Not getting the binary search algorithm correct

Question

I am trying to implement binary search and I did the following:

function bs(a,x) {
    // a : array to look into
    // x : number to find
    let mpoint = Math.floor(a.length / 2);
    if(x >= a[mpoint]) {
        if(x == a[mpoint]) { return mpoint;}
        else {
            return bs([...a].slice(mpoint,a.length), x)
        }
    }else {
        if(x == a[mpoint]) {return mpoint;}
        else {
            return bs([...a].slice(0,mpoint),x)
        }
    }
}


bs([ 2, 3, 4, 10, 40 ], 10)

But I get an incorrect index as a result. What am I doing incorrectly?

The if inside the else makes no sense since there is no way x can be equal in that section. — epascarello
– epascarello, Commented Jun 17, 2019 at 13:03
The index is correct but apparently it is of the sliced array. So add the mpoint. — mighty_dev
– mighty_dev, Commented Jun 17, 2019 at 13:05
The [...a] spread literal is superfluous, slice creates a new array anyway — Bergi
– Bergi, Commented Jun 17, 2019 at 13:07

Shuki Avraham · Accepted Answer · 2019-06-17 13:03:28Z

4

Try to change:

return bs([...a].slice(mpoint,a.length), x)

to:

return bs([...a].slice(mpoint,a.length), x) + mpoint

answered Jun 17, 2019 at 13:03

Shuki Avraham

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