Question

Create a method to accept a integer array as parameter and count the number of positive, negative and zeroes in it.

Answer 1

Answer:

zero and positive and negative

Answer 2

Input:  

arr[] = {2, -1, 5, 6, 0, -3}

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.

1. 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;

}

Output:

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