LeetCode 423: Reconstruct Original Digits from English Solution

Master LeetCode problem 423 (Reconstruct Original Digits from English), a medium 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.

423. Reconstruct Original Digits from English

Medium

Hash Table Math String

Problem Explanation

Explanation

To solve this problem, we can analyze the unique characters present in each number's English representation and use that information to reconstruct the original digits.

We can count the occurrences of unique characters in the input string s and then infer the count of each digit based on the presence of unique characters. For example, the digit '0' is uniquely represented by the character 'z', so if we find 'z' in the input string, we know there must be a '0' in the output. Similarly, the digit '2' is uniquely represented by the character 'w', and so on.

By identifying unique characters for each digit, we can determine the count of each digit in the output string.

Solution Code

class Solution {
    public String originalDigits(String s) {
        int[] count = new int[10];
        for (char ch : s.toCharArray()) {
            if (ch == 'z') count[0]++;
            if (ch == 'w') count[2]++;
            if (ch == 'x') count[6]++;
            if (ch == 'g') count[8]++;
            if (ch == 'u') count[4]++;
            if (ch == 'f') count[5]++; // also contains '4'
            if (ch == 'h') count[3]++; // also contains '8'
            if (ch == 's') count[7]++; // also contains '6'
            if (ch == 'i') count[9]++; // also contains '5', '6', '8'
            if (ch == 'o') count[1]++; // also contains '0', '2', '4'
        }
        
        count[5] -= count[4];
        count[3] -= count[8];
        count[7] -= count[6];
        count[9] -= (count[5] + count[6] + count[8]);
        count[1] -= (count[0] + count[2] + count[4]);
        
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < 10; i++) {
            for (int j = 0; j < count[i]; j++) {
                sb.append(i);
            }
        }
        
        return sb.toString();
    }
}

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