给定两个大小为N 的数组A[]和B[] ，任务是计算对的最大数量，其中每对包含来自每个数组的一个，使得A[i] > B[i] 。此外，阵列 A 可以重新排列任意次数。
例子：
Input: A[] = {20, 30, 50}, B[]= {60, 40, 25}
Output: 2
Explanation:
Initially:
A[0] = 20 < B[0] = 60
A[1] = 30 < B[1] = 40
A[2] = 50 > B[2] = 25
Clearly, this arrangement has only 1 value such that A[i] > B[i].
This array A[] when rearranged to {20, 50, 30}:
A[0] = 20 < B[0] = 60
A[1] = 50 > B[1] = 40
A[2] = 30 > B[2] = 25
2 values follow the condition A[i] > B[i] which is the maximum for these set of arrays.
Input: A[] = {10, 3, 7, 5, 8}, B[] = {8, 6, 2, 5, 9}
Output: 4
Explanation:
Initially:
A[0] = 10 > B[0] = 8
A[1] = 3 < B[1] = 6
A[2] = 7 > B[2] = 2
A[3] = 5 = B[3] = 5
A[4] = 8 < B[4] = 9
Clearly, this arrangement has only 2 values such that A[i] > B[i].
This array A[] when rearranged to {10, 8, 5, 7, 3}:
A[0] = 10 > B[0] = 8
A[1] = 8 > B[1] = 6
A[2] = 5 > B[2] = 2
A[3] = 7 > B[3] = 5
A[4] = 3 < B[4] = 9
4 values follow the condition A[i] > B[i] which is the maximum for these set of arrays.
方法：思路是使用堆的概念。由于 B[] 的排列在问题中无关紧要，我们可以对两个数组执行最大堆。在执行 max heap 并将值存储在两个不同的堆中后，遍历对应于 A[] 和 B[] 的堆以计算满足给定条件A[i] > B[i]的索引数量。
下面是上述方法的实现：
C++14
// C++ program to find the maximum count of
// values that follow the given condition
#include
using namespace std;
// Function to find the maximum count of
// values that follow the given condition
int check(int A[], int B[], int N)
{
// Initializing the max-heap for the array A[]
priority_queue pq1,pq2;
// Adding the values of A[] into max heap
for (int i = 0; i < N; i++) {
pq1.push(A[i]);
}
// Adding the values of B[] into max heap
for (int i = 0; i < N; i++) {
pq2.push(B[i]);
}
// Counter variable
int c = 0;
// Loop to iterate through the heap
for (int i = 0; i < N; i++) {
// Comparing the values at the top.
// If the value of heap A[] is greater,
// then counter is incremented
if (pq1.top()>pq2.top()) {
c++;
pq1.pop();
pq2.pop();
}
else {
if (pq2.size() == 0) {
break;
}
pq2.pop();
}
}
return (c);
}
// Driver code
int main()
{
int A[] = { 10, 3, 7, 5, 8 };
int B[] = { 8, 6, 2, 5, 9 };
int N = sizeof(A)/sizeof(A[0]);
cout<<(check(A, B, N));
}
// This code is contributed by mohit kumar 29 Java
// Java program to find the maximum count of
// values that follow the given condition
import java.util.*;
public class GFG {
// Function to find the maximum count of
// values that follow the given condition
static int check(int A[], int B[], int N)
{
// Initializing the max-heap for the array A[]
PriorityQueue pq1
= new PriorityQueue(
Collections.reverseOrder());
// Initializing the max-heap for the array B[]
PriorityQueue pq2
= new PriorityQueue(
Collections.reverseOrder());
// Adding the values of A[] into max heap
for (int i = 0; i < N; i++) {
pq1.add(A[i]);
}
// Adding the values of B[] into max heap
for (int i = 0; i < N; i++) {
pq2.add(B[i]);
}
// Counter variable
int c = 0;
// Loop to iterate through the heap
for (int i = 0; i < N; i++) {
// Comparing the values at the top.
// If the value of heap A[] is greater,
// then counter is incremented
if (pq1.peek().compareTo(pq2.peek()) == 1) {
c++;
pq1.poll();
pq2.poll();
}
else {
if (pq2.size() == 0) {
break;
}
pq2.poll();
}
}
return (c);
}
// Driver code
public static void main(String args[])
{
int A[] = { 10, 3, 7, 5, 8 };
int B[] = { 8, 6, 2, 5, 9 };
int N = A.length;
System.out.println(check(A, B, N));
}
} Python3
# Python3 program to find the maximum count of
# values that follow the given condition
import heapq
# Function to find the maximum count of
# values that follow the given condition
def check(A, B,N):
# Initializing the max-heap for the array A[]
pq1 = []
pq2 = []
# Adding the values of A[] into max heap
for i in range(N):
heapq.heappush(pq1,-A[i])
# Adding the values of B[] into max heap
for i in range(N):
heapq.heappush(pq2,-B[i])
# Counter variable
c = 0
# Loop to iterate through the heap
for i in range(N):
# Comparing the values at the top.
# If the value of heap A[] is greater,
# then counter is incremented
if -pq1[0] > -pq2[0]:
c += 1
heapq.heappop(pq1)
heapq.heappop(pq2)
else:
if len(pq2) == 0:
break
heapq.heappop(pq2)
return (c)
# Driver code
A = [ 10, 3, 7, 5, 8 ]
B = [ 8, 6, 2, 5, 9 ]
N = len(A)
print(check(A, B, N))
# This code is contributed by apurva rajC#
// C# program to find the maximum count of
// values that follow the given condition
using System;
using System.Collections.Generic;
class GFG{
// Function to find the maximum count of
// values that follow the given condition
static int check(int[] A, int[] B, int N)
{
// Initializing the max-heap for the array A[]
List pq1 = new List();
// Initializing the max-heap for the array B[]
List pq2 = new List();
// Adding the values of A[] into max heap
for(int i = 0; i < N; i++)
{
pq1.Add(A[i]);
}
// Adding the values of B[] into max heap
for(int i = 0; i < N; i++)
{
pq2.Add(B[i]);
}
pq1.Sort();
pq1.Reverse();
pq2.Sort();
pq2.Reverse();
// Counter variable
int c = 0;
// Loop to iterate through the heap
for(int i = 0; i < N; i++)
{
// Comparing the values at the top.
// If the value of heap A[] is greater,
// then counter is incremented
if (pq1[0] > pq2[0])
{
c++;
pq1.RemoveAt(0);
pq2.RemoveAt(0);
}
else
{
if (pq2.Count == 0)
{
break;
}
pq2.RemoveAt(0);
}
}
return c;
}
// Driver code
static public void Main()
{
int[] A = { 10, 3, 7, 5, 8 };
int[] B = { 8, 6, 2, 5, 9 };
int N = A.Length;
Console.WriteLine(check(A, B, N));
}
}
// This code is contributed by avanitrachhadiya2155 Javascript
4时间复杂度： O(N * log(N))
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