| 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 | The function uses backtracking to try different digit assignments for letters, respecting: |
| 12 | - Each letter maps to a unique digit |
| 13 | - Leading digits cannot be zero |
| 14 | - The mathematical equation must be satisfied |
| 15 | """ |
| 16 | # Edge Case: Empty or None puzzle |
| 17 | if not puzzle: |
| 18 | return None |
| 19 | |
| 20 | # Parse the puzzle |
| 21 | parts = puzzle.replace("==", "=").replace(" ", "").split("=") |
| 22 | |
| 23 | # Edge Case: Invalid puzzle format (no equals sign or too many parts) |
| 24 | if len(parts) != 2: |
| 25 | return None |
| 26 | |
| 27 | left_side, right_side = parts[0], parts[1] |
| 28 | |
| 29 | # Extract all terms (numbers being added) |
| 30 | terms = left_side.replace("+", " ").split() |
| 31 | |
| 32 | # Edge Case: No terms on the left side |
| 33 | if not terms: |
| 34 | return None |
| 35 | |
| 36 | # Get all unique letters in the puzzle |
| 37 | letters = set() |
| 38 | for term in terms + [right_side]: |
| 39 | letters.update(term) |
| 40 | |
| 41 | # Edge Case: No letters in puzzle |
| 42 | if not letters: |
| 43 | return None |
| 44 | |
| 45 | # Get leading letters (first letter of each term) |
| 46 | leading_letters = {term[0] for term in terms + [right_side]} |
| 47 | |
| 48 | # Convert to lists for indexing |
| 49 | letters = list(letters) |
| 50 | |
| 51 | # Edge Case: More than 10 unique letters (impossible to assign unique digits) |
| 52 | if len(letters) > 10: |
| 53 | return None |
| 54 | |
| 55 | # Try all possible digit assignments using backtracking |
| 56 | def backtrack(index, used_digits, letter_to_digit): |
| 57 | """ |
| 58 | Recursively try digit assignments for letters. |
| 59 | |
| 60 | Args: |
| 61 | index: Current letter index we're assigning a digit to |
| 62 | used_digits: Set of digits already assigned |
| 63 | letter_to_digit: Current mapping of letters to digits |
| 64 | |
| 65 | Returns: |
| 66 | A valid letter-to-digit mapping if found, otherwise None |
| 67 | """ |
| 68 | # Base case: All letters have been assigned digits |
| 69 | if index == len(letters): |
| 70 | # Check if the equation is satisfied |
| 71 | # Convert terms to numbers |
| 72 | try: |
| 73 | term_values = [] |
| 74 | for term in terms: |
| 75 | num = 0 |
| 76 | for char in term: |
| 77 | num = num * 10 + letter_to_digit[char] |
| 78 | term_values.append(num) |
| 79 | |
| 80 | # Convert result to number |
| 81 | result_value = 0 |
| 82 | for char in right_side: |
| 83 | result_value = result_value * 10 + letter_to_digit[char] |
| 84 | |
| 85 | # Check if equation holds |
| 86 | if sum(term_values) == result_value: |
| 87 | return letter_to_digit.copy() |
| 88 | except KeyError: |
| 89 | # This shouldn't happen if we've assigned all letters, but just in case |
| 90 | return None |
| 91 | |
| 92 | return None |
| 93 | |
| 94 | # Try assigning each unused digit to the current letter |
| 95 | current_letter = letters[index] |
| 96 | |
| 97 | for digit in range(10): |
| 98 | # Skip if digit is already used |
| 99 | if digit in used_digits: |
| 100 | continue |
| 101 | |
| 102 | # Skip if assigning 0 to a leading letter |
| 103 | if digit == 0 and current_letter in leading_letters: |
| 104 | continue |
| 105 | |
| 106 | # Try this assignment |
| 107 | letter_to_digit[current_letter] = digit |
| 108 | used_digits.add(digit) |
| 109 | |
| 110 | # Recurse to the next letter |
| 111 | result = backtrack(index + 1, used_digits, letter_to_digit) |
| 112 | if result is not None: |
| 113 | return result |
| 114 | |
| 115 | # Backtrack |
| 116 | used_digits.remove(digit) |
| 117 | del letter_to_digit[current_letter] |
| 118 | |
| 119 | return None |
| 120 | |
| 121 | # Start the backtracking process |
| 122 | solution = backtrack(0, set(), {}) |
| 123 | return solution |
| 124 | |
| 125 | # Handled Edge Cases: Empty/None puzzle, invalid puzzle format, no terms on left side, no letters in puzzle, more than 10 unique letters |
| 126 | # Handled Edge Cases: All letters assigned, equation validation, leading zero constraint, digit uniqueness constraint |