Lemonade Change

Lemonade Change

Greedy — always return the largest bill first when making change.

Easy

Time O(n) Space O(1)

May 16, 2026 3 min read

Problem#

Customers pay $5,$ 10, or $20 for a$ 5 lemonade. You start with no change. Return true iff you can give exact change to every customer.

Examples#

[5,5,5,10,20] → true.
[5,5,10,10,20] → false.

Constraints#

1 <= bills.length <= 10^5.

Approach#

Track counts of $5 and$ 10 bills. On $10: need a$ 5. On $20: prefer one$ 10 + one $5; else three$ 5.

Solution#

class Solution {
public:
    bool lemonadeChange(vector<int>& bills) {
        int five = 0, ten = 0;
        for (int b : bills) {
            if (b == 5) ++five;
            else if (b == 10) {
                if (!five) return false;
                --five; ++ten;
            } else {
                if (ten && five) { --ten; --five; }
                else if (five >= 3) five -= 3;
                else return false;
            }
        }
        return true;
    }
};

func lemonadeChange(bills []int) bool {
    five, ten := 0, 0
    for _, b := range bills {
        switch b {
        case 5:
            five++
        case 10:
            if five == 0 {
                return false
            }
            five--
            ten++
        default:
            if ten > 0 && five > 0 {
                ten--
                five--
            } else if five >= 3 {
                five -= 3
            } else {
                return false
            }
        }
    }
    return true
}

from typing import List

class Solution:
    def lemonadeChange(self, bills: List[int]) -> bool:
        five, ten = 0, 0
        for b in bills:
            if b == 5:
                five += 1
            elif b == 10:
                if five == 0:
                    return False
                five -= 1
                ten += 1
            else:
                if ten > 0 and five > 0:
                    ten -= 1
                    five -= 1
                elif five >= 3:
                    five -= 3
                else:
                    return False
        return True

function lemonadeChange(bills) {
    let five = 0, ten = 0;
    for (const b of bills) {
        if (b === 5) {
            five++;
        } else if (b === 10) {
            if (five === 0) return false;
            five--;
            ten++;
        } else {
            if (ten > 0 && five > 0) {
                ten--;
                five--;
            } else if (five >= 3) {
                five -= 3;
            } else {
                return false;
            }
        }
    }
    return true;
}

class Solution {
    public boolean lemonadeChange(int[] bills) {
        int five = 0, ten = 0;
        for (int b : bills) {
            if (b == 5) {
                five++;
            } else if (b == 10) {
                if (five == 0) return false;
                five--;
                ten++;
            } else {
                if (ten > 0 && five > 0) {
                    ten--;
                    five--;
                } else if (five >= 3) {
                    five -= 3;
                } else {
                    return false;
                }
            }
        }
        return true;
    }
}

function lemonadeChange(bills: number[]): boolean {
    let five = 0, ten = 0;
    for (const b of bills) {
        if (b === 5) {
            five++;
        } else if (b === 10) {
            if (five === 0) return false;
            five--;
            ten++;
        } else {
            if (ten > 0 && five > 0) {
                ten--;
                five--;
            } else if (five >= 3) {
                five -= 3;
            } else {
                return false;
            }
        }
    }
    return true;
}

Editorial#

Preferring to give a $10 when possible preserves the more flexible$ 5s for cases that need them. This is the only safe choice — a $10 is useless for any non-$ 20 customer.

Complexity#

Time: O(n).
Space: O(1).

Problem#

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Approach#

Solution#

Editorial#

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