Leetcode #1307: Verbal Arithmetic Puzzle
In this guide, we solve Leetcode #1307 Verbal Arithmetic Puzzle in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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).
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Array, Math, String, Backtracking
Intuition
We must explore combinations of choices, but many branches can be pruned early.
Backtracking enumerates valid candidates while keeping the search space under control.
Approach
Use DFS to build candidates step by step, and backtrack when constraints are violated.
Pruning keeps the exploration practical for typical constraints.
Steps:
- Define the decision tree.
- DFS through choices and backtrack.
- Prune invalid paths early.
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'
Such that: "SEND" + "MORE" = "MONEY" , 9567 + 1085 = 10652
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
The time complexity is Exponential (worst case). The space complexity is O(depth).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.