# Javascript Big Combination Problem

A recent CodeSignal Challenge was to calculate 1000C500 (mod 1e9+7) and I got defeated =(

All my trials exceeded the time limit.. Here is the best JS solution by psr
, could anyone explain what happens in this line??? I learnt ES6 but got no idea about this syntax...

f[o = n + 1/k] = o in f


full solution for reference, please tell me to delete this if I violated any rule...

f = countWays = (n, k) => f[o = n + 1/k] = o in f
? f[o]
: k
? n && (f(--n, k) + f(n, k - 1)) % (1e9 + 7)
: 1


### Discussion Thanks Barbar's comments in StackOverflow, I understand the algorithm now.

The algorithm can be rewritten as follows:

f = nCk = (n, k) => {   //Declare both f and nCk as the same function
let o = n + 1/k         //o will be the key of function object f
f[o] = o in f           //Define f[o] based on a nested ternary expression
? f[o]              //Avoid recalculation if f has key o already
: k==0              //nC0 is always 1
? 1
: n<k           //nCk is improper and assigned 0 if n<k
? 0
: f(--n, k) //Do recursion nCk = (n-1)Ck + (n-1)C(k-1)
+ f(n, k - 1)
return f[o]             //Done!
}


Here goes an example of 5C2

f(n,k)  n   k   o   f[o]
f(5,2)  5   2   5.5 f[5.5]=f(4,2)+f(4,1)=10
f(4,2)  4   2   4.5 f[4.5]=f(3,2)+f(3,1)=6
f(3,2)  3   2   3.5 f[3.5]=f(2,2)+f(2,1)=3
f(2,2)  2   2   2.5 f[2.5]=f(1,2)+f(1,1)=1
f(1,2)  1   2   1.5 f[1.5]=f(0,2)+f(0,1)=0
f(0,2)  0   2   0.5 f[0.5]=0
f(0,1)  0   1   1   f=0
f(1,1)  1   1   2   f=f(0,1)+f(0,0)=1
f(0,0)  0   0   Inf f[Inf]=1 //Inf aka Infinity
f(2,1)  2   1   3   f=f(1,1)+f(1,0)=2
f(1,0)  1   0   Inf f[Inf]=1
f(3,1)  3   1   4   f=f(2,1)+f(2,0)=3
f(n,0)  n   0   Inf f[Inf]=1
f(4,1)  4   1   5   f=f(3,1)+f(3,0)=4


P.S. I got a few takeaways when investigating into this algorithm

1. Double declaration of function on the same line as a trick for recursion

2. Immediate use of a key with its value just assigned

3. Infinity can be used as a key of an object(!)

4. Syntax o in f checks if object f has the key o  