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 ? 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 <bits/stdc++.h>using namespace std;// Function to find number of pairsbool 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 listlong 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 codeint 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 themimport 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 pairsdef 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 listdef 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 codea = [-4, -2, 3, 3]k = 3n = len(a)# Function callprint(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 themusing 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)
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