937. Reorder Data in Log Files

time: O(nlogn), n is logs content length

space:(n)

class Solution {
    public String[] reorderLogFiles(String[] logs) {
        if(logs == null || logs.length == 0) return null;
        
        Arrays.sort(logs, (a,b) -> {
            String[] aStr = a.split(" ", 2); // string becomes 2 parts, [identifier, content]
            String[] bStr = b.split(" ", 2);
            boolean isADigit = Character.isDigit(aStr[1].charAt(0)); // check content is digit
            boolean isBDigit = Character.isDigit(bStr[1].charAt(0));
            
            if (!isADigit && !isBDigit) { // both string
                if (!aStr[1].equals(bStr[1])) { // if a String != b String return their compare
                    return aStr[1].compareTo(bStr[1]);
                } else {
                    return aStr[0].compareTo(bStr[0]); // a String == b String, use identifier to compare
                }
            } else if (isADigit && isBDigit) { // both digit, maintain original order
                return 0;
            } else if (!isADigit) { // -1 means a put first, so if a is char, put first
                return -1;
            } else {
                return 1;
            }
        });
        return logs;        
    }
    
}

/*
["identifier asdddsdsa fefwef", "identifier 1 2 3 4 4"]  

letter-logs before digital
sorted lexicographically by their contents, 
contents are the same, then sort them lexicographically by their identifiers.

[]
*/