Function printCharAtOddPos – CTS PATTERN

Print characters at odd positions (1-indexed) in a given string by fixing logical errors in the provided C function.

Understand the Problem

Problem Statement

You are required to correct the syntax of the given code without changing its logic. You can click on Run anytime to check the compilation/execution status of the program. You can use printf to debug your code. The submitted code should be logically/syntactically correct and pass all test cases. Do not write the main() function as it is not required.

Code Approach: For this question, you will need to correct the given implementation. We do not expect you to modify the approach or incorporate any additional library methods.

The function printCharAtOddPos(char *str) accepts a string str as the input. The function is supposed to print only the characters at the odd positions in the string str.

The function compiles fine but fails to print the desired result due to logical errors.

Your task is to fix the program so that it passes all test cases.

Constraints

String length can be from 0 to 1000 characters
String may contain letters, digits, and special characters
Odd positions are 1-indexed (1st, 3rd, 5th, etc.)
Memory should be handled properly (no leaks)
Function should work correctly for empty strings

Examples

Example 1

Input

Hello

Output

Hlo

Explanation

Characters at odd positions (1-indexed): Position 1 = 'H', Position 3 = 'l', Position 5 = 'o'. So output is 'Hlo'.

Example 2

Input

Programming

Output

Porgmig

Explanation

Characters at odd positions: Position 1 = 'P', Position 3 = 'o', Position 5 = 'r', Position 7 = 'g', Position 9 = 'm', Position 11 = 'g'. So output is 'Porgmig'.

Example 3

Input

Output

Explanation

Single character string has only one character at position 1 (odd position), so output is 'A'.

Example 4

Input

Output

Explanation

Empty string has no characters, so no output is produced.

Solution

#include <stdio.h>
#include <string.h>

void printCharAtOddPos(char *str)
{
    int index = 0, len = strlen(str);
    while (index < len)
    {
        if (index % 2 == 0)
            printf("%c", str[index]);
        index += 1;
    }
}

Time:O(n) - where n is the length of the string. We iterate through each character exactly once.

Space:O(1) - We use only a constant amount of extra space regardless of input size.

Approach:

Step-by-step explanation:

Calculate string length using strlen()
Initialize index to 0 to start from the beginning of the string
Loop through each character while index is less than string length
Check if current index is even (0, 2, 4...) using index % 2 == 0
If condition is true, print the character at that index using printf()
Increment index to move to next character

Why this works: In 0-indexed arrays, even indices (0, 2, 4...) correspond to odd positions (1, 3, 5...) in 1-indexed counting, which is what the problem requires.

Visual Explanation

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