Problem

Given a column title as appear in an Excel sheet, return its corresponding column number. For example:

1
2
3
4
5
6
7
8
9

A -> 1
B -> 2
C -> 3
...
Z -> 26
AA -> 27
AB -> 28
...
AAA -> 703

Examples

Example 1:

1
2

Input: columnTitle = "A"
Output: 1

Example 2:

1
2

Input: columnTitle = "AB"
Output: 28

Example 3:

1
2

Input: columnTitle = "ZY"
Output: 701

Similar problem

Excel Sheet Column Title

Solution

Method 1 - Using Hashmap

Here is the approach:

  1. Create a HashMap: Map each character (‘A’ to ‘Z’) to its corresponding integer value (1 to 26).
  2. Initialize Result: Start with a result variable initialized to 0.
  3. Process Each Character: Iterate through each character of the input string from left to right.
  4. Calculate Value:
    • Use the HashMap to get the numerical value of the character.
    • Update the result using the formula for converting base-26 to base-10: result = result * 26 + (character_value).
  5. Return Result: The final value of result after processing all characters is the column number.

Code

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16

class Solution {
    public int titleToNumber(String columnTitle) {
        HashMap<Character, Integer> charToValue = new HashMap<>();
        
        // Create the mapping from 'A' to 'Z'
        for (char ch = 'A'; ch <= 'Z'; ch++) {
            charToValue.put(ch, ch - 'A' + 1);
        }
        
        int ans = 0;
        for (int i = 0; i < columnTitle.length(); i++) {
            ans = ans * 26 + charToValue.get(columnTitle.charAt(i));
        }
        return ans;
    }
}
1
2
3
4
5
6
7
8

class Solution:
    def titleToNumber(self, columnTitle: str) -> int:
        char_to_value = {chr(i + ord('A')): i + 1 for i in range(26)}
        
        ans: int = 0
        for char in columnTitle:
            ans = ans * 26 + char_to_value[char]
        return ans
1

- 🧺 Space complexity: `O(1)`, since the HashMap size is fixed with 26 entries regardless of the input size.

Method 2 - Simple multiplication by 26

Actually using a hashmap is not necessary here. Here is the approach:

  1. Initialize Result: Start with a result variable initialized to 0.
  2. Process Each Character: Iterate through each character of the input string from left to right.
  3. Calculate Value:
    • Convert the character to its corresponding numerical value (e.g., ‘A’ to 1, ‘B’ to 2, etc.).
    • Update the result using the formula for converting base-26 to base-10: result = result * 26 + (character_value).
  4. Return Result: The final value of result after processing all characters is the column number.

Code

1
2
3
4
5
6
7
8
9

 class Solution {
     public int titleToNumber(String columnTitle) {
         int ans = 0;
         for (int i = 0; i < columnTitle.length(); i++) {
             ans = ans * 26 + (columnTitle.charAt(i) - 'A' + 1);
         }
         return ans;
     }
 }
1
2
3
4
5
6

class Solution:
    def titleToNumber(self, columnTitle: str) -> int:
        ans: int = 0
        for char in columnTitle:
            ans = ans * 26 + (ord(char) - ord('A') + 1)
        return ans

Complexity

  • ⏰ Time complexity: O(n), where n is the length of the input string, as we process each character once.
  • 🧺 Space complexity: O(1)