Verbal Arithmetic Puzzle — LeetCode 1307 Python Solution
- Problem
- #1307
- Pattern
- Backtracking
- Reading time
- 17 min
- Source
- leetcode.com
The problem
Given an equation, represented by words on the left side and the result on the right side. You need to check if the equation is solvable under the following rules: Each character is decoded as one digit (0 - 9).
Example
- Input
- words = ["SEND","MORE"], result = "MONEY"
- Output
- true
- Explanation
- Map 'S'-> 9, 'E'->5, 'N'->6, 'D'->7, 'M'->1, 'O'->0, 'R'->8, 'Y'->'2'
Python solution
class Solution:
def isAnyMapping(
self, words, row, col, bal, letToDig, digToLet, totalRows, totalCols
):
# If traversed all columns.
if col == totalCols:
return bal == 0
# At the end of a particular column.
if row == totalRows:
return bal % 10 == 0 and self.isAnyMapping(
words, 0, col + 1, bal // 10, letToDig, digToLet, totalRows, totalCols
)
w = words[row]
# If the current string 'w' has no character in the ('col')th index.
if col >= len(w):
return self.isAnyMapping(
words, row + 1, col, bal, letToDig, digToLet, totalRows, totalCols
)
# Take the current character in the variable letter.
letter = w[len(w) - 1 - col]
# Create a variable 'sign' to check whether we have to add it or subtract it.
if row < totalRows - 1:
sign = 1
else:
sign = -1
# If we have a prior valid mapping, then use that mapping.
# The second condition is for the leading zeros.
if letter in letToDig and (
letToDig[letter] != 0
or (letToDig[letter] == 0 and len(w) == 1)
or col != len(w) - 1
):
return self.isAnyMapping(
words,
row + 1,
col,
bal + sign * letToDig[letter],
letToDig,
digToLet,
totalRows,
totalCols,
)
# Choose a new mapping.
else:
for i in range(10):
# If 'i'th mapping is valid then select it.
if digToLet[i] == "-" and (
i != 0 or (i == 0 and len(w) == 1) or col != len(w) - 1
):
digToLet[i] = letter
letToDig[letter] = i
# Call the function again with the new mapping.
if self.isAnyMapping(
words,
row + 1,
col,
bal + sign * letToDig[letter],
letToDig,
digToLet,
totalRows,
totalCols,
):
return True
# Unselect the mapping.
digToLet[i] = "-"
if letter in letToDig:
del letToDig[letter]
# If nothing is correct then just return false.
return False
def isSolvable(self, words, result):
# Add the string 'result' in the list 'words'.
words.append(result)
# Initialize 'totalRows' with the size of the list.
totalRows = len(words)
# Find the longest string in the list and set 'totalCols' with the size of that string.
totalCols = max(len(word) for word in words)
# Create a HashMap for the letter to digit mapping.
letToDig = {}
# Create a list for the digit to letter mapping.
digToLet = ["-"] * 10
return self.isAnyMapping(
words, 0, 0, 0, letToDig, digToLet, totalRows, totalCols
)Complexity
| Measure | Complexity |
|---|---|
| Time | Exponential (worst case) |
| Space | O(depth) auxiliary |
Pattern: Backtracking
Build candidates one choice at a time and abandon a branch the moment it cannot work. LeetCode 1307. Verbal Arithmetic Puzzle is filed here on both counts: the reference solution below belongs to the algorithm family this hub collects, and LeetCode tags it Backtracking.
The backtracking guide has the Python template for the pattern and the 105 LeetCode problems that use it.
Related problems
Frequently asked questions
- How hard is LeetCode 1307. Verbal Arithmetic Puzzle?
- LeetCode 1307. Verbal Arithmetic Puzzle is rated Hard on LeetCode.
- What topics does LeetCode 1307. Verbal Arithmetic Puzzle cover?
- LeetCode 1307. Verbal Arithmetic Puzzle is tagged Array, Math, String and Backtracking on LeetCode.