  printf("%d ", arr[i]);
 }
 printf("\n");
}

// Driver program
int main()
{
 int arr[] = { 2, -1, 5, 6, 0, -3 };
 int n;
 n = sizeof(arr) / sizeof(arr[0]);

 printArray(arr, n);

 printf("Count of Positive elements = %d\n",
  countPositiveNumbers(arr, n));
 printf("Count of Negative elements = %d\n",
  countNegativeNumbers(arr, n));

 return 0;
}

Output:

Array: 2 -1 5 6 0 -3 
Count of Positive elements = 3
Count of Negative elements = 2

Time complexity: O(n) where n is size of the given array

Auxiliary space: O(1)