Excel Sheet Column Title

Problem Description

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
• …

Examples

Constraints

• 1 ≤ columnNumber ≤ 2³¹ - 1

Base-26 Conversion with a Shift

Excel column titles are essentially a base-26 numbering system, but with a twist: they are 1-indexed (A=1) rather than 0-indexed. This means that 26 maps to Z, and 27 starts the next digit place (AA).

In standard base conversion, we find the remainder using modulo % 26 and then divide the number. Because this system is 1-indexed, we must subtract 1 from the number before performing each modulo operation:

1. columnNumber -= 1
2. remainder = columnNumber % 26
3. Convert remainder to a character: (char) ('A' + remainder)
4. columnNumber /= 26
5. Repeat until the number becomes 0, then reverse the accumulated characters.

Solution: Base-26 Conversion

class Solution {
    public String convertToTitle(int columnNumber) {
        StringBuilder sb = new StringBuilder();

        while (columnNumber > 0) {
            columnNumber--; // adjust to 0-indexed
            int remainder = columnNumber % 26;
            sb.append((char) ('A' + remainder));
            columnNumber /= 26;
        }

        return sb.reverse().toString();
    }
}

class Solution:
    def convertToTitle(self, columnNumber: int) -> str:
        result = []

        while columnNumber > 0:
            columnNumber -= 1  # adjust to 0-indexed
            remainder = columnNumber % 26
            result.append(chr(ord('A') + remainder))
            columnNumber //= 26

        return "".join(reversed(result))

#include <string>
#include <algorithm>

class Solution {
public:
    std::string convertToTitle(int columnNumber) {
        std::string result = "";

        while (columnNumber > 0) {
            columnNumber--; // adjust to 0-indexed
            int remainder = columnNumber % 26;
            result += (char)('A' + remainder);
            columnNumber /= 26;
        }

        std::reverse(result.begin(), result.end());
        return result;
    }
};

Complexity Analysis:

• Time Complexity: O(log₂₆ N) where N is columnNumber. The loop divides the number by 26 at each step.
• Space Complexity: O(1) auxiliary space (excluding the output string).

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