problem statement
In this problem you have to count how many of the characters of screen exist in dithered, to do this you can loop over the elements of screen and then for each element of screen search in dithered whether that character is there or not, O( n^3 ). But I do this O( n^2 ), since characters are only uppercase letters, at first I mark which characters are there in dithered, and then loop over screen and for each character it takes O( 1 ) to check whether it is in dithered or not.
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