Friday programming quiz
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maybe I've found an alternative inputs.Select(x => new { OriginalValue = x, Distance = Math.Abs(x - inputs.Average()) }).OrderBy(a => a.Distance).First().OriginalValue;
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I'm not familiar with linq nor lambda, but I really like this solution over the original suggested one. Am I right that t is an implied class that contains 2 fields? 1. It doesn't take work to understand the logic. 2. Rather than N^4 loops, this has N^2. (Like you said, not efficient, but much better than the original.)
Actually, if I was doing it again I would use Daniel Grunwald trick to calculate the average only once:
int closest = (from avg in new [] { inputs.Average() }
from i in inputs
orderby Math.Abs(i - avg)
select i).FirstOrDefault();Using the lambda format (as per the rules :) ) I believe this translates to:
int closest = new[] { inputs.Average() }
.SelectMany(avg => inputs.OrderBy(i => Math.Abs(i - avg)))
.FirstOrDefault();I believe this solution will iterate over the original collection twice - once for the average, once for the ordering, but have no idea what the big O number is for the ordering.
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I promised a programming quiz a couple weeks ago, so here's one: Given an array of integers, select the integer in the array that is the closest to the average of all integers in the array; you must use nothing but a single LINQ lambda statement. If more than one integer is the closest, then any of the closest integers will do. For example, the following code should assign "14" to the int "closest":
int[] inputs = {1, 2, 3, 5, -1, 7, 145, -33, 22, 14};
int closest = // put a LINQ lambda expression hereIf you're up for actually figuring out your own solution, ignore the rest of this post. OR If you're lazy and unethical, feel free to copy-paste the sample solution in the small text below and claim it as your own. inputs.Where(x => Math.Abs(x - inputs.Average()) == inputs.Select(y => Math.Abs(y - inputs.Average())).Min()).First();
We saw fairly eye to eye.. Here was my first go:
int closest = inputs.FirstOrDefault(num => Math.Abs((Math.Abs(num) - (int)inputs.Average())) == inputs.Min(diff => Math.Abs(Math.Abs(diff) - (int)(inputs.Average()))));
And with some visual optimization
int closest = inputs.FirstOrDefault(num => Math.Abs(num - inputs.Average()) == inputs.Min(diff => Math.Abs(diff - inputs.Average())));
[EDIT] And thinking a bit more about it
int closest = inputs.Select(input => new
{
input,
diff = Math.Abs(input - inputs.Average())
}).OrderBy(x => x.diff).FirstOrDefault().input; -
Actually, if I was doing it again I would use Daniel Grunwald trick to calculate the average only once:
int closest = (from avg in new [] { inputs.Average() }
from i in inputs
orderby Math.Abs(i - avg)
select i).FirstOrDefault();Using the lambda format (as per the rules :) ) I believe this translates to:
int closest = new[] { inputs.Average() }
.SelectMany(avg => inputs.OrderBy(i => Math.Abs(i - avg)))
.FirstOrDefault();I believe this solution will iterate over the original collection twice - once for the average, once for the ordering, but have no idea what the big O number is for the ordering.
I really can't judge LINQ, nor lambda statements because I don't have the experience with them. The original rule was "you must use nothing but a single LINQ lambda statement." Someone else complained that the original poster used two lambda statements. As far as I can tell, the original poster used two lambda variables in a single statement. My complaint about the original statement was that it was too hard to read. Then I complained about the efficiency of the original which would loop between N^3 and N^4 times. That was after taking the time to really understand what the first statement said. I came up with a set of C# commands that would find the answer in N*2 loops. Using two if statements and either one or three more statements per loop. Once I get over the "You define the variable, then you assign a statement's value to it" prejudice, your lambda statement is very readable, more elegant than my solution and with internal optimizations going on, your N*3 loops may be as fast or faster than my N*2 loops. I may even find the "Define the statement, then assign the variable" process enjoyable in time. Daniel still gets credit for first finding a good lambda expression. :)
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This little change you suggest puts the perfomance back to O(N^2) and not linear. inputs.Average() gets called for each element in the inputs array. Daniel's solution is the correct one. --- Adar Wesley