How do you solve Verbal Arithmetic Puzzle on LeetCode?

Verbal Arithmetic Puzzle (LeetCode #1307) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n!) time complexity and O(n) space complexity. Key insight: Each character must map to a unique digit.

What is the brute force solution for Verbal Arithmetic Puzzle?

The brute force approach for Verbal Arithmetic Puzzle is Brute Force, with O(n²) time complexity and O(1) space complexity. We can try every possible mapping of characters to digits and check if the equation holds. This is a straightforward method but can be very slow due to the large number of combinations. The optimal Optimal Solution approach improves this to O(n!).

What algorithm or data structure is used to solve Verbal Arithmetic Puzzle?

Verbal Arithmetic Puzzle uses the following concepts: Array, Math, String, Backtracking. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Verbal Arithmetic Puzzle?

Clearly explain your thought process while coding. Discuss edge cases and how your solution handles them. Practice backtracking problems to become comfortable with the technique.

What are common mistakes when solving Verbal Arithmetic Puzzle?

Not considering leading zeros when assigning digits. Failing to prune the search space effectively.

#1307

Verbal Arithmetic Puzzle

Hard

ArrayMathStringBacktrackingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n!)
Space	O(1)	O(n)

💡

Intuition

Time O(n!)Space O(n)

Using backtracking with pruning allows us to efficiently explore valid digit assignments while avoiding unnecessary checks. We can eliminate invalid configurations early, which speeds up the process.

⚙️

Algorithm

3 steps

1Step 1: Create a mapping of characters to digits and a set to track used digits.
2Step 2: Use backtracking to assign digits to characters, ensuring no leading zeros.
3Step 3: Check if the current assignment satisfies the equation. If it does, return true; if not, backtrack.

solution.py25 lines

1def is_solvable(words, result):
2    unique_chars = set(''.join(words) + result)
3    if len(unique_chars) > 10:
4        return False
5    char_to_digit = {}
6    return backtrack(words, result, char_to_digit, set())
7
8
9def backtrack(words, result, char_to_digit, used_digits):
10    if len(char_to_digit) == len(set(''.join(words) + result)):
11        words_sum = sum(int(''.join(str(char_to_digit[c]) for c in word)) for word in words)
12        result_num = int(''.join(str(char_to_digit[c]) for c in result))
13        return words_sum == result_num
14    for digit in range(10):
15        if digit in used_digits:
16            continue
17        for char in set(''.join(words) + result):
18            if char not in char_to_digit:
19                char_to_digit[char] = digit
20                used_digits.add(digit)
21                if backtrack(words, result, char_to_digit, used_digits):
22                    return True
23                del char_to_digit[char]
24                used_digits.remove(digit)
25    return False

ℹ

Complexity note: The time complexity is O(n!) due to the factorial growth of permutations for digit assignments, but pruning reduces the number of checks significantly.

1Each character must map to a unique digit.
2Leading zeros are not allowed for any word.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.