I had a feeling I could get this answer in constant time, by just 'mathing' the value out! So I got out my good old pen and paper, and started working through it!

The repeating pattern of the list made me start thinking of the long list in cycles. Where one cycle was each person in line going as many times as they are supposed to in that cycle. I thought I would be able to determine how long the list would need to be to 'index' into it with the soda number, based on how many cycles it took! And we can! Since the cycles lengths follow the patter of: P, 2P, 3P, etc. I knew this could be represented as a summation formula, and I was pretty sure it had an algebraic solution! After a bit of googling I was able to find it! Applied to this problem the formula is as follows:

P = Number of People in the Line
C = Number of Cycles
Length = P / 2 * C * (C+1)

From here, I wanted to kinda reverse that formula. I wanted to know how many cycles it would take to achieve a specific length. After re-arranging that formula I was able to use the quadratic equation to rearrange, and solve for C, given Length and P. The quadratic formula gives two answers, and we are able to throw out one since we can guarantee it is negative and not relevant to our problem.
So now we have this!

C = (-1 + sqrt(1 + * (8 / P) * Length)) / 2

So now if we take the soda bottle number input as the Length, and the number of people in line as P, we can get a discrete value of the number of cycles it will take!

To figure out which person got the bottle I really only needed to index into the last cycle. And know I was able to figure out which cycle that was!
From here it was just some index math, to figure out which person/index we were looking for!

I wanted to make sure this solution worked, so I also wrote an iterative version to compare! I wrote a quick main function that compares the two results and they were equivalent for everything I tested! I also timed the two versions to see the speed differences, and as expected the constant time solution is orders of magnitudes faster than the iterative solution!

Math Took: 2.316656937049712 Iter Took: 77.97191874496184

Here is my full rust solution and tests!

fnfind_index_iter(num_people:u64,soda_num:u64)->u64{letmutcurrent_group_size=num_people;letmuttotal_size=num_people;letmuttotal_groups=1;whilesoda_num>total_size{current_group_size+=num_people;total_size+=current_group_size;total_groups+=1;}letlocal_group_index=current_group_size-(total_size-soda_num)-1;local_group_index/(total_groups+0)}fnfind_index_math(num_people:usize,soda_num:u64)->usize{letsoda_index=soda_num-1;letnumber_of_groups=((-1.0+(1.0+(8.0/num_peopleasf64*soda_numasf64)).sqrt())/2.0).ceil();letnum_before_last_group=num_peopleasf64/2.0*number_of_groups*(number_of_groups-1.0);letlocal_group_index=soda_indexasf64-num_before_last_group;letans=((local_group_index/number_of_groups).floor())asusize;ans}pubfnwhos_soda<'a>(people:&'a[&str],soda_num:u64)->&'astr{people[find_index_math(people.len(),soda_num)]// people[find_index_iter(people.len() as u64, soda_num) as usize]}#[cfg(test)]modtests{usecrate::*;#[test]fnit_for_first_few_groups_with_3_names(){assert_eq!(whos_soda(&["a","b","c"],1),"a");assert_eq!(whos_soda(&["a","b","c"],2),"b");assert_eq!(whos_soda(&["a","b","c"],3),"c");assert_eq!(whos_soda(&["a","b","c"],4),"a");assert_eq!(whos_soda(&["a","b","c"],5),"a");assert_eq!(whos_soda(&["a","b","c"],6),"b");assert_eq!(whos_soda(&["a","b","c"],7),"b");assert_eq!(whos_soda(&["a","b","c"],8),"c");assert_eq!(whos_soda(&["a","b","c"],9),"c");}#[test]fnit_for_first_few_groups_with_5_names(){assert_eq!(whos_soda(&["a","b","c","d","e"],1),"a");assert_eq!(whos_soda(&["a","b","c","d","e"],2),"b");assert_eq!(whos_soda(&["a","b","c","d","e"],3),"c");assert_eq!(whos_soda(&["a","b","c","d","e"],4),"d");assert_eq!(whos_soda(&["a","b","c","d","e"],5),"e");assert_eq!(whos_soda(&["a","b","c","d","e"],6),"a");assert_eq!(whos_soda(&["a","b","c","d","e"],7),"a");assert_eq!(whos_soda(&["a","b","c","d","e"],8),"b");assert_eq!(whos_soda(&["a","b","c","d","e"],9),"b");assert_eq!(whos_soda(&["a","b","c","d","e"],10),"c");assert_eq!(whos_soda(&["a","b","c","d","e"],11),"c");assert_eq!(whos_soda(&["a","b","c","d","e"],12),"d");assert_eq!(whos_soda(&["a","b","c","d","e"],13),"d");assert_eq!(whos_soda(&["a","b","c","d","e"],14),"e");assert_eq!(whos_soda(&["a","b","c","d","e"],15),"e");}}pubfnmain(){letmutcurrent_timing_iter=0.0;letmutcurrent_timing_math=0.0;forpeoplein1..10{foriin1..2000000{usestd::time::Instant;letnow=Instant::now();letmath_ans=find_index_math(peopleasusize,i);letelapsed=now.elapsed();letmath_sec=(elapsed.as_secs()asf64)+(elapsed.subsec_nanos()asf64/1000_000_000.0);letnow=Instant::now();letiter_ans=find_index_iter(people,i)asusize;letelapsed=now.elapsed();letiter_sec=(elapsed.as_secs()asf64)+(elapsed.subsec_nanos()asf64/1000_000_000.0);current_timing_math+=math_sec;current_timing_iter+=iter_sec;assert_eq!(iter_ans,math_ans,"{}x{}",people,i);}}println!("Math Took: {} Iter Took: {}",current_timing_math,current_timing_iter);}

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This was one of my favorites in a while!

I had a feeling I could get this answer in constant time, by just 'mathing' the value out! So I got out my good old pen and paper, and started working through it!

The repeating pattern of the list made me start thinking of the long list in cycles. Where one cycle was each person in line going as many times as they are supposed to in that cycle. I thought I would be able to determine how long the list would need to be to 'index' into it with the

`soda number`

, based on how many cycles it took! And we can! Since the cycles lengths follow the patter of: P, 2P, 3P, etc. I knew this could be represented as a summation formula, and I was pretty sure it had an algebraic solution! After a bit of googling I was able to find it! Applied to this problem the formula is as follows:From here, I wanted to kinda reverse that formula. I wanted to know how many cycles it would take to achieve a specific length. After re-arranging that formula I was able to use the quadratic equation to rearrange, and solve for C, given Length and P. The quadratic formula gives two answers, and we are able to throw out one since we can guarantee it is negative and not relevant to our problem.

So now we have this!

So now if we take the soda bottle number input as the Length, and the number of people in line as P, we can get a discrete value of the number of cycles it will take!

To figure out which person got the bottle I really only needed to index into the last cycle. And know I was able to figure out which cycle that was!

From here it was just some index math, to figure out which person/index we were looking for!

I wanted to make sure this solution worked, so I also wrote an iterative version to compare! I wrote a quick main function that compares the two results and they were equivalent for everything I tested! I also timed the two versions to see the speed differences, and as expected the constant time solution is orders of magnitudes faster than the iterative solution!

Here is my full rust solution and tests!