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

• Difficulty Level : Hard
• Last Updated : 13 May, 2022

Given an array arr[] of size N and an integer K, the task is to find the Kth 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 â‰¤ 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 Kth 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 ``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

 ``

Output:

`-6`

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