Count the triplets such that A[i] < B[j] < C[k]

Given three array A[], B[] and C[] of N integers each. The task is to find the count of triplets (A[i], B[j], C[k]) such that A[i] < B[j] < C[k].

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 

Approach: Sort all the given arrays. Now fix an element say X in array B[] and for each X, the answer will be the product of the count of elements in array A[] which are less than X and the count of elements in array C[] which are greater than X. We can compute both of these counts using modified binary search.
Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
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 Kundu

Python 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 ChitraNayal

C#

// 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 AnkitRai01

PHP

<?php
// PHP implementation of the approach

// Function to return the count
// of elements in arr[] which are
// less than the given key
function countLessThan(&$arr, $n, $key)
{
    $l = 0;
    $r = $n - 1;
    $index = -1;

    // Modified binary search
    while ($l <= $r)
    {
        $m = intval(($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
function countGreaterThan(&$arr, $n, $key)
{
    $l = 0;
    $r = $n - 1;
    $index = -1;

    // Modified binary search
    while ($l <= $r)
    {
        $m = intval(($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
function countTriplets($n, &$a, &$b, &$c)
{
    // Sort all three arrays
    sort($a);
    sort($b);
    sort($c);

    $count = 0;

    // Iterate for all the elements of array B
    for ($i = 0; $i < $n; ++$i)
    {
        $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
$a = array( 1, 5 );
$b = array( 2, 4 );
$c = array( 3, 6 );
$size = sizeof($a);

echo countTriplets($size, $a, $b, $c);

// This code is contributed by ChitraNayal
?>

Javascript

<script>

// JavaScript implementation of the approach


// Function to return the count
    // of elements in arr[] which are
    // less than the given key
function countLessThan(arr,n,key)
{
    let l = 0, r = n - 1;
        let index = -1;

        // Modified binary search
        while (l <= r)
        {
            let m = Math.floor((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
function countGreaterThan(arr,n,key)
{
    let l = 0, r = n - 1;
        let index = -1;

        // Modified binary search
        while (l <= r)
        {
           let m = Math.floor((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
function countTriplets(n,a,b,c)
{
    // Sort all three arrays
        a.sort(function(e,f){return e-f;}) ;
        b.sort(function(e,f){return e-f;}) ;
        c.sort(function(e,f){return e-f;}) ;


        let count = 0;

        // Iterate for all the elements of array B
        for (let i = 0; i < n; ++i)
        {
            let current = b[i];


            // Count of elements in A[]
            // which are less than the
            // chosen element from B[]
            let low = countLessThan(a, n, current);

            // Count of elements in C[]
            // which are greater than the
            // chosen element from B[]
            let high = countGreaterThan(c, n, current);

            // Update the count
            count += (low * high);
        }

        return count;
}

// Driver code
let a=[1, 5 ];
let b=[2, 4];
let c=[3, 6 ];
let size = a.length;
document.write(countTriplets(size, a, b, c));


// This code is contributed by avanitrachhadiya2155

</script>

Output:

3

Time Complexity: O(nlog(n))
Auxiliary Space: O(1)