| 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 with constraint propagation to find a valid digit assignment |
| 12 | where each letter represents a unique digit, no leading digit is zero, and the equation holds. |
| 13 | |
| 14 | Handled Edge Cases: |
| 15 | - Invalid puzzle format (no '==') |
| 16 | - Words with invalid characters |
| 17 | - Leading zeros in multi-digit words |
| 18 | - No solution exists |
| 19 | - Empty or malformed puzzle string |
| 20 | """ |
| 21 | # Edge Case: Empty or malformed puzzle string |
| 22 | if not puzzle or "==" not in puzzle: |
| 23 | return None |
| 24 | |
| 25 | # Parse the puzzle |
| 26 | left_side, right_side = puzzle.replace(" ", "").split("==") |
| 27 | addends = left_side.split("+") |
| 28 | result = right_side |
| 29 | |
| 30 | # Edge Case: Invalid puzzle format (no addends or result) |
| 31 | if not addends or not result: |
| 32 | return None |
| 33 | |
| 34 | # Collect all unique letters |
| 35 | letters = set() |
| 36 | for word in addends + [result]: |
| 37 | # Edge Case: Words with invalid characters |
| 38 | if not word.isalpha(): |
| 39 | return None |
| 40 | letters.update(word) |
| 41 | |
| 42 | # Edge Case: More than 10 unique letters (impossible to assign unique digits) |
| 43 | if len(letters) > 10: |
| 44 | return None |
| 45 | |
| 46 | # Identify first letters (can't be zero) |
| 47 | first_letters = {word[0] for word in addends + [result]} |
| 48 | |
| 49 | # Convert words to lists of characters for easier processing |
| 50 | addend_chars = [list(word) for word in addends] |
| 51 | result_chars = list(result) |
| 52 | |
| 53 | # Prepare variables for backtracking |
| 54 | letters_list = list(letters) |
| 55 | digits = [None] * len(letters_list) # digits[i] corresponds to letters_list[i] |
| 56 | used_digits = [False] * 10 |
| 57 | |
| 58 | def is_valid_assignment(): |
| 59 | """Check if current digit assignment satisfies the equation.""" |
| 60 | # Build number mappings |
| 61 | letter_to_digit = {letters_list[i]: digits[i] for i in range(len(letters_list)) if digits[i] is not None} |
| 62 | |
| 63 | # Check for leading zeros |
| 64 | for word in addends + [result]: |
| 65 | # Edge Case: Leading zeros in multi-digit words |
| 66 | if len(word) > 1 and letter_to_digit[word[0]] == 0: |
| 67 | return False |
| 68 | |
| 69 | # Convert addends and result to numbers |
| 70 | try: |
| 71 | addend_nums = [] |
| 72 | for word_chars in addend_chars: |
| 73 | num = 0 |
| 74 | for char in word_chars: |
| 75 | num = num * 10 + letter_to_digit[char] |
| 76 | addend_nums.append(num) |
| 77 | |
| 78 | result_num = 0 |
| 79 | for char in result_chars: |
| 80 | result_num = result_num * 10 + letter_to_digit[char] |
| 81 | |
| 82 | return sum(addend_nums) == result_num |
| 83 | except KeyError: |
| 84 | # Not all letters have been assigned yet |
| 85 | return False |
| 86 | |
| 87 | def backtrack(index): |
| 88 | """Backtracking function to assign digits to letters.""" |
| 89 | # Base case: all letters have been assigned |
| 90 | if index == len(letters_list): |
| 91 | if is_valid_assignment(): |
| 92 | return True |
| 93 | return False |
| 94 | |
| 95 | letter = letters_list[index] |
| 96 | |
| 97 | # Try each digit from 0 to 9 |
| 98 | for digit in range(10): |
| 99 | # Skip if digit is already used |
| 100 | if used_digits[digit]: |
| 101 | continue |
| 102 | |
| 103 | # Skip if this is a first letter and digit is 0 |
| 104 | # Edge Case: Leading digit of a multi-digit number must not be zero |
| 105 | if letter in first_letters and digit == 0: |
| 106 | continue |
| 107 | |
| 108 | # Assign digit |
| 109 | digits[index] = digit |
| 110 | used_digits[digit] = True |
| 111 | |
| 112 | # Recurse |
| 113 | if backtrack(index + 1): |
| 114 | return True |
| 115 | |
| 116 | # Backtrack |
| 117 | digits[index] = None |
| 118 | used_digits[digit] = False |
| 119 | |
| 120 | return False |
| 121 | |
| 122 | # Start backtracking |
| 123 | if backtrack(0): |
| 124 | # Build and return the solution dictionary |
| 125 | solution = {letters_list[i]: digits[i] for i in range(len(letters_list))} |
| 126 | return solution |
| 127 | else: |
| 128 | # Edge Case: No solution exists |
| 129 | return None |