给定一个由 N 个正元素组成的数组，找到四元组的数量 (i, j, k, m)，使得 i < j < k < m 使得积a i a j a k a m是可能的最大值。
例子：
Input : N = 7, arr = {1, 2, 3, 3, 3, 3, 5}
Output : 4
Explanation
The maximum quadruple product possible is 135, which can be
achieved by the following quadruples {i, j, k, m} such that aiajakam = 135:
1) a3a4a5a7
2) a3a4a6a7
3) a4a5a6a7
4) a3a5a6a7
Input : N = 4, arr = {1, 5, 2, 1}
Output : 1
Explanation
The maximum quadruple product possible is 10, which can be
achieved by the following quadruple {1, 2, 3, 4} as a1a2a3a4 = 10蛮力：O(n 4 )
生成所有可能的四元组并计算四元组，得出最大乘积。
优化方案：
很容易看出，四个最大数的乘积将是最大值。因此，现在可以将问题简化为找到选择四个最大元素的方法的数量。为此，我们维护一个频率数组，用于存储数组中每个元素的频率。
假设最大元素是X，频率为F X ，那么如果该元素的频率>= 4，则最适合选择X, X, X 四个元素，因为给定最大乘积和数量这样做的方法是F X C 4
如果频率小于 4，则选择它的方式数为 1，现在所需的元素数为 4 – F X 。对于第二个元素，比如说 Y，方法的数量是： F X C剩余选择。剩余的选择表示在选择第一个元素后我们需要选择的附加元素的数量。如果在任何时候，residual_choices = 0，则表示选择了四元组，因此我们可以停止算法。
C++
// CPP program to find the number of Quadruples
// having maximum product
#include
using namespace std;
// Returns the number of ways to select r objects
// out of available n choices
int NCR(int n, int r)
{
int numerator = 1;
int denominator = 1;
// ncr = (n * (n - 1) * (n - 2) * .....
// ... (n - r + 1)) / (r * (r - 1) * ... * 1)
while (r > 0) {
numerator *= n;
denominator *= r;
n--;
r--;
}
return (numerator / denominator);
}
// Returns the number of quadruples having maximum product
int findWays(int arr[], int n)
{
// stores the frequency of each element
map count;
if (n < 4)
return 0;
for (int i = 0; i < n; i++) {
count[arr[i]]++;
}
// remaining_choices denotes the remaining
// elements to select inorder to form quadruple
int remaining_choices = 4;
int ans = 1;
// traverse the elements of the map in reverse order
for (auto iter = count.rbegin(); iter != count.rend(); ++iter) {
int number = iter->first;
int frequency = iter->second;
// If Frequency of element < remaining choices,
// select all of these elements, else select only
// the number of elements required
int toSelect = min(remaining_choices, frequency);
ans = ans * NCR(frequency, toSelect);
// Decrement remaining_choices acc to the number
// of the current elements selected
remaining_choices -= toSelect;
// if the quadruple is formed stop the algorithm
if (!remaining_choices) {
break;
}
}
return ans;
}
// Driver Code
int main()
{
int arr[] = { 1, 2, 3, 3, 3, 5 };
int n = sizeof(arr) / sizeof(arr[0]);
int maxQuadrupleWays = findWays(arr, n);
cout << maxQuadrupleWays;
return 0;
} Java
// Java program to find the number of Quadruples
// having maximum product
import java.util.*;
class Solution
{
// Returns the number of ways to select r objects
// out of available n choices
static int NCR(int n, int r)
{
int numerator = 1;
int denominator = 1;
// ncr = (n * (n - 1) * (n - 2) * .....
// ... (n - r + 1)) / (r * (r - 1) * ... * 1)
while (r > 0) {
numerator *= n;
denominator *= r;
n--;
r--;
}
return (numerator / denominator);
}
// Returns the number of quadruples having maximum product
static int findWays(int arr[], int n)
{
// stores the frequency of each element
HashMap count= new HashMap();
if (n < 4)
return 0;
for (int i = 0; i < n; i++) {
count.put(arr[i],(count.get(arr[i])==null?0🙁int)count.get(arr[i])));
}
// remaining_choices denotes the remaining
// elements to select inorder to form quadruple
int remaining_choices = 4;
int ans = 1;
// Getting an iterator
Iterator hmIterator = count.entrySet().iterator();
while (hmIterator.hasNext()) {
Map.Entry mapElement = (Map.Entry)hmIterator.next();
int number =(int) mapElement.getKey();
int frequency =(int)mapElement.getValue();
// If Frequency of element < remaining choices,
// select all of these elements, else select only
// the number of elements required
int toSelect = Math.min(remaining_choices, frequency);
ans = ans * NCR(frequency, toSelect);
// Decrement remaining_choices acc to the number
// of the current elements selected
remaining_choices -= toSelect;
// if the quadruple is formed stop the algorithm
if (remaining_choices==0) {
break;
}
}
return ans;
}
// Driver Code
public static void main(String args[])
{
int arr[] = { 1, 2, 3, 3, 3, 5 };
int n = arr.length;
int maxQuadrupleWays = findWays(arr, n);
System.out.print( maxQuadrupleWays);
}
}
//contributed by Arnab Kundu Python3
# Python3 program to find
# the number of Quadruples
# having maximum product
from collections import defaultdict
# Returns the number of ways
# to select r objects out of
# available n choices
def NCR(n, r):
numerator = 1
denominator = 1
# ncr = (n * (n - 1) *
# (n - 2) * .....
# ... (n - r + 1)) /
# (r * (r - 1) * ... * 1)
while (r > 0):
numerator *= n
denominator *= r
n -= 1
r -= 1
return (numerator // denominator)
# Returns the number of
# quadruples having
# maximum product
def findWays(arr, n):
# stores the frequency
# of each element
count = defaultdict (int)
if (n < 4):
return 0
for i in range (n):
count[arr[i]] += 1
# remaining_choices denotes
# the remaining elements to
# select inorder to form quadruple
remaining_choices = 4
ans = 1
# traverse the elements of
# the map in reverse order
for it in reversed(sorted(count.keys())):
number = it
frequency = count[it]
# If Frequency of element <
# remaining choices, select
# all of these elements,
# else select only the
# number of elements required
toSelect = min(remaining_choices,
frequency)
ans = ans * NCR(frequency,
toSelect)
# Decrement remaining_choices
# acc to the number of the
# current elements selected
remaining_choices -= toSelect
# if the quadruple is
# formed stop the algorithm
if (not remaining_choices):
break
return ans
# Driver Code
if __name__ == "__main__":
arr = [1, 2, 3, 3, 3, 5]
n = len(arr)
maxQuadrupleWays = findWays(arr, n)
print (maxQuadrupleWays)
# This code is contributed by ChitranayalJavascript
输出：
1时间复杂度： O(NlogN)，其中 N 是数组的大小。