LeetCode //C - 168. Excel Sheet Column Title

168. Excel Sheet Column Title

Given an integer columnNumber, return its corresponding column title as it appears in an Excel sheet.

For example:

A -> 1
B -> 2
C -> 3
…
Z -> 26
AA -> 27
AB -> 28
…

Example 1:

Input: columnNumber = 1
Output: “A”

Example 2:

Input: columnNumber = 28
Output: “AB”

Example 3:

Input: columnNumber = 701
Output: “ZY”

Constraints:

• $1<=columnNumber<=2^{31}-1$

From: LeetCode
Link: 168. Excel Sheet Column Title

Solution:

Ideas:

1. Memory Allocation: We allocate memory for the result string. Since column numbers are large, we assume a reasonable size (8 characters) to cover most cases.

2. Loop through the column number: While columnNumber is greater than 0:

• Decrement columnNumber by 1 to shift it to a 0-based index.
• Compute the remainder when columnNumber is divided by 26. This gives us the position in the alphabet.
• Convert this remainder to the corresponding character by adding it to the ASCII value of ‘A’.
• Store the character in the result array and increment the index.
• Update columnNumber by integer division by 26.

3. String Reversal: After forming the result, the string is reversed because characters are added from the least significant digit to the most significant digit.

4. Return the Result: The result is returned after null-terminating it.

Code:

char* convertToTitle(int columnNumber) {
    char* result = (char*)malloc(8 * sizeof(char));  // Allocate memory for the result
    int index = 0;
    
    while (columnNumber > 0) {
        columnNumber--;  // Adjust to 0-indexed
        int remainder = columnNumber % 26;
        result[index++] = 'A' + remainder;
        columnNumber /= 26;
    }
    
    result[index] = '\0';  // Null-terminate the string
    
    // Reverse the string
    int len = strlen(result);
    for (int i = 0; i < len / 2; i++) {
        char temp = result[i];
        result[i] = result[len - 1 - i];
        result[len - 1 - i] = temp;
    }
    
    return result;
}