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Daily Challenge #275 - Casino Chips

thepracticaldev profile image staff ・1 min read

You are given three piles of casino chips: white, green and black chips in the form of an array. Each day you need exactly two chips of different colors to play at the casino. You can chose any color, but you are not allowed to use two chips of the same color in a single day.

You will be given an array representing the number of chips of each color and your task is to return the maximum number of days you can play.

solve([1,1,1]) = 1, because after you pick on day one, there will be only one chip left
solve([1,2,1] = 2, you can pick twice; you pick two chips on day one then on day two
solve([4,1,1]) = 2

Brute force is not the way to go here. Look for a simplifying mathematical approach.


Good luck!

This challenge comes from KenKamau on CodeWars. Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!

Want to propose a challenge idea for a future post? Email with your suggestions!

Discussion (37)

masterroshan profile image
TJ Johnson • Edited

This should do it

def solve(chips):
  max_chips = max(chips)
  remaining_count = sum(chips)
  return min(max_chips, remaining_count)
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EDIT: first solution does not work for all possible test cases, this should

def solve(chips):
    return min(sum(chips) >> 1, sum(chips[1:]))
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lopert profile image

Doesn't this fail the [12,12,12] input?

masterroshan profile image
TJ Johnson

The output of solve([12,12,12]) is 12. But you don't have to take my word for it, you can always throw it in the shell and see for yourself.

masterroshan profile image
TJ Johnson

ah I see the correct output should be 18

Thread Thread
lopert profile image

Yes exactly! I just stumbled upon your solution while looking through the comments for what values I should be outputting, and comparing answers to my code.

Your updated solution works great! I'm not sure I get the "math" side of it, as my solution is definitely brute force (recursively remove tokens until we can't anymore).

I get that your code sorts, then returns either the total number of chips / 2 OR the total number of chips in the smaller stacks, whichever is smaller.

Math-wise, I'll take a stab at understanding it.

If we have 1 larger / infinite stack, then we are limited by the other two. We pair a chip from the smaller stacks with one from the large stack everyday, and that's our total. That's the second half of the return statement.

However, if we have stacks that are somewhat even, we can rotate through and deplete them evenly. We code this by taking the total number of chips, and dividing them by the daily chip requirement (we can use the bitwise operator >> here since the daily chip requirement is 2).

Thread Thread
masterroshan profile image
TJ Johnson • Edited

I approached it like this: There's a case where it's better to split the highest stack in half to divvy between the lower 2 stacks, this only happens when the lower 2 stacks add up to be greater than the highest stack, then there are these "leftovers" in the stacks that you can combine.
I ended up with something looking like
high/2 + (mid + low)/2
which can be reduced mathematically to
(high + mid + low)/2
There was a point where it stopped making sense and I was just reducing the math and logic. This was a solution that helped me

arnoldarn profile image

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dora63366916 profile image
Dora • Edited

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3642066 profile image
David Augustus

worked on mobile

ava41452156 profile image

I agree with you. Gambling is very interesting but I personally prefer they have great welcome offer and and they support ma favorite pay pal;)

qm3ster profile image
Mihail Malo

"constant time" in Javascript

const solve = chips => {
  const [low, mid, high] = [...chips].sort()
  const extra = low & 1
  const rings = low - extra
  const tower = mid - rings
  const last = (high !== mid) & extra
  return rings * 3 + tower + last
  1. first we take whole rings, (0-1, 1-2, 2-0)
  2. then we take a the stack of lone pairs ("tower")
  3. finally we see if remainder from rings that's not in the "tower" can last us one more day with remainder of tower.
bubbler4 profile image

APL (using Dyalog APL):

      Solve 1 1 1
      Solve 1 2 1
      Solve 4 1 1
      Solve 12 12 12
      Solve 7 4 10

(Use the bookmarklet on this post to see the code with APL font and syntax highlighting.)

Explanation: Two possible answers are the sum of two lower values, and half of the entire sum (floored). We need to take the lower of the two.

