What I've done today is Spiral primes in Rust and First Bad Version in JavaScript.

Math

Problem

Spiral primes

Problem 58

Starting with 1 and spiralling anticlockwise in the following way, a square spiral with side length 7 is formed.

37 36 35 34 33 32 31 38 17 16 15 14 13 30 39 18 5 4 3 12 29 40 19 6 1 2 11 28 41 20 7 8 9 10 27 42 21 22 23 24 25 26 43 44 45 46 47 48 49

It is interesting to note that the odd squares lie along the bottom right diagonal, but what is more interesting is that 8 out of the 13 numbers lying along both diagonals are prime; that is, a ratio of 8/13 ≈ 62%.

If one complete new layer is wrapped around the spiral above, a square spiral with side length 9 will be formed. If this process is continued, what is the side length of the square spiral for which the ratio of primes along both diagonals first falls below 10%?

Solution

Nothing to say.

Implementation

extern crate primal;

use primal::Sieve;

fn main() {
    let sieve: Sieve = Sieve::new(1_000_000_000);
    let mut cur: usize = 1;
    let mut spiral: usize = 1;
    let mut prime: usize = 0;
    let bound = 0.1;
    let ans: usize;
    loop {
        let ratio = prime as f64 / (spiral * 4 - 3) as f64;
        println!("{}, {}", spiral, ratio);
        if ratio < bound && spiral != 1 {
            ans = spiral * 2 - 1;
            break;
        }
        for _i in 0..4 {
            cur += spiral * 2;
            if sieve.is_prime(cur) {
                prime += 1;
            }
        }
        spiral += 1;
    }
    println!("Answer is {}", ans);
}

Algorithm

Problem

278. First Bad Version

You are a product manager and currently leading a team to develop a new product. Unfortunately, the latest version of your product fails the quality check. Since each version is developed based on the previous version, all the versions after a bad version are also bad.

Suppose you have n versions [1, 2, ..., n] and you want to find out the first bad one, which causes all the following ones to be bad.

You are given an API bool isBadVersion(version) which will return whether version is bad. Implement a function to find the first bad version. You should minimize the number of calls to the API.

Example:

Given n = 5, and version = 4 is the first bad version.

call isBadVersion(3) -> false
call isBadVersion(5) -> true
call isBadVersion(4) -> true

Then 4 is the first bad version. 

Solution

Simple binary search.

Implementation

var solution = function(isBadVersion) {
  /**
   * @param {integer} n Total versions
   * @return {integer} The first bad version
   */
  return function(n) {
    let begin = 1;
    let end = n;
    while (begin < end) {
      let mid = Math.floor((begin+end)/2);
      if (isBadVersion(mid)) end = mid;
      else begin = mid + 1;
    }
    return begin;
  };
};

;D