if your dictionary is sorted, you could create a "map" of your dictionary with binary search: e.g. search for "b", "c" etc (most implementations return the position of the query in the dict., even when query is not found). it is log(n). so m * log(n) for the map and m * 100 for the word selection.

your solution would be m * n.

In reply to Re: improving the efficiency of a script by lima1
in thread improving the efficiency of a script by sulfericacid

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