Serve Customer Queue with Different Notes in Python

Suppose we have an array called notes represents different rupee notes holding by customers in a queue. They are all waiting to buy tickets worth Rs 50. Here possible notes are [50, 100 and 200]. We have to check whether we can sell tickets to the people in order or not, initially we have Rs 0 in our hand.

So, if the input is like notes = [50, 50, 100, 100], then the output will be True for first two we do not need to return anything, but now we have two Rs 50 notes. So for last two we can return them Rs 50 notes and sell all tickets in order.

To solve this, we will follow these steps −

• freq := an empty map
• i := 0
• while i < size of notes, do
  • if notes[i] is same as 50, then
    • freq[50] := freq[50] + 1
  • otherwise when notes[i] is same as 100, then
    • freq[100] := freq[100] + 1
    • if freq[50] is 0, then
      • come out from loop
    • freq[50] := freq[50] - 1
  • otherwise,
    • if freq[100] > 0 and freq[50] > 0, then
      • freq[100] := freq[100] - 1
      • freq[50] := freq[50] - 1
    • otherwise when freq[50] >= 3, then
      • freq[50] := freq[50] - 3
    • otherwise,
      • come out from loop
  • i := i + 1
• if i is same as size of notes , then
  • return True
• return False

Example

Let us see the following implementation to get better understanding −

 Live Demo

from collections import defaultdict
def solve(notes):
   freq = defaultdict(int)
   i = 0
   while i < len(notes):
      if notes[i] == 50:
         freq[50] += 1
      elif notes[i] == 100:
         freq[100] += 1
         if freq[50] == 0:
            break
         freq[50] -= 1
      else:
         if freq[100] > 0 and freq[50] > 0:
            freq[100] -= 1
            freq[50] -= 1
         elif freq[50] >= 3:
            freq[50] -= 3
         else:
            break
      i += 1
   if i == len(notes):
      return True
   return False
notes = [50, 50, 100, 100]
print(solve(notes))

Input

[50, 50, 100, 100]

Output

True