LeetCode 1307: Verbal Arithmetic Puzzle Solution

Master LeetCode problem 1307 (Verbal Arithmetic Puzzle), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1307. Verbal Arithmetic Puzzle

Hard

Array Math String Backtracking

Problem Explanation

Explanation:

To solve this problem, we can use backtracking to try out different mappings of characters to digits. At each step, we assign a digit to a character and recursively check if the current assignment leads to a valid solution. We backtrack if the assignment doesn't work out.

Here are the detailed steps:

Create a mapping to store the mapping of characters to digits.
Implement a recursive backtracking function that tries out different mappings.
Iterate through all the characters in the equation and update the mapping.
Check if the current mapping is valid by verifying the sum of words equals the result.
If the mapping is valid, return true. Otherwise, backtrack and try another mapping.
Finally, return false if no valid mapping is found.

Time Complexity:

The time complexity of this solution is O(10^N), where N is the number of different characters used in the expression.

Space Complexity:

The space complexity is O(1) since we are using a fixed size mapping array.

Solution Code

class Solution {
    public boolean isSolvable(String[] words, String result) {
        return backtrack(words, result, new int[26], 0, 0, 0);
    }

    private boolean backtrack(String[] words, String result, int[] mapping, int index, int sum, int carry) {
        if (index == result.length()) {
            for (String word : words) {
                int wordSum = 0;
                for (int i = 0; i < word.length(); i++) {
                    wordSum = wordSum * 10 + mapping[word.charAt(i) - 'A'];
                }
                if (wordSum != sum) {
                    return false;
                }
            }
            return true;
        }

        if (index < result.length()) {
            int resDigit = result.charAt(index) - 'A';
            if (mapping[resDigit] != 0) {
                return backtrack(words, result, mapping, index + 1, sum, carry);
            }
            for (int d = 0; d < 10; d++) {
                if (d == 0 && resDigit == result.charAt(0) - 'A') {
                    continue;
                }
                if (mapping[resDigit] == 0 && mapping[d] == 0) {
                    mapping[resDigit] = d;
                    if (backtrack(words, result, mapping, index + 1, sum + d, carry)) {
                        return true;
                    }
                    mapping[resDigit] = 0;
                }
            }
        }

        return false;
    }
}

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