Say I got a set of 10 random numbers between 0 and 100.
An operator gives me also a random number between 0 and 100. Then I got to find the number in the set that is
Someone tagged this question Mathematica, so here's a Mathematica answer:
set = {1,10,34,39,69,89,94,96,98,100};
opno = 45;
set[[Flatten[
Position[set - opno, i_ /; Abs[i] == Min[Abs[set - opno]]]]]]
It works when there are multiple elements of set equally distant from the operator number.
input
, create another array of the same size Math.abs(input[i] - operatorNumber)
k
)input[k]
NB
function closestTo(number, set) {
var closest = set[0];
var prev = Math.abs(set[0] - number);
for (var i = 1; i < set.length; i++) {
var diff = Math.abs(set[i] - number);
if (diff < prev) {
prev = diff;
closest = set[i];
}
}
return closest;
}
How about this:
1) Put the set into a binary tree.
2) Insert the operator number into the tree
3) Return the Operators parent
python example:
#!/usr/bin/env python
import random
from operator import itemgetter
sample = random.sample(range(100), 10)
pivot = random.randint(0, 100)
print 'sample: ', sample
print 'pivot:', pivot
print 'closest:', sample[
sorted(
map(lambda i, e: (i, abs(e - pivot)), range(10), sample),
key=itemgetter(1)
)[1][0]]
# sample: [61, 2, 3, 85, 15, 18, 19, 8, 66, 4]
# pivot: 51
# closest: 66
if set is ordered, do a binary search to find the value, (or the 2 values) that are closest. Then distinguish which of 2 is closest by ... subtracting?
If set is not ordered, just iterate through the set, (Sorting it would itself take more than one pass), and for each member, check to see if the difference is smaller than the smallest difference you have seen so far, and if it is, record it as the new smallest difference, and that number as the new candidate answer. .
public int FindClosest(int targetVal, int[] set)
{
int dif = 100, cand = 0;
foreach(int x in set)
if (Math.Abs(x-targetVal) < dif)
{
dif = Math.Abs(x-targetVal);
cand = x;
}
return cand;
}