Leecode swipe questions and record LCP 18 breakfast combination

topic

Xiaokou chooses a breakfast stand in the autumn market , One dimensional integer array staple The price of each staple food was recorded , One dimensional integer array drinks The price of each drink is recorded in . Small bucks plan to choose a staple food and a drink , And it doesn't cost more than x element . Please return the total number of purchase options for small deduction .

Be careful ： The answer needs to be 1e9 + 7 (1000000007) Take the mold for the bottom , Such as ： The initial result of calculation is ：1000000008, Please return 1

Example 1：

Input ：staple = [10,20,5], drinks = [5,5,2], x = 15

Output ：6

explain ： Little button 6 A purchase plan , The corresponding subscripts of the selected staple food and beverage in the array are respectively ：
The first 1 Kind of plan ：staple[0] + drinks[0] = 10 + 5 = 15;
The first 2 Kind of plan ：staple[0] + drinks[1] = 10 + 5 = 15;
The first 3 Kind of plan ：staple[0] + drinks[2] = 10 + 2 = 12;
The first 4 Kind of plan ：staple[2] + drinks[0] = 5 + 5 = 10;
The first 5 Kind of plan ：staple[2] + drinks[1] = 5 + 5 = 10;
The first 6 Kind of plan ：staple[2] + drinks[2] = 5 + 2 = 7.

Example 2：

Input ：staple = [2,1,1], drinks = [8,9,5,1], x = 9

Output ：8

explain ： Little button 8 A purchase plan , The corresponding subscripts of the selected staple food and beverage in the array are respectively ：
The first 1 Kind of plan ：staple[0] + drinks[2] = 2 + 5 = 7;
The first 2 Kind of plan ：staple[0] + drinks[3] = 2 + 1 = 3;
The first 3 Kind of plan ：staple[1] + drinks[0] = 1 + 8 = 9;
The first 4 Kind of plan ：staple[1] + drinks[2] = 1 + 5 = 6;
The first 5 Kind of plan ：staple[1] + drinks[3] = 1 + 1 = 2;
The first 6 Kind of plan ：staple[2] + drinks[0] = 1 + 8 = 9;
The first 7 Kind of plan ：staple[2] + drinks[2] = 1 + 5 = 6;
The first 8 Kind of plan ：staple[2] + drinks[3] = 1 + 1 = 2;

Tips ：

1 <= staple.length <= 10^5
1 <= drinks.length <= 10^5
1 <= staple[i],drinks[i] <= 10^5
1 <= x <= 2*10^5

answer

public class Solution {
    
    public int breakfastNumber(int[] staple, int[] drinks, int x)
    {
    
        Arrays.sort(staple);
        Arrays.sort(drinks);
        int count = 0;
        int j = drinks.length - 1;
        for(int i = 0;i < staple.length; i++)
        {
    
            if(staple[i] >= x) break;
            for(;j >= 0; j--)
            {
    
                if(staple[i] + drinks[j] <= x)
                {
    
                    count = (count + j + 1) % 1000000007;
                    break;
                }
            }
        }
        return count;
    }
}