该算法查找前缀的时间复杂度

Question

This algorithm i wrote to check whether one string is a prefix for another string in the array. The complexity is O(n*(n-1)*k), where n is the number of strings in the array, and K is the largest length of the string. Am I right here in complexity analysis?

public static void isPrefix(String[] strs){

    for(int i=0; i<strs.length; i++){
        for(int j=i+1; j<strs.length; j++){
            String a = strs[i];
            String b = strs[j];

            if(!commonStr(a,b).isEmpty()){
                System.out.println(a + "->" + b);
            }
        }
    }
}

private static String commonStr(String a, String b){
    int smaller = Math.min(a.length(), b.length());
    for(int k=0; k<smaller; k++){
        if(a.charAt(k) != b.charAt(k)){
            return "";
        }
    }
    return a.substring(0,smaller);
}

public static void main(String[] args){
    String[] strs = {"ab", "abc", "cde", "abef"};
    isPrefix(strs);
}

Answer 1

You're right, I believe. It's just that K is not exactly that. But roughly speaking it's OK to say so.

Also it is K * n * (n-1) / 2 as you don't inspect all
ordered couples of strings (you inspect only half of them).
In your example, you inspect 6 couples, not 12.

Note that if your strings are say between 1 million and 2 million chars long,
but your n is just say 20 or 50 or 100, then K prevails and this estimate
is to be interpreted with care. Typically one would expect n >> K though,
I guess that's what you had in mind too.