860. Lemonade Change

Edited by Charles

Easy · Python · Greedy

Problem

Read the original problem on LeetCode.

The official page contains the problem statement, examples, and constraints.

Explanation, examples & other approaches

Core idea

Track the counts of $5 and $10 bills. Give change as each customer arrives, using a $10 bill before spending three $5 bills when possible.

Original study notes follow, including intermediate snippets and earlier attempts.

Note:

1. Iterate the list bills.
   1. If the current note is $5, collect it.
   2. If the current note is $10, check if we have at least one $5 note.
      1. If yes, collect it and dock one $5 note.
      2. If no, return False.
   3. If the current note is $20, check if we have at least one $5 note and one $10 note first.
      1. If yes, collect it and dock one $5 note and one $10 note.
      2. If no, check if we have at least 3 $5 notes.
         1. If yes, collect it and dock three $5 notes.
         2. If no, return False.
   4. The cases above include all scenarios for the first note.

class Solution:
    def lemonadeChange(self, bills: List[int]) -> bool:
        cash = {5: 0, 10: 0, 20:0}
        n = len(bills)

        for i in range(n):
            if bills[i] == 5:
                cash[5] += 1

            if bills[i] == 10:
                if cash[5] >= 1:
                    cash[5] -= 1
                    cash[10] += 1
                else:
                    return False

            if bills[i] == 20:
                if cash[10] >= 1 and cash[5] >= 1:
                    cash[20] += 1
                    cash[10] -= 1
                    cash[5] -= 1
                elif cash[10] == 0 and cash[5] >= 3:
                    cash[20] += 1
                    cash[5] -= 3
                else:
                    return False

        return True

Study record

Field	Value
Attempts	1
Mastery	❚
The Latest Attempt	January 28, 2026

Python solution

class Solution:
    def lemonadeChange(self, bills: List[int]) -> bool:
        five = ten = 0
        for b in bills:
            if b == 5:
                five += 1
            elif b == 10:
                five -= 1
                ten += 1

            # b = 20 case
            elif ten > 0:
                five -= 1
                ten -= 1
            else:
                five -= 3

            if five < 0:
                return False
        return True

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