solution and notes for problem 273

problems/273/paxtonfitzpatrick.md

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+# [Problem 273: Integer to English Words](https://leetcode.com/problems/integer-to-english-words/description/?envType=daily-question)
+
+## Initial thoughts (stream-of-consciousness)
+
+- oh this one looks like fun as well. Interesting that it's labeled hard... on first glance it seems like it'll require a larger-than-usual amount of coding, but nothing particularly complex conceptually
+- I think there's gonna be a few different reusable components I'll need:
+  - a mapping from digits to their english word when in the ones place (e.g., `{'1': 'One', '2': 'Two', ...}`)
+    - this will also give me the word representation of the hundreds place by just adding 'Hundred' after it
+  - a mapping from digits to their english words when in the tens place (e.g., `{'2': 'Twenty', '3': 'Thirty', ...}`)
+    - note: I'll need to handle the teens separately. While parsing the number, if I run into a 1 in the tens place, I'll use a specific helper function/mapping that consumes both the tens and ones place to get the correct word (e.g., '13' ➡️ 'Thirteen')
+    - I think I'll define these mappings on the class so they can be reused across test cases without needing to redefine them
+    - another note: it looks like the output is expected to be in title case, so I'll need to remember to set things up that way
+  - a helper function that takes a sequence of 3 numbers and uses the mappings above to convert them to their combined english words (e.g., f('123') ➡️ 'One Hundred Twenty Three')
+- and then my main function will:
+  - convert the input number to a string
+  - split it into groups of 3 digits, with the $\lt$ 3-letter group at the beginning if its length isn't divisible by 3
+    - I can do both of the above steps using the ',' string format specifier to convert the int to a string with commas, and then splitting it into a list on the commas
+  - process each of those groups independently using the helper function above, and add its appropriate "suffix" (e.g., 'Thousand', 'Million', etc.)
+    - the constraints say `num` can be at most $2^{31} - 1$, which is 2,147,483,647, so I'll only need to handle up to 'Billion'
+  - join the groups together and return the result
+- I don't foresee any major issues with this approach, so I'm gonna go ahead and start implementing it
+
+## Refining the problem, round 2 thoughts
+
+- any edge cases to consider?
+  - only potential "gotcha" I can think of is that typically 0s should lead to an empty string, or no aded output (e.g., 410000 ➡️ 'Four Hundred Ten Thousand'; 5002 ➡️ 'Five Thousand Two'), except for when the full number itself is 0, in which case the output should be 'Zero'. But I'll just add a check for this at the beginning of the main function
+- what would the time and space complexity of this be?
+  - I think the only part that scales with the input is looping over the 3-digit groups to process them, so the number of groups I need to process will increase by 1 as `num` increases by a factor of 1,000. I think this would mean both the time and space complexities are $O(\log n)$
+
+## Attempted solution(s)
+
+```python
+class Solution:
+    ones_mapping = {
+        '0': '',
+        '1': 'One',
+        '2': 'Two',
+        '3': 'Three',
+        '4': 'Four',
+        '5': 'Five',
+        '6': 'Six',
+        '7': 'Seven',
+        '8': 'Eight',
+        '9': 'Nine'
+    }
+    tens_mapping = {
+        '2': 'Twenty',
+        '3': 'Thirty',
+        '4': 'Forty',
+        '5': 'Fifty',
+        '6': 'Sixty',
+        '7': 'Seventy',
+        '8': 'Eighty',
+        '9': 'Ninety'
+    }
+    teens_mapping = {
+        '10': 'Ten',
+        '11': 'Eleven',
+        '12': 'Twelve',
+        '13': 'Thirteen',
+        '14': 'Fourteen',
+        '15': 'Fifteen',
+        '16': 'Sixteen',
+        '17': 'Seventeen',
+        '18': 'Eighteen',
+        '19': 'Nineteen'
+    }
+    group_suffixes = ('', ' Thousand', ' Million', ' Billion')
+
+    def numberToWords(self, num: int) -> str:
+        if num == 0:
+            return 'Zero'
+        result = ''
+        for digit_group, suffix in zip(
+            reversed(f'{num:,}'.split(',')),
+            self.group_suffixes
+        ):
+            if digit_group != '000':
+                group_words = self._process_digit_group(digit_group) + suffix
+                result = f'{group_words} {result}'
+        return result.strip()
+
+    def _process_digit_group(self, digits):
+        match len(digits):
+            case 1:
+                return self.ones_mapping[digits]
+            case 2:
+                if digits[0] == '1':
+                    return self.teens_mapping[digits]
+                return (
+                    f'{self.tens_mapping[digits[0]]} '
+                    f'{self.ones_mapping[digits[1]]}'
+                ).rstrip()
+        if digits[0] == '0':
+            group_words = ''
+        else:
+            group_words = self.ones_mapping[digits[0]] + ' Hundred'
+        match digits[1]:
+            case '0':
+                return f'{group_words} {self.ones_mapping[digits[2]]}'.strip()
+            case '1':
+                return f'{group_words} {self.teens_mapping[digits[1:]]}'.lstrip()
+        return (
+            f'{group_words} {self.tens_mapping[digits[1]]} '
+            f'{self.ones_mapping[digits[2]]}'
+        ).strip()
+```
+
+![](https://github.com/user-attachments/assets/71d3b83a-56de-48a2-8209-f787c0034b4d)