https://github.com/tatsuyafujisaki/hackerrank-10-days-of-javascript-solutions

Solutions to 10 Days of JavaScript on HackerRank
https://github.com/tatsuyafujisaki/hackerrank-10-days-of-javascript-solutions

10-days-of-javascript hackerrank javascript

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Solutions to 10 Days of JavaScript on HackerRank

• Host: GitHub
• URL: https://github.com/tatsuyafujisaki/hackerrank-10-days-of-javascript-solutions
• Owner: tatsuyafujisaki
• Created: 2020-02-08T07:42:56.000Z (almost 5 years ago)
• Default Branch: main
• Last Pushed: 2021-01-07T13:36:15.000Z (about 4 years ago)
• Last Synced: 2024-12-08T21:55:44.232Z (about 2 months ago)
• Topics: 10-days-of-javascript, hackerrank, javascript
• Language: JavaScript
• Homepage: https://www.hackerrank.com/domains/tutorials/10-days-of-javascript
• Size: 57.6 KB
• Stars: 2
• Watchers: 0
• Forks: 0
• Open Issues: 0
• Metadata Files:
  • Readme: README.md

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README

        
# Explanation of `Day 6: Bitwise Operators`

* `a & b` is at least `k - 2` because:

  * If `k - 2` is even, the least significant bit (LSB) of `k - 2` is `0`. Assign `a = k - 2`.

  * If `k - 1` is even, LSB of `k - 1` is 0. Assign `a = k - 1`.

  * Either way, LSB of `a + 1` is `1`. Assign `b = a + 1`. So, `a & b == a`.

* `a & b` can be `k - 1` in the following case:

  * Given the constraints `a < b` and `a & b < k`, for `a & b == k - 1` to hold, `a = k - 1`.

  * For `((k - 1) & b) === k - 1` to hold, `b` is the same as `k - 1` except one bit, which is positioned at `a`'s any bit of 0, is 1.

  * To be precise, the exception bit of `b` must be positioned at `a`'s LSB of `0` to minimize the chance of `b` exceeding `n`. Such `b` is `(k - 1) | k`.

* In conclusion, `a & b = k - 1` when `a = k - 1` and `b = ((k - 1) | k) <= n`. Otherwise, `a & b = k - 2`.

A naive implementation of the above logic is:

```js

function getMaxLessThanK(n, k) {

  let a = k - 1;

  let b = (k - 1) | k;

  if (n < b) {

    a = k - 2;

    b = a + 1;

  }

  return a & b;

}

```

The shorter implementation is:

```js

function getMaxLessThanK(n, k) {

  return ((k - 1) | k) <= n ? (k - 1) : (k - 2);

}

```