LeetCode #860 — EASY

Lemonade Change

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

At a lemonade stand, each lemonade costs $5. Customers are standing in a queue to buy from you and order one at a time (in the order specified by bills). Each customer will only buy one lemonade and pay with either a $5, $10, or $20 bill. You must provide the correct change to each customer so that the net transaction is that the customer pays $5.

Note that you do not have any change in hand at first.

Given an integer array bills where bills[i] is the bill the i^th customer pays, return true if you can provide every customer with the correct change, or false otherwise.

Example 1:

Input: bills = [5,5,5,10,20]
Output: true
Explanation: 
From the first 3 customers, we collect three $5 bills in order.
From the fourth customer, we collect a $10 bill and give back a $5.
From the fifth customer, we give a $10 bill and a $5 bill.
Since all customers got correct change, we output true.

Example 2:

Input: bills = [5,5,10,10,20]
Output: false
Explanation: 
From the first two customers in order, we collect two $5 bills.
For the next two customers in order, we collect a $10 bill and give back a $5 bill.
For the last customer, we can not give the change of $15 back because we only have two $10 bills.
Since not every customer received the correct change, the answer is false.

Constraints:

1 <= bills.length <= 10⁵
bills[i] is either 5, 10, or 20.

Step 01

Brute Force Baseline

Problem summary: At a lemonade stand, each lemonade costs $5. Customers are standing in a queue to buy from you and order one at a time (in the order specified by bills). Each customer will only buy one lemonade and pay with either a $5, $10, or $20 bill. You must provide the correct change to each customer so that the net transaction is that the customer pays $5. Note that you do not have any change in hand at first. Given an integer array bills where bills[i] is the bill the ith customer pays, return true if you can provide every customer with the correct change, or false otherwise.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Greedy

Example 1

[5,5,5,10,20]

Example 2

[5,5,10,10,20]

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #860: Lemonade Change
class Solution {
    public boolean lemonadeChange(int[] bills) {
        int five = 0, ten = 0;
        for (int v : bills) {
            switch (v) {
                case 5 -> ++five;
                case 10 -> {
                    ++ten;
                    --five;
                }
                case 20 -> {
                    if (ten > 0) {
                        --ten;
                        --five;
                    } else {
                        five -= 3;
                    }
                }
            }
            if (five < 0) {
                return false;
            }
        }
        return true;
    }
}

// Accepted solution for LeetCode #860: Lemonade Change
func lemonadeChange(bills []int) bool {
	five, ten := 0, 0
	for _, v := range bills {
		if v == 5 {
			five++
		} else if v == 10 {
			ten++
			five--
		} else {
			if ten > 0 {
				ten--
				five--
			} else {
				five -= 3
			}
		}
		if five < 0 {
			return false
		}
	}
	return true
}

# Accepted solution for LeetCode #860: Lemonade Change
class Solution:
    def lemonadeChange(self, bills: List[int]) -> bool:
        five = ten = 0
        for v in bills:
            if v == 5:
                five += 1
            elif v == 10:
                ten += 1
                five -= 1
            else:
                if ten:
                    ten -= 1
                    five -= 1
                else:
                    five -= 3
            if five < 0:
                return False
        return True

// Accepted solution for LeetCode #860: Lemonade Change
impl Solution {
    pub fn lemonade_change(bills: Vec<i32>) -> bool {
        let (mut five, mut ten) = (0, 0);
        for bill in bills.iter() {
            match bill {
                5 => {
                    five += 1;
                }
                10 => {
                    five -= 1;
                    ten += 1;
                }
                _ => {
                    if ten != 0 {
                        ten -= 1;
                        five -= 1;
                    } else {
                        five -= 3;
                    }
                }
            }

            if five < 0 {
                return false;
            }
        }
        true
    }
}

// Accepted solution for LeetCode #860: Lemonade Change
function lemonadeChange(bills: number[]): boolean {
    let [five, ten] = [0, 0];
    for (const x of bills) {
        switch (x) {
            case 5:
                five++;
                break;
            case 10:
                five--;
                ten++;
                break;
            case 20:
                if (ten) {
                    ten--;
                    five--;
                } else {
                    five -= 3;
                }
                break;
        }

        if (five < 0) {
            return false;
        }
    }
    return true;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.