(⌊2÷⍨+/)⌊(+/-⌈/)  ⍝ Input: an array of 3 numbers
        ⌊         ⍝ Take minimum of two sides...
                  ⍝  Choice 1:
          +/      ⍝   Sum (reduction / by addition +)
            -     ⍝   Minus
             ⌈/   ⍝   Max (reduction / by max-of-two ⌈)
                  ⍝  Choice 2:
     +/           ⍝   Sum
  2÷⍨             ⍝   Divided by 2
 ⌊                ⍝   Floor of the above
nocnica profile image
Nočnica Fee

Nice explanation!

qm3ster profile image
Mihail Malo


fn solve(mut chips: [u64; 3]) -> u64 {
    let [low, mid, high] = chips;
    let extra = low & 1;
    let rings = low - extra;
    let tower = mid - rings;
    let last = (high != mid) as u64 & extra;
    rings * 3 + tower + last

fn main() {
    assert_eq!(solve([1, 1, 1]), 1);
    assert_eq!(solve([1, 2, 1]), 2);
    assert_eq!(solve([4, 1, 1]), 2);
    dbg!(solve([8, 1, 4]));
    dbg!(solve([7, 4, 10]));
    dbg!(solve([12, 12, 12]));
    dbg!(solve([1, 23, 2]));

Look at it go!

Explaination in

saviourcode profile image
Sourabh Choure

In C with O(1):

#include <stdio.h>

int solve(int a,int b,int c)
    int Days;
    (a>=b)? (b>=c? (Days=c,c-=c):(Days=b,b-=b)):(a>=c? (Days=c,c-=c):(Days=a,a-=a));

    Days+=(a==0) ? (b>=c? c:b):((b==0) ? (a>=c? c:a):((a>=b) ? b:a));

    return Days;

int main(void)
    printf("Days: %d \n",solve(4,1,1));
    printf("Days: %d \n",solve(8,1,4));
    printf("Days: %d \n",solve(7,4,10));
    printf("Days: %d \n",solve(12,12,12));
    printf("Days: %d \n",solve(1,23,2));
    return 0;
harrisgeo88 profile image
Harris Geo 👨🏻‍💻

My Javascript solution

const solve = (chips) => {
  chips.sort((a,b) => a - b)
  let days = 0

  while(days >= 0) {
    if (chips[0] > 0) {
    } else if (chips[1] > 0 && chips[2] > 0) {
    } else {
      break; // no chips left! exit the loop

  return days
sommilee profile image
Lee Seng

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fusleor profile image
Fusleo Rigaton

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nicoray86 profile image
NicoRay86 • Edited

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tomwhite1122 profile image

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qasius profile image

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casinomentor profile image

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lucifarjamse00 profile image

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jillpattison6 profile image

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juanrodriguezarc profile image
Juan Rodríguez
function solve(arr){
    const [a,b,c] = arr.sort((a,b)=> b - a);
    const [x,y] = [a, b+c];
    const z = y > x ? Math.floor((x-y)/2): 0;
    return y + z;

a3 profile image

Go #GoLang

package main

import (

func main() {
    fmt.Println("Hello, playground");
    conins := []int{12,2,1}
    res := play(conins)

func play(c []int) int {
    if(c[0] == c[1] && c[1]==c[2] && c[0] == c[2]) {
        return int(math.Floor(float64(c[0])*1.5))
    return int(math.Min(float64(c[2]), float64(c[0]+c[1])));
minhtn12706743 profile image

code php

function solve(array $arr)
$maxDay = 0;
$checkDuplicateValue = array_unique($arr);
if($arr !== $checkDuplicateValue) {
$maxDay = $arr[0];
return $maxDay;
$maxDay = $arr[0] + 1;
return $maxDay;

savagepixie profile image


const solve = xs => {
  if (xs[0] === xs[1] && xs[0] === xs[2]) return Math.floor(xs[0] * 1.5)
  return Math.min(xs[2], xs[1] + xs[0])
3642066 profile image
David Augustus

its all flash
it sucks

kaus2595 profile image
shiaz profile image

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