Discussion on: Coding Puzzles: Week of 4/8

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Tuesday (7 KYU): Find the stray number

def stray(arr):
    for item in arr:
        if arr.count(item) == 1:
            return item

Joshua Gilless • Apr 9 '19

Thanks for this! I don't really know C++, but I figured I'd give it a shot:

int stray(std::vector<int> numbers) {
    int n1, n2, n3;
    for (int i = 2; i < numbers.size(); i++) {
      n1 = numbers[i];
      n2 = numbers[i - 1];
      n3 = numbers[i - 2];
      if (n1 == n2 && n1 != n3) {
        return n3;
      } else if (n1 == n3 && n1 != n2) {
        return n2;
      } else if (n1 != n2 && n2 == n3) {
        return n1;
      }
    }
};

Joshua Gilless • Apr 9 '19 • Edited

@laurieontech did something really cool in JS with a bitwise XOR (her answer is above), so I figured I'd update this C++ answer with a bitwise XOR since I love it!

int stray(std::vector<int> numbers) {
    int r = numbers[0];
    for (int i = 1; i < numbers.size(); i++)
    {
        r ^= numbers[i];
    }
    return r;
};

russ • Apr 9 '19

Looks a bit messy, but I was going for something that might not be too inefficient with a large input array

def stray(arr):
    count = {}
    for index, i in enumerate(arr):
        count[i] = count.setdefault(i, 0) + 1
        if index >= 2 and len(count.keys()) > 1:
            break
    return next(k for k, v in count.items() if v == 1)

Jelle Bekker • Apr 10 '19

in c# using linq:

public static int Stray(int[] numbers)
{
    return numbers.Aggregate((x, y) => x ^ y);
}

or:

public static int Stray(int[] numbers)
{
    return numbers.GroupBy(x => x).Single(num => num.Count() == 1).Key;
}

Ahmed Musallam • Apr 10 '19 • Edited

here my cheap solution:

function stray(array) {
    var sorted = array.sort((a,b) => a - b)
    if (sorted[0] === sorted[1]) return sorted.pop();
    else return sorted.shift()
}

or shorter and more unreadable:

function stray(array) {
    var sorted = array.sort((a,b) => a - b)
    return sorted[0] === sorted[1] ? sorted.pop() : sorted.shift();
}

Courtney • Apr 10 '19 • Edited

This is the first I'm finding this, excited to play along! Here is yesterday's (edited to make the colors show up):

function stray(numbers) {
  let num1 = numbers[0], num2;
  let dupOf1 = false;
  for(let i=1; i<numbers.length; i++) {
      if(numbers[i] === num1) {
        if(num2) return num2;
        else dupOf1 = true;
      }
      else if(numbers[i] === num2) return num1;
      else num2 = numbers[i];
    }
  return dupOf1 ? num2 : num1;
}

Jacob Evans • Apr 24 '19 • Edited

I don't think this is better than the bitwise solution but it's a different one lol

const stray =(n)=> n.filter((ele,_,ar) => ele !== ar[1]).pop()

Laurie • Apr 9 '19

function stray(arr) {
    return arr.reduce((a, b) => a ^ b)
}

Joshua Gilless • Apr 9 '19

XOR is a great idea, thanks!

Laurie • Apr 9 '19

Math for the win! :D

Jacob Evans • Apr 24 '19

Well I can't think of a better answer for this particular problem domain.

russ • Apr 9 '19

What is this necromancy?

Laurie • Apr 9 '19 • Edited

Haha, bitwise xor. Since it's immutable and commutative it'll reduce down to the stray!

I should add that this only works because it’s an odd number of elements in the array. An even number of matching elements cancel each other out to result in the “stray”.

russ • Apr 9 '19

Oh my, it even works in python!

def stray(arr):
    count = {}
    for index, i in enumerate(arr):
        count[i] = count.setdefault(i, 0) + 1
        if index >= 2 and len(count.keys()) > 1:
            break
    return next(k for k, v in count.items() if v == 1)

from functools import reduce

def intstray(arr: [int]):
    return reduce(lambda x,y: x ^ y, arr)

assert(intstray([1, 1, 2]) == 2)
#nope..!
#assert(intstray([1, 1, 2, 1]) == 2)
assert(intstray([17, 17, 3, 17, 17, 17, 17]) == 3)
assert(intstray([1, 2, 2]) == 1)
assert(stray(["bob", "bob", "bob", "steve", "bob"]) == "steve")

Laurie • Apr 9 '19

This is one of those moments when I wish gifs worked better on Dev. But yay!

Ali Spittel • Apr 9 '19

They should work in normal image markdown!

Laurie • Apr 9 '19

Whaaa?! How did I not know this! ...game changer

Md Imrul Hassan • Apr 10 '19

Same idea in Haskell:

import Data.Bits

stray :: [Int] -> Int
stray = foldl1 xor