给定三个数组A [] ， B []和C [] ，每个数组由N个整数组成。任务是找到三元组(A [i]，B [j]，C [k])的数量，以使A [i] 。
Input: A[] = {1, 5}, B[] = {2, 4}, C[] = {3, 6}
Output: 3
Triplets are (1, 2, 3), (1, 4, 6) and (1, 2, 6)
Input: A[] = {1, 1, 1}, B[] = {2, 2, 2}, C[] = {3, 3, 3}
Output: 27
方法：对所有给定的数组进行排序。现在在数组B []中固定一个元素X ，对于每个X ，答案将是数组A []中小于X的元素数与数组C []中较大的元素数的乘积比X。我们可以使用修改后的二进制搜索来计算这两个计数。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the count
// of elements in arr[] which are
// less than the given key
int countLessThan(int arr[], int n, int key)
{
int l = 0, r = n - 1;
int index = -1;
// Modified binary search
while (l <= r) {
int m = (l + r) / 2;
if (arr[m] < key) {
l = m + 1;
index = m;
}
else {
r = m - 1;
}
}
return (index + 1);
}
// Function to return the count
// of elements in arr[] which are
// greater than the given key
int countGreaterThan(int arr[], int n, int key)
{
int l = 0, r = n - 1;
int index = -1;
// Modified binary search
while (l <= r) {
int m = (l + r) / 2;
if (arr[m] <= key) {
l = m + 1;
}
else {
r = m - 1;
index = m;
}
}
if (index == -1)
return 0;
return (n - index);
}
// Function to return the count
// of the required triplets
int countTriplets(int n, int* a, int* b, int* c)
{
// Sort all three arrays
sort(a, a + n);
sort(b, b + n);
sort(c, c + n);
int count = 0;
// Iterate for all the elements of array B
for (int i = 0; i < n; ++i) {
int current = b[i];
int a_index = -1, c_index = -1;
// Count of elements in A[]
// which are less than the
// chosen element from B[]
int low = countLessThan(a, n, current);
// Count of elements in C[]
// which are greater than the
// chosen element from B[]
int high = countGreaterThan(c, n, current);
// Update the count
count += (low * high);
}
return count;
}
// Driver code
int main()
{
int a[] = { 1, 5 };
int b[] = { 2, 4 };
int c[] = { 3, 6 };
int size = sizeof(a) / sizeof(a[0]);
cout << countTriplets(size, a, b, c);
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function to return the count
// of elements in arr[] which are
// less than the given key
static int countLessThan(int arr[], int n, int key)
{
int l = 0, r = n - 1;
int index = -1;
// Modified binary search
while (l <= r)
{
int m = (l + r) / 2;
if (arr[m] < key)
{
l = m + 1;
index = m;
}
else
{
r = m - 1;
}
}
return (index + 1);
}
// Function to return the count
// of elements in arr[] which are
// greater than the given key
static int countGreaterThan(int arr[], int n, int key)
{
int l = 0, r = n - 1;
int index = -1;
// Modified binary search
while (l <= r)
{
int m = (l + r) / 2;
if (arr[m] <= key)
{
l = m + 1;
}
else
{
r = m - 1;
index = m;
}
}
if (index == -1)
return 0;
return (n - index);
}
// Function to return the count
// of the required triplets
static int countTriplets(int n, int a[], int b[], int c[])
{
// Sort all three arrays
Arrays.sort(a) ;
Arrays.sort(b);
Arrays.sort(c);
int count = 0;
// Iterate for all the elements of array B
for (int i = 0; i < n; ++i)
{
int current = b[i];
// Count of elements in A[]
// which are less than the
// chosen element from B[]
int low = countLessThan(a, n, current);
// Count of elements in C[]
// which are greater than the
// chosen element from B[]
int high = countGreaterThan(c, n, current);
// Update the count
count += (low * high);
}
return count;
}
// Driver code
public static void main(String args[])
{
int a[] = { 1, 5 };
int b[] = { 2, 4 };
int c[] = { 3, 6 };
int size = a.length;
System.out.println(countTriplets(size, a, b, c));
}
}
// This code is contributed by Arnab KunduPython 3
# Python 3 implementation of the approach
# Function to return the count
# of elements in arr[] which are
# less than the given key
def countLessThan(arr, n, key):
l = 0
r = n - 1
index = -1
# Modified binary search
while (l <= r):
m = (l + r) // 2
if (arr[m] < key) :
l = m + 1
index = m
else :
r = m - 1
return (index + 1)
# Function to return the count
# of elements in arr[] which are
# greater than the given key
def countGreaterThan(arr, n, key):
l = 0
r = n - 1
index = -1
# Modified binary search
while (l <= r) :
m = (l + r) // 2
if (arr[m] <= key) :
l = m + 1
else :
r = m - 1
index = m
if (index == -1):
return 0
return (n - index)
# Function to return the count
# of the required triplets
def countTriplets(n, a, b, c):
# Sort all three arrays
a.sort
b.sort()
c.sort()
count = 0
# Iterate for all the elements of array B
for i in range(n):
current = b[i]
a_index = -1
c_index = -1
# Count of elements in A[]
# which are less than the
# chosen element from B[]
low = countLessThan(a, n, current)
# Count of elements in C[]
# which are greater than the
# chosen element from B[]
high = countGreaterThan(c, n, current)
# Update the count
count += (low * high)
return count
# Driver code
if __name__ == "__main__":
a = [ 1, 5 ]
b = [ 2, 4 ]
c = [ 3, 6 ]
size = len(a)
print( countTriplets(size, a, b, c))
# This code is contributed by ChitraNayalC#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the count
// of elements in arr[] which are
// less than the given key
static int countLessThan(int []arr, int n, int key)
{
int l = 0, r = n - 1;
int index = -1;
// Modified binary search
while (l <= r)
{
int m = (l + r) / 2;
if (arr[m] < key)
{
l = m + 1;
index = m;
}
else
{
r = m - 1;
}
}
return (index + 1);
}
// Function to return the count
// of elements in arr[] which are
// greater than the given key
static int countGreaterThan(int []arr, int n, int key)
{
int l = 0, r = n - 1;
int index = -1;
// Modified binary search
while (l <= r)
{
int m = (l + r) / 2;
if (arr[m] <= key)
{
l = m + 1;
}
else
{
r = m - 1;
index = m;
}
}
if (index == -1)
return 0;
return (n - index);
}
// Function to return the count
// of the required triplets
static int countTriplets(int n, int []a, int []b, int []c)
{
// Sort all three arrays
Array.Sort(a) ;
Array.Sort(b);
Array.Sort(c);
int count = 0;
// Iterate for all the elements of array B
for (int i = 0; i < n; ++i)
{
int current = b[i];
// Count of elements in A[]
// which are less than the
// chosen element from B[]
int low = countLessThan(a, n, current);
// Count of elements in C[]
// which are greater than the
// chosen element from B[]
int high = countGreaterThan(c, n, current);
// Update the count
count += (low * high);
}
return count;
}
// Driver code
public static void Main()
{
int []a = { 1, 5 };
int []b = { 2, 4 };
int []c = { 3, 6 };
int size = a.Length;
Console.WriteLine(countTriplets(size, a, b, c));
}
}
// This code is contributed by AnkitRai01PHP
输出：
3