ReconstructOriginalDigitsFromEnglish423.java

/**
 * Given a non-empty string containing an out-of-order English representation
 * of digits 0-9, output the digits in ascending order.
 * 
 * Note:
 * 1. Input contains only lowercase English letters.
 * 2. Input is guaranteed to be valid and can be transformed to its original
 * digits. That means invalid inputs such as "abc" or "zerone" are not permitted.
 * 3. Input length is less than 50,000.
 * 
 * Example 1:
 * Input: "owoztneoer"
 * Output: "012"
 * 
 * Example 2:
 * Input: "fviefuro"
 * Output: "45"
 */


public class ReconstructOriginalDigitsFromEnglish423 {

    /**
     * https://leetcode.com/problems/reconstruct-original-digits-from-english/discuss/91207/one-pass-O(n)-JAVA-Solution-Simple-and-Clear
     */
    public String originalDigits(String s) {
        int[] count = new int[10];
        for (int i = 0; i < s.length(); i++){
            char c = s.charAt(i);
            if (c == 'z') count[0]++;
            if (c == 'w') count[2]++;
            if (c == 'x') count[6]++;
            if (c == 's') count[7]++; //7-6
            if (c == 'g') count[8]++;
            if (c == 'u') count[4]++; 
            if (c == 'f') count[5]++; //5-4
            if (c == 'h') count[3]++; //3-8
            if (c == 'i') count[9]++; //9-8-5-6
            if (c == 'o') count[1]++; //1-0-2-4
        }
        count[7] -= count[6];
        count[5] -= count[4];
        count[3] -= count[8];
        count[9] = count[9] - count[8] - count[5] - count[6];
        count[1] = count[1] - count[0] - count[2] - count[4];
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i <= 9; i++){
            for (int j = 0; j < count[i]; j++){
                sb.append(i);
            }
        }
        return sb.toString();
    }

}