860. Lemonade Change¶

Intuition¶

The problem requires determining if we can provide the correct change for each customer in a line, given that customers will pay with a \$5, \$10, or \$20 bill. The key is to ensure that for each \$10 or \$20 bill received, we have enough smaller bills (preferably \$5 bills) to give back the correct change.

Approach¶

Initialize two counters, five and ten, to keep track of the number of \$5 and \$10 bills we have.
Iterate through the array of bills:
If the bill is \$5, increment the five counter.
If the bill is \$10, check if we have at least one \$5 bill to give as change. If so, decrement the five counter and increment the ten counter. If not, return false.
If the bill is \$20, prioritize giving one \$10 and one \$5 as change (if possible). If not, try to give three \$5 bills. If neither is possible, return false.
If we can successfully provide change for all customers, return true.

Complexity¶

Time complexity: O(n)
We only loop through the array of bills once.
Space complexity: O(1)
Only a constant amount of extra space is used for the five and ten counters.

Code¶

class Solution {
    public boolean lemonadeChange(int[] bills) {
        int five = 0, ten = 0;

        for (int bill : bills) {
            if (bill == 5) {
                five++;
            } else if (bill == 10) {
                if (five == 0) return false;
                five--;
                ten++;
            } else { // bill == 20
                if (ten > 0 && five > 0) {
                    ten--;
                    five--;
                } else if (five >= 3) {
                    five -= 3;
                } else {
                    return false;
                }
            }
        }
        return true;
    }
}