# 最好的選擇是不要浪費(fèi)餅干
class Solution:
    def findContentChildren(self, g: List[int], s: List[int]) -> int:
        # 先對胃口值和餅干尺寸排序
        g.sort()
        s.sort()
        g_l = len(g)
        g_index = 0
        s_l = len(s)
        s_index = 0
        # 計數(shù)
        count = 0
        # 終止條件：孩子數(shù) 和 餅干數(shù)是否在條件內(nèi)
        while g_index < g_l and s_index < s_l:
            # 胃口小于餅干
            if g[g_index] <= s[s_index]:
                # 餅干被消耗
                count += 1
                g_index += 1
                s_index += 1
            # 胃口大于餅干
            else:
                # 尋求更多的餅干滿足胃口
                s_index += 1
        return count

class Solution:
    def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
        # 判斷是否為空
        if not intervals:
            return 0
        # 對end值進(jìn)行升序排序
        intervals.sort(key = lambda x: x[1])
        # 維護(hù)一個最小值
        end_pos = intervals[0][1]
        # 只有單個區(qū)間時無重疊?。?！因此定義為1
        count = 1
        # 終止條件
        for i in range(1, len(intervals)):
            # 判斷是否連續(xù)
            if end_pos <= intervals[i][0]:
                count += 1
                end_pos = intervals[i][1]
        return len(intervals) - count

class Solution:
    def lemonadeChange(self, bills: List[int]) -> bool:
        five, ten, twenty = 0, 0, 0
        for bill in bills:
            if bill == 5:
                five += 1
            if bill == 10:
                # 是否可以找回
                if five <= 0:
                    return False
                # 收下 10 元
                ten += 1
                # 找回 5 元
                five -= 1
            if bill == 20:
                # 是否可以找回一張5元和一張10元
                if five > 0 and ten > 0:
                    five -= 1
                    ten -= 1
                    twenty += 1
                # 是否可以找回三張 5 元
                elif five >= 3:
                    five -= 3
                    twenty += 1
                else:
                    return False
        return True