Data Modelling & AI Data Structure & Algorithm

Kth smallest number in array formed by product of any two elements from two arrays

23 June 2025

0

Given two sorted arrays A[] and B[] consisting of N and M integers respectively, the task is to find the K^th smallest number in the array formed by the product of all possible pairs from array A[] and B[] respectively.

Examples:

Input: A[] = {1, 2, 3}, B[] = {-1, 1}, K = 4
Output: 1
Explanation: The array formed by the product of any two numbers from array A[] and B[] respectively is prod[] = {-3, -2, -1, 1, 2, 3}. Hence, the 4th smallest integer in the prod[] array is 1.

Input: A[] = {-4, -2, 0, 3}, B[] = {1, 10}, K = 7
Output: 3

Approach: The given problem can be solved with the help of Binary Search over all possible values of products. The approach discussed here is very similar to the approach discussed in this article. Below are the steps to follow:

Create a function check(key), which returns whether the number of elements smaller than the key in the product array is more than K or not. It can be done using the two-pointer technique similar to the algorithm discussed in the article here.
Perform a binary search over the range [INT_MIN, INT_MAX] to find the smallest number ans such that the number of elements smaller than ans in the product array is at least K.
After completing the above steps, print the value of ans as the result.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
#define int long long
 
// Function to check if count of elements
// greater than req in product array are
// more than K or not
bool check(int req, vector<int> posA,
           vector<int> posB, vector<int> negA,
           vector<int> negB, int K)
{
    // Stores the count of numbers less
    // than or equal to req
    int cnt = 0;
 
    // Case with both elements of A[] and
    // B[] are negative
    int first = 0;
    int second = negB.size() - 1;
 
    // Count number of pairs formed from
    // array A[] and B[] with both elements
    // negative and there product <= req
    while (first < negA.size()) {
        while (second >= 0
               && negA[first]
                          * negB[second]
                      <= req)
            second--;
 
        // Update cnt
        cnt += negB.size() - second - 1;
        first++;
    }
 
    // Case with both elements of A[] and
    // B[] are positive
    first = 0;
    second = posB.size() - 1;
 
    // Count number of pairs formed from
    // array A[] and B[] with both elements
    // positive and there product <= req
    while (first < posA.size()) {
        while (second >= 0
               && posA[first]
                          * posB[second]
                      > req)
            second--;
 
        // Update cnt
        cnt += second + 1;
        first++;
    }
 
    // Case with elements of A[] and B[]
    // as positive and negative respectively
    first = posA.size() - 1;
    second = negB.size() - 1;
 
    // Count number of pairs formed from
    // +ve integers of A[] and -ve integer
    // of array B[] product <= req
    while (second >= 0) {
        while (first >= 0
               && posA[first]
                          * negB[second]
                      <= req)
            first--;
 
        // Update cnt
        cnt += posA.size() - first - 1;
        second--;
    }
 
    // Case with elements of A[] and B[]
    // as negative and positive respectively
    first = negA.size() - 1;
    second = posB.size() - 1;
 
    // Count number of pairs formed from
    // -ve and +ve integers from A[] and
    // B[] with product <= req
    for (; first >= 0; first--) {
        while (second >= 0
               && negA[first]
                          * posB[second]
                      <= req)
            second--;
 
        // Update cnt
        cnt += posB.size() - second - 1;
    }
 
    // Return Answer
    return (cnt >= K);
}
 
// Function to find the Kth smallest
// number in array formed by product of
// any two elements from A[] and B[]
int kthSmallestProduct(vector<int>& A,
                       vector<int>& B,
                       int K)
{
    vector<int> posA, negA, posB, negB;
 
    // Loop to iterate array A[]
    for (const auto& it : A) {
        if (it >= 0)
            posA.push_back(it);
        else
            negA.push_back(it);
    }
 
    // Loop to iterate array B[]
    for (const auto& it : B)
        if (it >= 0)
            posB.push_back(it);
        else
            negB.push_back(it);
 
    // Stores the lower and upper bounds
    // of the binary search
    int l = LLONG_MIN, r = LLONG_MAX;
 
    // Stores the final answer
    int ans;
 
    // Find the kth smallest integer
    // using binary search
    while (l <= r) {
 
        // Stores the mid
        int mid = (l + r) / 2;
 
        // If the number of elements
        // greater than mid in product
        // array are more than K
        if (check(mid, posA, posB,
                  negA, negB, K)) {
            ans = mid;
            r = mid - 1;
        }
        else {
            l = mid + 1;
        }
    }
 
    // Return answer
    return ans;
}
 
// Driver Code
int32_t main()
{
    vector<int> A = { -4, -2, 0, 3 };
    vector<int> B = { 1, 10 };
    int K = 7;
 
    cout << kthSmallestProduct(A, B, K);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
 
// Function to check if count of elements
// greater than req in product array are
// more than K or not
static boolean check(int req, Vector<Integer> posA,
           Vector<Integer> posB, Vector<Integer> negA,
           Vector<Integer> negB, int K)
{
   
    // Stores the count of numbers less
    // than or equal to req
    int cnt = 0;
 
