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LCP 18. 早餐组合

题目描述

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

解法

方法一

Python3

class Solution:
    def breakfastNumber(self, staple: List[int], drinks: List[int], x: int) -> int:
        res, n = 0, len(drinks)
        drinks.sort()
        for s in staple:
            remain = x - s
            if remain >= drinks[0]:
                left, right = 0, n - 1
                while left < right:
                    mid = (left + right + 1) >> 1
                    if drinks[mid] <= remain:
                        left = mid
                    else:
                        right = mid - 1
                res = (res + left + 1) % 1000000007
        return res

Java

class Solution {
    public int breakfastNumber(int[] staple, int[] drinks, int x) {
        int res = 0, n = drinks.length;
        Arrays.sort(drinks);
        for (int s : staple) {
            int remain = x - s;
            if (remain >= drinks[0]) {
                int left = 0, right = n - 1;
                while (left < right) {
                    int mid = (left + right + 1) >>> 1;
                    if (drinks[mid] <= remain) {
                        left = mid;
                    } else {
                        right = mid - 1;
                    }
                }
                res = (res + left + 1) % 1000000007;
            }
        }
        return res;
    }
}

C++

class Solution {
public:
    int breakfastNumber(vector<int>& staple, vector<int>& drinks, int x) {
        int res = 0, n = drinks.size();
        sort(drinks.begin(), drinks.end());
        for (int s : staple) {
            int remain = x - s;
            if (remain >= drinks[0]) {
                int left = 0, right = n - 1;
                while (left < right) {
                    int mid = left + right + 1 >> 1;
                    if (drinks[mid] <= remain)
                        left = mid;
                    else
                        right = mid - 1;
                }
                res = (res + left + 1) % 1000000007;
            }
        }
        return res;
    }
};

Go

func breakfastNumber(staple []int, drinks []int, x int) int {
	res, n := 0, len(drinks)
	sort.Ints(drinks)
	for _, s := range staple {
		remain := x - s
		if remain >= drinks[0] {
			left, right := 0, n-1
			for left < right {
				mid := (left + right + 1) >> 1
				if drinks[mid] <= remain {
					left = mid
				} else {
					right = mid - 1
				}
			}
			res = (res + left + 1) % 1000000007
		}
	}
	return res
}