Algorithm

1. Read n: Accept an integer.
2. Test n % 2 == 0: The % operator gives the remainder of integer division.
3. Print Even: Use this result when the remainder is zero.
4. Print Odd: Use this result when the remainder is not zero.
5. End: Only one of the two results is printed.

Explanation

The condition directly checks divisibility by 2. If the remainder is zero, n can be written as 2 multiplied by an integer. In a flowchart, place this condition inside a decision diamond. Label the true branch Yes and the false branch No. Each branch displays its result and reaches End.

Examples

For n = 8, 8 % 2 is 0, so the output is Even. For n = 7, the remainder is 1, so the output is Odd. Zero is even because 0 = 2 × 0. For n = −3, C gives a remainder of −1; it is non-zero, so the number is still correctly reported as Odd.

Important point

Do not test oddness only by n % 2 == 1 when negative inputs are allowed. A negative odd number can give remainder −1. Testing for a zero remainder and using the other branch for Odd handles both signs. The program also checks that reading the input succeeds before using n. Input values must fit int.

Time and space complexity

One remainder and one comparison are enough. There is no loop, so time is O(1). Only a fixed number of variables is required, giving O(1) extra space.

Reference table

Diagram or flowchart

C program and output

/* CodeBhavya | PPS AR25 | Learning example */
#include <stdio.h>
int main(void)
{
    int n;
    if (scanf("%d", &n) != 1) { puts("Invalid input"); return 1; }
    puts(n % 2 == 0 ? "Even" : "Odd");
    return 0;
}

-3

Odd

Key points

• An integer is even when its remainder on division by 2 is zero; otherwise it is odd.
• Test for a zero remainder, not for remainder equal to one.