小扣在秋日市集选择了一家早餐摊位,一维整型数组 staple
中记录了每种主食的价格,一维整型数组 drinks
中记录了每种饮料的价格。小扣的计划选择一份主食和一款饮料,且花费不超过 x
元。请返回小扣共有多少种购买方案。
注意:答案需要以 1e9 + 7 (1000000007)
为底取模,如:计算初始结果为:1000000008
,请返回 1
示例 1:
输入:
staple = [10,20,5], drinks = [5,5,2], x = 15
输出:
6
解释:小扣有 6 种购买方案,所选主食与所选饮料在数组中对应的下标分别是:
第 1 种方案:staple[0] + drinks[0] = 10 + 5 = 15;
第 2 种方案:staple[0] + drinks[1] = 10 + 5 = 15;
第 3 种方案:staple[0] + drinks[2] = 10 + 2 = 12;
第 4 种方案:staple[2] + drinks[0] = 5 + 5 = 10;
第 5 种方案:staple[2] + drinks[1] = 5 + 5 = 10;
第 6 种方案:staple[2] + drinks[2] = 5 + 2 = 7。
示例 2:
输入:
staple = [2,1,1], drinks = [8,9,5,1], x = 9
输出:
8
解释:小扣有 8 种购买方案,所选主食与所选饮料在数组中对应的下标分别是:
第 1 种方案:staple[0] + drinks[2] = 2 + 5 = 7;
第 2 种方案:staple[0] + drinks[3] = 2 + 1 = 3;
第 3 种方案:staple[1] + drinks[0] = 1 + 8 = 9;
第 4 种方案:staple[1] + drinks[2] = 1 + 5 = 6;
第 5 种方案:staple[1] + drinks[3] = 1 + 1 = 2;
第 6 种方案:staple[2] + drinks[0] = 1 + 8 = 9;
第 7 种方案:staple[2] + drinks[2] = 1 + 5 = 6;
第 8 种方案:staple[2] + drinks[3] = 1 + 1 = 2;
提示:
1 <= staple.length <= 10^5
1 <= drinks.length <= 10^5
1 <= staple[i],drinks[i] <= 10^5
1 <= x <= 2*10^5
pass 56 / 65
class Solution: def breakfastNumber(self, staple: List[int], drinks: List[int], x: int) -> int: staple.sort() drinks.sort() res=0 for i in range(len(staple)-1,-1,-1): if staple[i]+drinks[-1]<=x: res+=(i+1)*len(drinks) break if staple[i]>=x: continue for j in range(len(drinks)-1,-1,-1): if staple[i]+drinks[j]<=x: res+=j+1 break return int(res%(1000000007))
ac
class Solution: def breakfastNumber(self, staple: List[int], drinks: List[int], x: int) -> int: staple.sort() drinks.sort() res=0 i,j=0,len(drinks)-1 while i<len(staple) and j>=0: if staple[i]+drinks[j]>x: j-=1 else: res+=j+1 i+=1 return int(res%(1000000007))