What is the time complexity of Verbal Arithmetic Puzzle?

The optimal solution has a time complexity of O(10^n).

What data structures are used to solve Verbal Arithmetic Puzzle?

This problem uses Array, Math, String, Backtracking. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Verbal Arithmetic Puzzle?

Insight 1: Backtracking is necessary because we need to explore different digit assignments to letters and revert if a particular assignment doesn't lead to a valid solution. Insight 2: Avoiding leading zeros significantly reduces the search space and must be checked explicitly.

Which companies ask LeetCode 1307. Verbal Arithmetic Puzzle in interviews?

This problem is frequently asked at Atlassian, Wells Fargo.

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Verbal Arithmetic Puzzle - LeetCode 1307 Solution

Verbal Arithmetic Puzzle - Complete Solution Guide

Verbal Arithmetic Puzzle is LeetCode problem 1307, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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). No two characters can map to the same digit. Each words[i] and result are decoded as one number without leading zeros. Sum of numbers on the left side ( words ) will equal to the number on the right side ( result ). Return true if the equation is solvable, otherwise return false . Exampl

Detailed Explanation

The Verbal Arithmetic Puzzle problem asks us to determine if a given equation, where words represent numbers, can be solved by assigning digits (0-9) to the letters such that each letter corresponds to a unique digit, and the sum of the numbers represented by the words equals the number represented by the result. The numbers must not have leading zeros.

Solution Approach

The provided solution uses a backtracking algorithm to explore all possible digit assignments. It iterates through each column (digit place) of the equation from right to left. For each column, it calculates the sum of the digits corresponding to the letters in the words and checks if the sum modulo 10 matches the digit assigned to the letter in the result. Leading zero constraints are explicitly checked before assigning a digit. The algorithm maintains a mapping of characters to digits and a boolean array indicating which digits have already been used.

Step-by-Step Algorithm

Step 1: Reverse all words and the result string to facilitate column-wise processing, starting from the least significant digit.
Step 2: Identify all characters that appear as the leading character in any word or the result. These characters cannot be assigned the digit 0.
Step 3: Initialize a mapping to store character-digit assignments and a boolean array to track used digits.
Step 4: Implement a recursive function `solve(col, carry)` to process the equation column by column. `col` represents the current column being processed, and `carry` represents the carry-over from the previous column.
Step 5: In the `solve` function, iterate through each word using `process_word(col, word_idx, current_sum)`. If the character at the current column is already mapped, add its assigned digit to the `current_sum`. Otherwise, try all unassigned digits for the character, respecting the leading zero constraint. Recurse with the updated `current_sum`.
Step 6: In the `process_result` function, check if the sum of the digits from the words, along with the carry, equals the digit in the result string for the current column. If the character representing the result's digit is already assigned, check if it matches the calculated digit. If not, try assigning the calculated digit, respecting the leading zero and digit uniqueness constraints. If all conditions are met recurse deeper.
Step 7: If a valid assignment is found for all columns and the final carry is 0, return `true`. Otherwise, backtrack by resetting the assigned digit and marking it as unused. If no valid assignment is found, return `false`.

Key Insights

Insight 1: Backtracking is necessary because we need to explore different digit assignments to letters and revert if a particular assignment doesn't lead to a valid solution.
Insight 2: Avoiding leading zeros significantly reduces the search space and must be checked explicitly.
Insight 3: Reversing the words and result simplifies the summation process by processing columns from right to left, similar to manual addition.
Insight 4: Optimizing search by identifying letters that cannot be zero (leading characters) early

Complexity Analysis

Time Complexity: O(10^n)

Space Complexity: O(n)

Topics

This problem involves: Array, Math, String, Backtracking.

Companies

Asked at: Atlassian, Wells Fargo.

Verbal Arithmetic Puzzle | LeetCode 1307

Hard

Array Math String Backtracking

Time: O(10^n)Space: O(n)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Set;

class Solution {
    private String[] words;
    private String result;
    private Map<Character, Integer> mapping = new HashMap<>();
    private boolean[] used_digits = new boolean[10];
    private Set<Character> leading_chars = new HashSet<>();

    public boolean isSolvable(String[] words, String result) {
        // Optimization: The result cannot be shorter than the longest word.
        for (String word : words) {
            if (word.length() > result.length()) {
                return false;
            }
        }

        this.words = new String[words.length];
        for (int i = 0; i < words.length; i++) {
            this.words[i] = new StringBuilder(words[i]).reverse().toString();
        }
        this.result = new StringBuilder(result).reverse().toString();
        this.mapping = new HashMap<>();
        this.used_digits = new boolean[10];
        this.leading_chars = new HashSet<>();

        for (String word : words) {
            if (word.length() > 1) {
                this.leading_chars.add(word.charAt(word.length() - 1));
            }
        }
        if (result.length() > 1) {
            this.leading_chars.add(result.charAt(result.length() - 1));
        }

        return solve(0, 0);
    }

    private boolean solve(int col, int carry) {
        // Base case: Processed all columns up to the length of the result.
        if (col == result.length()) {
            return carry == 0;
        }

        // Recursively process each word for the current column.
        return process_word(col, 0, carry);
    }

    private boolean process_word(int col, int word_idx, int current_sum) {
        // Base case for this helper: All words processed for this column.
        if (word_idx == words.length) {
            return process_result(col, current_sum);
        }

        String word = words[word_idx];

