Find Kth number from sorted array formed by multiplying any two numbers in the array

Given an array arr[] of size N and an integer K, the task is to find the K^th number from the product array.
Note: A product array prod[] of an array is a sorted array of size (N*(N-1))/2 in which each element is formed as prod[k] = arr[i] * arr[j], where 0 ? i < j < N.
Examples:

Input: arr[] = {-4, -2, 3, 3}, K = 3 
Output: -6 
Final prod[] array = {-12, -12, -6, -6, 8, 9} 
where prod[K] = -6
Input: arr[] = {5, 4, 3, 2, -1, 0, 0}, K = 20 
Output: 15 

Naive Approach: Generate the prod[] array by iterating the given array twice and then sort the prod[] array and find the K^th element from the array. 
Time Complexity: O(N² * log(N))
Efficient Approach: The number of negative, zero, and positive pairs can be easily determined, so you can tell whether the answer is negative, zero, or positive. If the answer is negative, it is possible to measure the number of pairs that are greater than or equal to K by selecting a negative number and a positive number one by one, so the answer is obtained using a binary search. The answer is exactly the same when the answer is positive, but consider choosing the same element twice, and subtracting it will count each pair exactly twice.
Below is the implementation of the above approach:

C++

// C++ implementation to find the
// Kth number in the list formed
// from product of any two numbers
// in the array and sorting them

#include <bits/stdc++.h>
using namespace std;

// Function to find number of pairs
bool check(long long x,
           vector<int>& pos,
           vector<int>& neg, int k)
{
    long long pairs = 0;

    int p = neg.size() - 1;
    int nn = neg.size() - 1;
    int pp = pos.size() - 1;

    // Negative and Negative
    for (int i = 0; i < neg.size(); i++) {
        while (p >= 0 and neg[i] * neg[p] <= x)
            p--;

        // Add Possible Pairs
        pairs += min(nn - p, nn - i);
    }

    // Positive and Positive
    p = 0;
    for (int i = pos.size() - 1; i >= 0; i--) {
        while (p < pos.size() and pos[i] * pos[p] <= x)
            p++;

        // Add Possible pairs
        pairs += min(p, i);
    }

    // Negative and Positive
    p = pos.size() - 1;
    for (int i = neg.size() - 1;
         i >= 0;
         i--) {
        while (p >= 0 and neg[i] * pos[p] <= x)
            p--;

        // Add Possible pairs
        pairs += pp - p;
    }

    return (pairs >= k);
}

// Function to find the kth
// element in the list
long long kth_element(int a[],
                      int n, int k)
{
    vector<int> pos, neg;

    // Separate Positive and
    // Negative elements
    for (int i = 0; i < n; i++) {
        if (a[i] >= 0)
            pos.push_back(a[i]);
        else
            neg.push_back(a[i]);
    }

    // Sort the Elements
    sort(pos.begin(), pos.end());
    sort(neg.begin(), neg.end());

    long long l = -1e18,
              ans = 0, r = 1e18;

    // Binary search
    while (l <= r) {
        long long mid = (l + r) >> 1;
        if (check(mid, pos, neg, k)) {
            ans = mid;
            r = mid - 1;
        }
        else
            l = mid + 1;
    }

    // Return the required answer
    return ans;
}

// Driver code
int main()
{
    int a[] = { -4, -2, 3, 3 }, k = 3;

    int n = sizeof(a) / sizeof(a[0]);

    // Function call
    cout << kth_element(a, n, k);

    return 0;
}

Java

// Java implementation to find the
// Kth number in the list formed
// from product of any two numbers
// in the array and sorting them
import java.util.*;

class GFG
{

    // Function to find number of pairs
    static boolean check(int x, Vector pos, Vector neg, int k)
    {
        int pairs = 0;

        int p = neg.size() - 1;
        int nn = neg.size() - 1;
        int pp = pos.size() - 1;

        // Negative and Negative
        for (int i = 0; i < neg.size(); i++)
        {
            while ((p >= 0) && ((int)neg.get(i) *
                    (int)neg.get(p) <= x))
                p--;

