| 1 | 1 | def solve(puzzle: str) -> dict[str, int] | None: |
| 2 | | pass |
| 2 | """ |
| 3 | Solve an alphametics puzzle. |
| 4 | |
| 5 | Args: |
| 6 | puzzle: A string representing the alphametics puzzle (e.g., "SEND + MORE == MONEY") |
| 7 | |
| 8 | Returns: |
| 9 | A dictionary mapping letters to digits that solves the puzzle, or None if no solution exists. |
| 10 | |
| 11 | Example: |
| 12 | >>> solve("SEND + MORE == MONEY") |
| 13 | {'S': 9, 'E': 5, 'N': 6, 'D': 7, 'M': 1, 'O': 0, 'R': 8, 'Y': 2} |
| 14 | """ |
| 15 | # Edge Case: Empty or None puzzle |
| 16 | if not puzzle: |
| 17 | return None |
| 18 | |
| 19 | # Parse the puzzle |
| 20 | parts = puzzle.replace("==", "=").replace(" ", "").split("=") |
| 21 | |
| 22 | # Edge Case: Invalid puzzle format (no equals sign or too many parts) |
| 23 | if len(parts) != 2: |
| 24 | return None |
| 25 | |
| 26 | right_side = parts[1] |
| 27 | left_parts = parts[0].replace("+", " ").split() |
| 28 | |
| 29 | # Edge Case: No terms on the left side |
| 30 | if not left_parts: |
| 31 | return None |
| 32 | |
| 33 | # Extract all unique letters |
| 34 | letters = set() |
| 35 | for part in left_parts + [right_side]: |
| 36 | letters.update(part) |
| 37 | |
| 38 | # Edge Case: No letters in puzzle |
| 39 | if not letters: |
| 40 | return None |
| 41 | |
| 42 | # Extract first letters (can't be zero) |
| 43 | first_letters = {word[0] for word in left_parts + [right_side]} |
| 44 | |
| 45 | # Convert to sorted list for consistent ordering |
| 46 | letters = list(letters) |
| 47 | |
| 48 | # Edge Case: More than 10 unique letters (impossible to solve) |
| 49 | if len(letters) > 10: |
| 50 | return None |
| 51 | |
| 52 | # Generate all possible digit assignments |
| 53 | def backtrack(index, used_digits, letter_to_digit): |
| 54 | """ |
| 55 | Recursively try digit assignments using backtracking. |
| 56 | |
| 57 | Args: |
| 58 | index: Current index in the letters list |
| 59 | used_digits: Set of digits already assigned |
| 60 | letter_to_digit: Dictionary mapping letters to digits |
| 61 | |
| 62 | Returns: |
| 63 | Solution dictionary if found, else None |
| 64 | """ |
| 65 | # Base case: all letters assigned |
| 66 | if index == len(letters): |
| 67 | # Check if the equation is satisfied |
| 68 | left_value = 0 |
| 69 | for word in left_parts: |
| 70 | num = 0 |
| 71 | for char in word: |
| 72 | num = num * 10 + letter_to_digit[char] |
| 73 | left_value += num |
| 74 | |
| 75 | right_value = 0 |
| 76 | for char in right_side: |
| 77 | right_value = right_value * 10 + letter_to_digit[char] |
| 78 | |
| 79 | if left_value == right_value: |
| 80 | return letter_to_digit.copy() |
| 81 | return None |
| 82 | |
| 83 | # Try assigning each unused digit to the current letter |
| 84 | letter = letters[index] |
| 85 | |
| 86 | # Edge Case: First letter of a word cannot be zero |
| 87 | if letter in first_letters: |
| 88 | start_digit = 1 |
| 89 | else: |
| 90 | start_digit = 0 |
| 91 | |
| 92 | for digit in range(start_digit, 10): |
| 93 | if digit not in used_digits: |
| 94 | # Make the assignment |
| 95 | letter_to_digit[letter] = digit |
| 96 | used_digits.add(digit) |
| 97 | |
| 98 | # Recurse to the next letter |
| 99 | result = backtrack(index + 1, used_digits, letter_to_digit) |
| 100 | if result is not None: |
| 101 | return result |
| 102 | |
| 103 | # Backtrack |
| 104 | used_digits.remove(digit) |
| 105 | del letter_to_digit[letter] |
| 106 | |
| 107 | return None |
| 108 | |
| 109 | # Start the backtracking search |
| 110 | solution = backtrack(0, set(), {}) |
| 111 | return solution |
| 112 | # Handled Edge Cases: Empty/None puzzle, invalid format, no terms on left side, |
| 113 | # no letters, more than 10 unique letters, first letter being zero |