        // If the current word is shorter than the column index, move to the next word.
        if (col >= word.length()) {
            return process_word(col, word_idx + 1, current_sum);
        }

        char char_ = word.charAt(col);

        if (mapping.containsKey(char_)) {
            int new_sum = current_sum + mapping.get(char_);
            return process_word(col, word_idx + 1, new_sum);
        } else {
            // Character is unassigned, try all unused digits.
            for (int digit = 0; digit < 10; digit++) {
                if (!used_digits[digit]) {
                    if (digit == 0 && leading_chars.contains(char_)) {
                        continue;
                    }

                    mapping.put(char_, digit);
                    used_digits[digit] = true;

                    if (process_word(col, word_idx + 1, current_sum + digit)) {
                        return true;
                    }

                    used_digits[digit] = false;
                    mapping.remove(char_);
                }
            }
            return false;
        }
    }

    private boolean process_result(int col, int words_sum) {
        char char_ = result.charAt(col);

        if (mapping.containsKey(char_)) {
            if (mapping.get(char_) == words_sum % 10) {
                return solve(col + 1, words_sum / 10);
            } else {
                return false;
            }
        } else {
            // Result character is unassigned; its digit is determined by the sum.
            int digit = words_sum % 10;
            if (!used_digits[digit]) {
                if (digit == 0 && leading_chars.contains(char_)) {
                    return false;
                }

                mapping.put(char_, digit);
                used_digits[digit] = true;

                if (solve(col + 1, words_sum / 10)) {
                    return true;
                }

                used_digits[digit] = false;
                mapping.remove(char_);
            }

            return false;
        }
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 1307: Verbal Arithmetic Puzzle - Detailed explanation and solution approaches.

Topics

Array Math String Backtracking

Companies

AtlassianWells Fargo

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:1307

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Reverse all words and the result string to facilitate column-wise processing, starting from the least significant digit.
2
Step 2: Identify all characters that appear as the leading character in any word or the result. These characters cannot be assigned the digit 0.
3
Step 3: Initialize a mapping to store character-digit assignments and a boolean array to track used digits.
4
Step 4: Implement a recursive function `solve(col, carry)` to process the equation column by column. `col` represents the current column being processed, and `carry` represents the carry-over from the previous column.
5
Step 5: In the `solve` function, iterate through each word using `process_word(col, word_idx, current_sum)`. If the character at the current column is already mapped, add its assigned digit to the `current_sum`. Otherwise, try all unassigned digits for the character, respecting the leading zero constraint. Recurse with the updated `current_sum`.
6
Step 6: In the `process_result` function, check if the sum of the digits from the words, along with the carry, equals the digit in the result string for the current column. If the character representing the result's digit is already assigned, check if it matches the calculated digit. If not, try assigning the calculated digit, respecting the leading zero and digit uniqueness constraints. If all conditions are met recurse deeper.
7
Step 7: If a valid assignment is found for all columns and the final carry is 0, return `true`. Otherwise, backtrack by resetting the assigned digit and marking it as unused. If no valid assignment is found, return `false`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(10^n)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is exponential because, for each unassigned character, we try up to 10 digits. The space complexity is linear because the data structures used (mapping, used_digits, reversed strings) grow linearly with the number of unique characters.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Constraint Satisfaction Problem (CSP)

The problem can be formulated as a CSP and solved using CSP solvers. This involves defining variables (characters), domains (0-9), and constraints (uniqueness, equation sum).

Time: Depends on the CSP solver's efficiency, but generally better than naive backtracking with pruning techniquesSpace: O(n), where n is the number of variables

Trade-offs: CSP solvers can be more efficient, especially for larger problems, but require more setup and might not be readily available in all programming languages. They often use techniques like arc consistency to prune the search space more effectively.

✨

Key Insights

1Insight 1: Backtracking is necessary because we need to explore different digit assignments to letters and revert if a particular assignment doesn't lead to a valid solution.
2Insight 2: Avoiding leading zeros significantly reduces the search space and must be checked explicitly.
3Insight 3: Reversing the words and result simplifies the summation process by processing columns from right to left, similar to manual addition.
4Insight 4: Optimizing search by identifying letters that cannot be zero (leading characters) early

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Leading zeros. The solution explicitly checks if a leading character is assigned 0. If so, it rejects the assignment.
Edge case 2: Result string shorter than the longest word. The solution checks this early and returns `false` immediately.
Edge case 3: No possible mapping to satisfy the equation. The backtracking algorithm explores all possible assignments and returns `false` if none of them work.
Edge case 4: Empty or Null input. The code assumes non-empty, valid inputs according to constraints given in the prompt. Additional checks could be added to handle these scenarios for robustness.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling leading zeros correctly. This can lead to incorrect results, especially when the leading character should be a non-zero digit.

✗Common mistake 2: Not backtracking properly when an assignment doesn't lead to a solution. This can cause the algorithm to explore invalid paths and miss valid solutions.

✗Common mistake 3: Not ensuring unique digit assignments. Two different characters shouldn't map to the same digit.

✗Common mistake 4: Overlooking the initial optimization that the result cannot be shorter than the longest input word.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Cryptography - Substitution ciphers can be seen as a simplified version of this problem, where characters are mapped to other characters (or digits in this case).

🏦Real-world scenario 2: Puzzle solving - This algorithm can be adapted to solve various logic puzzles that involve constraints and searching for valid assignments.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 18, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Verbal Arithmetic Puzzle"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether math techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?