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[COLOR=black]Given a list of [/COLOR][COLOR=black]N [/COLOR][COLOR=black]integers, the 3-sum problem is to determine whether there exists 3 integers in the list (not necessarily distinct) such that [/COLOR][COLOR=black]x + y + z = 0[/COLOR][COLOR=black]. Suppose that you are executing the brute force algorithm below on a computer that executes 1 billion operations … 
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[COLOR=#000000][/COLOR] [COLOR=#000000][/COLOR] [COLOR=#000000][/COLOR] [COLOR=#000000][/COLOR] [COLOR=#000000][/COLOR] [COLOR=#000000][int count(Node * list)[COLOR=windowtext][/COLOR][/COLOR] [COLOR=#000000]{ [/COLOR] [COLOR=#000000]if (list == Null) [/COLOR] [COLOR=#000000]return 0; [/COLOR] [COLOR=#000000]else [/COLOR] [COLOR=#000000]return 1 + count(list->next); [/COLOR] [COLOR=#000000]} [/COLOR] [COLOR=#000000]Write a recurrence relation for the function count.][COLOR=windowtext][/COLOR][/COLOR]
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[COLOR=#000000][/COLOR] [COLOR=#000000][/COLOR] [COLOR=#000000][/COLOR] [COLOR=#000000][function exp(k, n) [/COLOR] [COLOR=#000000]{ [/COLOR] [COLOR=#000000]power := 1; [/COLOR] [COLOR=#000000]for i := 1 to n do [/COLOR] [COLOR=#000000]power := power * k; [/COLOR] [COLOR=#000000]return(power) [/COLOR] [COLOR=black]}] [/COLOR]
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