LeetCode #1307 — HARD

Verbal Arithmetic Puzzle

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

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.

Example 1:

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

Example 2:

Input: words = ["SIX","SEVEN","SEVEN"], result = "TWENTY"
Output: true
Explanation: Map 'S'-> 6, 'I'->5, 'X'->0, 'E'->8, 'V'->7, 'N'->2, 'T'->1, 'W'->'3', 'Y'->4
Such that: "SIX" + "SEVEN" + "SEVEN" = "TWENTY" ,  650 + 68782 + 68782 = 138214

Example 3:

Input: words = ["LEET","CODE"], result = "POINT"
Output: false
Explanation: There is no possible mapping to satisfy the equation, so we return false.
Note that two different characters cannot map to the same digit.

Constraints:

  • 2 <= words.length <= 5
  • 1 <= words[i].length, result.length <= 7
  • words[i], result contain only uppercase English letters.
  • The number of different characters used in the expression is at most 10.
Patterns Used
🔀Backtracking→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math · Backtracking

Example 1

["SEND","MORE"]
"MONEY"

Example 2

["SIX","SEVEN","SEVEN"]
"TWENTY"

Example 3

["LEET","CODE"]
"POINT"
Step 02

Core Insight

What unlocks the optimal approach

  • Use Backtracking and pruning to solve this problem.
  • If you set the values of some digits (from right to left), the other digits will be constrained.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Largest constraint values
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1307: Verbal Arithmetic Puzzle
class Solution {
    private boolean isAnyMapping(List<String> words, int row, int col, int bal,
        HashMap<Character, Integer> letToDig, char[] digToLet, int totalRows, int 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
                && isAnyMapping(
                    words, 0, col + 1, bal / 10, letToDig, digToLet, totalRows, totalCols));
        }

        String w = words.get(row);

        // If the current string 'w' has no character in the ('col')th index.
        if (col >= w.length()) {
            return isAnyMapping(words, row + 1, col, bal, letToDig, digToLet, totalRows, totalCols);
        }

        // Take the current character in the variable letter.
        char letter = w.charAt(w.length() - 1 - col);

        // Create a variable 'sign' to check whether we have to add it or subtract it.
        int sign = (row < totalRows - 1) ? 1 : -1;

        // If we have a prior valid mapping, then use that mapping.
        // The second condition is for the leading zeros.
        if (letToDig.containsKey(letter)
            && (letToDig.get(letter) != 0 || (letToDig.get(letter) == 0 && w.length() == 1)
                || col != w.length() - 1)) {

            return isAnyMapping(words, row + 1, col, bal + sign * letToDig.get(letter), letToDig,
                digToLet, totalRows, totalCols);

        } else {
            // Choose a new mapping.
            for (int i = 0; i < 10; i++) {
                // If 'i'th mapping is valid then select it.
                if (digToLet[i] == '-'
                    && (i != 0 || (i == 0 && w.length() == 1) || col != w.length() - 1)) {
                    digToLet[i] = letter;
                    letToDig.put(letter, i);

                    // Call the function again with the new mapping.
                    if (isAnyMapping(words, row + 1, col, bal + sign * letToDig.get(letter),
                            letToDig, digToLet, totalRows, totalCols)) {
                        return true;
                    }

                    // Unselect the mapping.
                    digToLet[i] = '-';
                    letToDig.remove(letter);
                }
            }
        }

        // If nothing is correct then just return false.
        return false;
    }

    public boolean isSolvable(String[] wordsArr, String result) {
        // Add the string 'result' in the list 'words'.
        List<String> words = new ArrayList<>();
        for (String word : wordsArr) {
            words.add(word);
        }
        words.add(result);

        int totalRows = words.size();

        // Find the longest string in the list and set 'totalCols' with the size of that string.
        int totalCols = 0;
        for (String word : words) {
            if (totalCols < word.length()) {
                totalCols = word.length();
            }
        }

        // Create a HashMap for the letter to digit mapping.
        HashMap<Character, Integer> letToDig = new HashMap<>();

        // Create a char array for the digit to letter mapping.
        char[] digToLet = new char[10];
        for (int i = 0; i < 10; i++) {
            digToLet[i] = '-';
        }

        return isAnyMapping(words, 0, 0, 0, letToDig, digToLet, totalRows, totalCols);
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n!)
Space
O(n)

Approach Breakdown

EXHAUSTIVE
O(nⁿ) time
O(n) space

Generate every possible combination without any filtering. At each of n positions we choose from up to n options, giving nⁿ total candidates. Each candidate takes O(n) to validate. No pruning means we waste time on clearly invalid partial solutions.

BACKTRACKING + PRUNING
O(n!) time
O(n) space

Backtracking explores a decision tree, but prunes branches that violate constraints early. Worst case is still factorial or exponential, but pruning dramatically reduces the constant factor in practice. Space is the recursion depth (usually O(n) for n-level decisions).

Shortcut: Backtracking time = size of the pruned search tree. Focus on proving your pruning eliminates most branches.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.

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