            // Add Possible Pairs
            pairs += Math.min(nn - p, nn - i);
        }

        // Positive and Positive
        p = 0;
        for (int i = pos.size() - 1; i >= 0; i--)
        {
            while ((p < pos.size()) && ((int)pos.get(i) *
                    (int)pos.get(p) <= x))
                p++;

            // Add Possible pairs
            pairs += Math.min(p, i);
        }

        // Negative and Positive
        p = pos.size() - 1;
        for (int i = neg.size() - 1;
            i >= 0; i--) {
            while ((p >= 0) && ((int)neg.get(i) *
                    (int)pos.get(p) <= x))
                p--;

            // Add Possible pairs
            pairs += pp - p;
        }

        return (pairs >= k);
    }

    // Function to find the kth
    // element in the list
    static int kth_element(int a[], int n, int k)
    {
        Vector pos = new Vector();
        Vector neg = new Vector();;

        // Separate Positive and
        // Negative elements
        for (int i = 0; i < n; i++)
        {
            if (a[i] >= 0)
                pos.add(a[i]);
            else
                neg.add(a[i]);
        }

        // Sort the Elements
        //sort(pos.begin(), pos.end());
        //sort(neg.begin(), neg.end());
        Collections.sort(pos);
        Collections.sort(neg);

        int l = (int)-1e8, ans = 0, r = (int)1e8;

        // Binary search
        while (l <= r)
        {
            int mid = (l + r) >> 1;
            if (check(mid, pos, neg, k))
            {
                ans = mid;
                r = mid - 1;
            }
            else
                l = mid + 1;
        }

        // Return the required answer
        return ans;
    }

    // Driver code
    public static void main (String[] args)
    {
        int a[] = { -4, -2, 3, 3 }, k = 3;
        int n = a.length;

        // Function call
        System.out.println(kth_element(a, n, k));
    }
}

// This code is contributed by AnkitRai01

Python3

# Python3 implementation to find the
# Kth number in the list formed
# from product of any two numbers
# in the array and sorting them

# Function to find number of pairs
def check(x, pos, neg, k):
    pairs = 0

    p = len(neg) - 1
    nn = len(neg) - 1
    pp = len(pos) - 1

    # Negative and Negative
    for i in range(len(neg)):
        while (p >= 0 and neg[i] * neg[p] <= x):
            p -= 1

        # Add Possible Pairs
        pairs += min(nn - p, nn - i)

    # Positive and Positive
    p = 0
    for i in range(len(pos) - 1, -1, -1):
        while (p < len(pos) and pos[i] * pos[p] <= x):
            p += 1

        # Add Possible pairs
        pairs += min(p, i)

    # Negative and Positive
    p = len(pos) - 1
    for i in range(len(neg) - 1, -1, -1):
        while (p >= 0 and neg[i] * pos[p] <= x):
            p -= 1

        # Add Possible pairs
        pairs += pp - p

    return (pairs >= k)

# Function to find the kth
# element in the list
def kth_element(a, n, k):
    pos, neg = [],[]

    # Separate Positive and
    # Negative elements
    for i in range(n):
        if (a[i] >= 0):
            pos.append(a[i])
        else:
            neg.append(a[i])

    # Sort the Elements
    pos = sorted(pos)
    neg = sorted(neg)

    l = -10**18
    ans = 0
    r = 10**18

    # Binary search
    while (l <= r):
        mid = (l + r) >> 1
        if (check(mid, pos, neg, k)):
            ans = mid
            r = mid - 1
        else:
            l = mid + 1

    # Return the required answer
    return ans

# Driver code
a = [-4, -2, 3, 3]
k = 3

n = len(a)

# Function call
print(kth_element(a, n, k))

# This code is contributed by mohit kumar 29

C#

// C# implementation to find the
// Kth number in the list formed
// from product of any two numbers
// in the array and sorting them
using System;
using System.Collections.Generic;

class GFG
{

    // Function to find number of pairs
    static bool check(int x, List<int> pos, List<int> neg, int k)
    {
        int pairs = 0;

        int p = neg.Count - 1;
        int nn = neg.Count - 1;
        int pp = pos.Count - 1;

        // Negative and Negative
        for (int i = 0; i < neg.Count; i++)
        {
            while ((p >= 0) && ((int)neg[i] *
                    (int)neg[p] <= x))
                p--;

