最大化接收巧克力的人数

给定两个数组A[] ，由N 个整数组成，B[]由 M 种巧克力的口味值和一个整数X 组成，任务是根据条件找到可以收到巧克力的最大人数：一个人只能拥有一种巧克力，其口味值在[A[i] – X, A[i] + X] 范围内。
注意：巧克力一旦送给一个人，就不能再送给任何其他人。

例子：

Input: A[] = {90, 49, 20, 39, 60}, B[] = {14, 24, 82}, X = 15
Output: 3
Explanation: 
1st person can pick the 3rd chocolate as the value of the 3rd chocolate ( = 82 ) lies in the range [75 ( = 90 – 15 ), 105 ( = 90 + 15)]. 
2nd person can’t pick any chocolate because there is no chocolate with value in the range [34 ( = 49 – 15 ), 64 ( = 49 + 15). 
3rd person can pick the 1st chocolate as the value of the 1st chocolate lies in the range [5 ( = 20 – 15), 35 ( = 20 – 15)]. 
4th person can pick the 2nd chocolate because value of the 2nd chocolate lies in the range [ 24 ( = 39 – 15) and 54 ( = 39 – 15)]. 
5th person can’t pick any chocolate because there is no chocolate with value in the range [45 ( = 60 – 15), 75 ( = 60 – 15)]. 
Therefore, the total number of people receiving a chocolate is 3, which is the maximum possible.

Input: A[] = {2, 4, 6, 40, 50}, B[] = {38, 36}, X=13
Output: 2

方法：这个问题可以使用贪心技术/a> 和搜索来解决。这里的关键观察是对于任何^第i 个人，如果可能，分配具有该范围内的最小可能值的巧克力。否则，从结果中排除该人。请按照以下步骤操作：

• 按非递减顺序对给定数组A[]和B[] 进行排序。
• 初始化一个多重集以存储数组B[]的元素。
• 初始化变量count = 0 ，以存储收到巧克力的人数。
• 遍历数组 A[] 并使用二分搜索或 lower_bound()^{找到可以分配给每个第 i 个人的最小巧克力值。}检查该值是否在[a _i – X, a _i + X]范围内。
• 如果发现为真，则增加count并从 multiset 中删除此巧克力的值。
• 完成以上步骤后，打印count的值作为结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to count the maximum number
// of persons receiving a chocolate
int countMaxPersons(int* A, int n, int* B,
                    int m, int x)
{
 
    // Initialize count as 0
    int count = 0;
 
    // Sort the given arrays
    sort(A, A + n);
    sort(B, B + m);
 
    // Initialize a multiset
    multiset s;
 
    // Insert B[] array values into
    // the multiset
    for (int i = 0; i < m; i++)
        s.insert(B[i]);
 
    // Traverse elements in array A[]
    for (int i = 0; i < n; i++) {
        int val = A[i] - x;
 
        // Search for the lowest value in B[]
        auto it = s.lower_bound(val);
 
        // If found, increment count,
        // and delete from set
        if (it != s.end()
            && *it <= A[i] + x) {
            count++;
            s.erase(it);
        }
    }
 
    // Return the number of people
    return count;
}
 
// Driver Code
int main()
{
    int A[] = { 90, 49, 20, 39, 49 };
    int B[] = { 14, 24, 82 };
    int X = 15;
    int N = sizeof(A) / sizeof(A[0]);
    int M = sizeof(B) / sizeof(B[0]);
 
    // Function Call
    cout << countMaxPersons(A, N, B, M, X);
}

Python3

# Python3 program for the above approach
from bisect import bisect_left
 
# Function to count the maximum number
# of persons receiving a chocolate
def countMaxPersons(A, n, B, m, x):
 
    # Initialize count as 0
    count = 0
 
    # Sort the given arrays
    A = sorted(A)
    B = sorted(B)
 
    # Initialize a multiset
    s = []
 
    # Insert B[] array values into
    # the multiset
    for i in range(m):
        s.append(B[i])
 
    # Traverse elements in array A[]
    for i in range(n):
        val = A[i] - x
 
        # Search for the lowest value in B[]
        it = bisect_left(s,val)
 
        # If found, increment count,
        # and delete from set
        if (it != len(s) and it <= A[i] + x):
            count += 1
            del s[it]
 
    # Return the number of people
    return count
 
# Driver Code
if __name__ == '__main__':
 
    A = [90, 49, 20, 39, 49]
    B = [14, 24, 82]
    X = 15
    N = len(A)
    M = len(B)
 
    # Function Call
    print(countMaxPersons(A, N, B, M, X))
 
    # This code is contributed by mohit kumar 29

输出：

3

时间复杂度： O(N*log N)
辅助空间： O(M)

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