JS | 位元運算 Bitwise Operator

Fri, 14 Feb 2020 10:52:38 +0800

The following is my solution regarding a CodeSignal Problem addTwoDigits

1st Draft (61 Chars) makes use of reduce

addTwoDigits=n=>
(""+n).split("").reduce((x,y)=>
x+Number(y)
,0)

2nd Draft (49 Chars) deploys a recursive skill

f = addTwoDigits = n => n<10
? n
: n%10 + f(Math.floor(n/10))

It turns out I ignored the constraints that the number has only 2 digits… but I learnt a new trick from the best solutionsss

addTwoDigits = n => n%10 + n/10|0

After some researches, there are quite some magical uses for the | bitwise operator, e.g. x|0 can function as:

Math.floor(x) for postive x
parseInt(x) for string x
0 for many special x (NaN, Infinity, null, undefined, [], {} …)

JS | 組合算法 Combination Algorithm

Wed, 15 Jan 2020 00:48:47 +0800

This is originally a post in Dev.to

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

Thanks Barbar’s comments in StackOverflow, I understand the algorithm now.

I have rewritten the algorithm 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 a walkthrough 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[1]=0
f(1,1) 1 1 2 f[2]=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[3]=f(1,1)+f(1,0) //=2
f(1,0) 1 0 Inf f[Inf]=1
f(3,1) 3 1 4 f[4]=f(2,1)+f(2,0) //=3
f(n,0) n 0 Inf f[Inf]=1
f(4,1) 4 1 5 f[5]=f(3,1)+f(3,0) //=4

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

Double declaration of function on the same line as a trick for recursion
Immediate use of a key with its value just assigned
Infinity can be used as a key of an object(!)
Syntax o in f checks if object f has the key o

數學 Math | 奧數牧人巷

JS | 位元運算 Bitwise Operator

JS | 組合算法 Combination Algorithm