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:**

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

Below is the implementation of the above approach:

`// 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; ` `} ` |

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**Output:**

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

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