Selecting rational numbers with maximum sum

Question

I have n rational numbers. Out of that I have to select m numbers such that

sum of numerators of m numbers /sum denominators of m numbers is maximum.

e.g. if I have 3 numbers 1/1, 1/2, 2/4 and I have to select 2 numbers. Then combinations will be

If 1/1, 1/2  are used then 1+1/1+2 = 2/3
If 1/1, 2/4 are used then  1+2/1+4=3/5
If 1/2, 2/4 are used then  1+2/2+4=3/6=1/2

Maximum is 2/3

Suppose I have array of n integers specifying numerators, and other array of n integers of denominators. And number m. What will be strategy ?

The numbers in input need not be reduced rational number. e.g a number can be 4/6 and not necessarily 2/3.

EDIT: A brute force solution will be try all permutations by selecting m numbers from n. And then apply above formula to find result and then see which combination gives maximum result.

So I want to know if there is any mathematical formula or property or a smarted way than brute force way.

1/1 + 2/4 = 4/4 + 2/4 = 6/4 = 1 2/4 = 1 1/2 How did you get 3/5? — SedJ601, Apr 16 '17 at 05:23
@SedrickJefferson (sum of numerators) / (sum of denominators) — Andrew Williamson, Apr 16 '17 at 05:24
Please post a [mcve] and highlight exactly where you are having a problem. — Erwin Bolwidt, Apr 16 '17 at 05:35
@trashgod I want to write a program. So I need the logic/mathematical property. Instead of brute force. — Kaushik Lele, Apr 16 '17 at 05:42
I think this is a pretty decent competitive programming question, took me a few minutes of good hard thinking. But it should probably go on cs.stackexchange.com — abhipil, Apr 16 '17 at 05:48
See [_mediant_](https://en.wikipedia.org/wiki/Mediant_%28mathematics%29). — trashgod, Apr 16 '17 at 05:50
If you are looking for a logic/mathematical answer, you should ask over at https://math.stackexchange.com/ — Andreas, Apr 16 '17 at 05:50
@Andreas asked there as will. Added here from compute programming perspective — Kaushik Lele, Apr 16 '17 at 05:56
I think its not permissible to discuss on going contest's problems: [link](https://www.hackerrank.com/contests/w31/challenges/spanning-tree-fraction) — Sanket Makani, Apr 16 '17 at 07:12
@KaushikLele That you are asking for the logic of a problem of an ongoing contest. — Sanket Makani, Apr 16 '17 at 08:16
@SanketMakani not logic, but basic mathematical concept which I am not aware of. As people suggested above it it is more of maths related question — Kaushik Lele, Apr 16 '17 at 08:18
@KaushikLele Still It is the main logic behind solving the mentioned problem of the contest and I think you should not ask this. — Sanket Makani, Apr 16 '17 at 08:20
Let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/141809/discussion-between-kaushik-lele-and-sanket-makani). — Kaushik Lele, Apr 16 '17 at 08:20

score 1 · Answer 1 · answered Apr 16 '17 at 05:40

I'll have a go at it.

Since we want to maximise the sum of numerators divided by denominators, we should choose those numbers where the difference between the numerator and the denominator is maximum. This would ensure that the maximum sum of numerator is chosen for the minimum sum of denominator for m numbers, which would give us the maximum value of the fraction we need.

For ex,

nums - 1/1, 1/2, 2/4
diff - 0  , -1 , -2
max is 2/3 using 1/1 and 1/2

Therefore, 1/1 and 1/2, would give us the max value.

If there is a tie, we can simply choose the fraction which has numerically larger numerators and denominators since that would increase the ratio.

For ex,

nums - 1/1, 1/2, 2/4, 2/3
diff - 0  , -1 , -2 , -1
max is 3/4 using 1/1 and 2/3

Hope it made sense.

that looks like the logic. I will try more test cases and accept the answer soon. — Kaushik Lele, Apr 16 '17 at 06:30
Fails in this testcase : `1/1 , 1/5, 50/100` and we need to select two numbers then your approach will select numbers : `1/1 , 1/5` which gives answer `2/6` but correct answer can be achieved by selection `1/1, 50/100` numbers which gives answer `51/101 `. — Sanket Makani, Apr 16 '17 at 07:14

Selecting rational numbers with maximum sum

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