Woodstock Blog

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[Google] Max Prodcut of Strings That Have No Common Char



Given a dictionary of wrods,find the pair of word with following property:

  1. the two word don’t have same letter.

  2. the multiple of the two word’s length is maximum.

I give a simple O(nnk)(k is the average length of word) method.but i think there will be better one.


Best answer suggest as top comment:

Assuming the word is A-Z/a-z only, use a bitmap to set which letters it contains.

e.g. ca => 000….101

bb => 000…010

This is called Bitmask of string. Read [Question] Check string with no common letters (Bitmask).


Iterate over the words in decreasing order of length.

for each pair of words, AND the bitmaps.

Return the first pair that gives a 0 result.

This should be nk + nn

Be cautious

For the second part of the solution above, is this code going to work?

Arrays.sort(strs) in descending order;
for (int i = 0; i < strs.length; i++) {
    for (int j = 0; j < i; j++) {
        if (strs[i].bitmask & strs[j].bitmask == 0) {
            // this pair do not have common char
            // since strs in descending order, and i, j start from 0
            // the product of length should be max
            return i + ' ' + j;

Well, this is wrong. For example: {“ababa”, “aaa”, “bbb”, “cc”}, if we do longest-string to shorest-string, we would return “aaa”, “bbb” immediately when we found it. However, 5 * 2 > 3 * 3.

So, we have to find largest product using max-heap, like we did in [Google] Top N Values From Sum of 2 Arrays, “pop 1 and push 2”.

Another solution

DP: for each string, find the reversed set of char, and then find the max string using the reversed set. This idea is great, too, but less intuitive.

It is explained here.


not written