Daily Challenge #240 - ATM

There's an ATM with unlimited money in bills of 10, 20, 50, 100, 200, and 500 dollars. You're given an amount of money n to withdraw with 1<=n<=1500.

Try to find the minimal number of bills that must be used to cash out n, or output -1 if it's impossible.

Example:
solve(1250) => 4 bills (500x2, 50x1, 200x1)
solve 1500 => 3 bills ($500x3)

Tests:
solve(10)
solve(550)
solve(770)

Good luck~!

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This challenge comes from Sanan07 on CodeWars. Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!

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Comments

[peter279k]

Here is Python solution:

def solve(n):
    count = 0
    while n - 500 >= 0:
        n -= 500
        count += 1
    while n - 200 >= 0:
        n -= 200
        count += 1
    while n - 100 >= 0:
        n -= 100
        count += 1
    while n - 50 >= 0:
        n -= 50
        count += 1
    while n - 20 >= 0:
        n -= 20
        count += 1
    while n - 10 >= 0:
        n -= 10
        count += 1

    if n != 0:
        return -1

    return count

[James Sinclair]

A Python solution.

from functools import reduce

sum_divmod = lambda acc, amt: (acc[0] + acc[1] // amt, acc[1] % amt)
def solve(amt):
    count, _ = reduce(sum_divmod, (500, 200, 100, 50, 20, 10), (0, amt))
    return count or -1

[Vidit Sarkar]

Here is a C++ solution,

int solve(int n){
    if(n < 10 || n > 1500 || n%10 != 0)
        return -1;

    int count = 0;
    for(int dollar : {500,200,100,50,20,10}){
        count += n/dollar;
        n %= dollar;
    }

    return count;
}

[Patrick Tingen]

Progress 4GL solution:

FUNCTION solve RETURNS INTEGER (amount AS INTEGER):
  DEFINE VARIABLE bill   AS INTEGER EXTENT 6 INITIAL [500,200,100,50,20,10] NO-UNDO.
  DEFINE VARIABLE needed AS INTEGER NO-UNDO.
  DEFINE VARIABLE i      AS INTEGER NO-UNDO.

  DO i = 1 TO EXTENT(bill):
    needed = needed + TRUNCATE(amount / bill[i],0).
    amount = amount - (bill[i] * TRUNCATE(amount / bill[i],0)).
  END.

  RETURN needed.
END FUNCTION. 

MESSAGE 
  solve(1250) SKIP
  solve(1500)
  VIEW-AS ALERT-BOX INFORMATION BUTTONS OK.

[Jeff Jadulco]

const bills = [500, 200, 100, 50, 20, 10]

const solve = (money) => {
  if (money < 1 || money > 1500) return -1

  let numBills = 0

  bills.forEach(bill => {
    const div = Math.floor(money / bill)
    money -= div * bill
    numBills += div
  })

  return numBills > 0 ? numBills : -1
}

[mxb]

Implementation in Frink

// https://dev.to/thepracticaldev/daily-challenge-240-atm-109c
solve[amount] :=
{
  bills = [500, 200, 100, 50, 20, 10]
  count = 0

  if amount < 1 or amount > 1500
    return -1

  for b = bills
  {
    nbills = floor[amount / b]
    amount = amount - (nbills b)
    count = count + nbills
  }

  return count
}