            // Add Possible Pairs
            pairs += Math.Min(nn - p, nn - i);
        }

        // Positive and Positive
        p = 0;
        for (int i = pos.Count - 1; i >= 0; i--)
        {
            while ((p < pos.Count) && ((int)pos[i] *
                    (int)pos[p] <= x))
                p++;

            // Add Possible pairs
            pairs += Math.Min(p, i);
        }

        // Negative and Positive
        p = pos.Count - 1;
        for (int i = neg.Count - 1; i >= 0; i--)
        {
            while ((p >= 0) && ((int)neg[i] *
                    (int)pos[p] <= x))
                p--;

            // Add Possible pairs
            pairs += pp - p;
        }

        return (pairs >= k);
    }

    // Function to find the kth
    // element in the list
    static int kth_element(int []a, int n, int k)
    {
        List<int> pos = new List<int>();
        List<int> neg = new List<int>();;

        // Separate Positive and
        // Negative elements
        for (int i = 0; i < n; i++)
        {
            if (a[i] >= 0)
                pos.Add(a[i]);
            else
                neg.Add(a[i]);
        }

        // Sort the Elements
        //sort(pos.begin(), pos.end());
        //sort(neg.begin(), neg.end());
        pos.Sort();
        neg.Sort();

        int l = (int)-1e8, ans = 0, r = (int)1e8;

        // Binary search
        while (l <= r)
        {
            int mid = (l + r) >> 1;
            if (check(mid, pos, neg, k))
            {
                ans = mid;
                r = mid - 1;
            }
            else
                l = mid + 1;
        }

        // Return the required answer
        return ans;
    }

    // Driver code
    public static void Main(String[] args)
    {
        int []a = { -4, -2, 3, 3 };
        int k = 3;
        int n = a.Length;

        // Function call
        Console.WriteLine(kth_element(a, n, k));
    }
}

// This code is contributed by 29AjayKumar

Javascript

<script>
    // Javascript implementation to find the
    // Kth number in the list formed
    // from product of any two numbers
    // in the array and sorting them

    // Function to find number of pairs
    function check(x, pos, neg, k)
    {
        let pairs = 0;

        let p = neg.length - 1;
        let nn = neg.length - 1;
        let pp = pos.length - 1;

        // Negative and Negative
        for (let i = 0; i < neg.length; i++)
        {
            while ((p >= 0) && (neg[i] * neg[p] <= x))
                p--;

            // Add Possible Pairs
            pairs += Math.min(nn - p, nn - i);
        }

        // Positive and Positive
        p = 0;
        for (let i = pos.length - 1; i >= 0; i--)
        {
            while ((p < pos.length) && (pos[i] * pos[p] <= x))
                p++;

            // Add Possible pairs
            pairs += Math.min(p, i);
        }

        // Negative and Positive
        p = pos.length - 1;
        for (let i = neg.length - 1; i >= 0; i--)
        {
            while ((p >= 0) && (neg[i] * pos[p] <= x))
                p--;

            // Add Possible pairs
            pairs += pp - p;
        }

        return (pairs >= k);
    }

    // Function to find the kth
    // element in the list
    function kth_element(a, n, k)
    {
        let pos = [];
        let neg = [];

        // Separate Positive and
        // Negative elements
        for (let i = 0; i < n; i++)
        {
            if (a[i] >= 0)
                pos.push(a[i]);
            else
                neg.push(a[i]);
        }

        // Sort the Elements
        //sort(pos.begin(), pos.end());
        //sort(neg.begin(), neg.end());
        pos.sort(function(a, b){return a - b});
        neg.sort(function(a, b){return a - b});

        let l = -1e8, ans = 0, r = 1e8;

        // Binary search
        while (l <= r)
        {
            let mid = (l + r) >> 1;
            if (check(mid, pos, neg, k))
            {
                ans = mid;
                r = mid - 1;
            }
            else
                l = mid + 1;
        }

        // Return the required answer
        return ans;
    }

    let a = [ -4, -2, 3, 3 ];
    let k = 3;
    let n = a.length;

    // Function call
    document.write(kth_element(a, n, k));

// This code is contributed by divyesh072019.
</script>

Output:

-6

Time Complexity: O(n logn)

Space Complexity: O(1)