    // Case with both elements of A[] and
    // B[] are negative
    int first = 0;
    int second = negB.size() - 1;
 
    // Count number of pairs formed from
    // array A[] and B[] with both elements
    // negative and there product <= req
    while (first < negA.size()) {
        while (second >= 0
               && negA.elementAt(first)
                          * negB.elementAt(second)
                      <= req)
            second--;
 
        // Update cnt
        cnt += negB.size() - second - 1;
        first++;
    }
 
    // Case with both elements of A[] and
    // B[] are positive
    first = 0;
    second = posB.size() - 1;
 
    // Count number of pairs formed from
    // array A[] and B[] with both elements
    // positive and there product <= req
    while (first < posA.size()) {
        while (second >= 0
               && posA.elementAt(first)
                          * posB.elementAt(second)
                      > req)
            second--;
 
        // Update cnt
        cnt += second + 1;
        first++;
    }
 
    // Case with elements of A[] and B[]
    // as positive and negative respectively
    first = posA.size() - 1;
    second = negB.size() - 1;
 
    // Count number of pairs formed from
    // +ve integers of A[] and -ve integer
    // of array B[] product <= req
    while (second >= 0) {
        while (first >= 0
               && posA.elementAt(first)
                          * negB.elementAt(second)
                      <= req)
            first--;
 
        // Update cnt
        cnt += posA.size() - first - 1;
        second--;
    }
 
    // Case with elements of A[] and B[]
    // as negative and positive respectively
    first = negA.size() - 1;
    second = posB.size() - 1;
 
    // Count number of pairs formed from
    // -ve and +ve integers from A[] and
    // B[] with product <= req
    for (; first >= 0; first--) {
        while (second >= 0
               && negA.elementAt(first)
                          * posB.elementAt(second)
                      <= req)
            second--;
 
        // Update cnt
        cnt += posB.size() - second - 1;
    }
 
    // Return Answer
    return (cnt >= K);
}
 
// Function to find the Kth smallest
// number in array formed by product of
// any two elements from A[] and B[]
static int kthSmallestProduct(int[] A,
                       int[] B,
                       int K)
{
    Vector<Integer> posA = new Vector<>(); 
    Vector<Integer> negA = new Vector<>();
    Vector<Integer> posB = new Vector<>();
    Vector<Integer> negB = new Vector<>();
    // Loop to iterate array A[]
    for (int  it : A) {
        if (it >= 0)
            posA.add(it);
        else
            negA.add(it);
    }
 
    // Loop to iterate array B[]
    for (int it : B)
        if (it >= 0)
            posB.add(it);
        else
            negB.add(it);
 
    // Stores the lower and upper bounds
    // of the binary search
    int l = Integer.MIN_VALUE, r = Integer.MAX_VALUE;
 
    // Stores the final answer
    int ans=0;
 
    // Find the kth smallest integer
    // using binary search
    while (l <= r) {
 
        // Stores the mid
        int mid = (l + r) / 2;
 
        // If the number of elements
        // greater than mid in product
        // array are more than K
        if (check(mid, posA, posB,
                  negA, negB, K)) {
            ans = mid;
            r = mid - 1;
        }
        else {
            l = mid + 1;
        }
    }
 
    // Return answer
    return ans;
}
 
// Driver Code
public static void main(String[] args)
{
   int[] A = { -4, -2, 0, 3 };
    int[] B = { 1, 10 };
    int K = 7;
 
    System.out.print(kthSmallestProduct(A, B, K));
 
}
}
 
// This code is contributed by gauravrajput1 

Python3

# python program for the above approach
LLONG_MAX = 9223372036854775807
LLONG_MIN = -9223372036854775807
 
# Function to check if count of elements
# greater than req in product array are
# more than K or not
def check(req, posA, posB, negA, negB, K):
 
    # Stores the count of numbers less
    # than or equal to req
    cnt = 0
 
    # Case with both elements of A[] and
    # B[] are negative
    first = 0
    second = len(negB) - 1
 
    # Count number of pairs formed from
    # array A[] and B[] with both elements
    # negative and there product <= req
    while (first < len(negA)):
        while (second >= 0 and negA[first] * negB[second] <= req):
            second -= 1
 
        # Update cnt
        cnt += len(negB) - second - 1
        first += 1
 
    # Case with both elements of A[] and
    # B[] are positive
    first = 0
    second = len(posB) - 1
 
    # Count number of pairs formed from
    # array A[] and B[] with both elements
    # positive and there product <= req
    while (first < len(posA)):
        while (second >= 0 and posA[first] * posB[second] > req):
            second -= 1
 
        # Update cnt
        cnt += second + 1
        first += 1
 
    # Case with elements of A[] and B[]
    # as positive and negative respectively
    first = len(posA) - 1
    second = len(negB) - 1
 
    # Count number of pairs formed from
    # +ve integers of A[] and -ve integer
    # of array B[] product <= req
    while (second >= 0):
        while (first >= 0 and posA[first] * negB[second] <= req):
            first -= 1
 
        # Update cnt
        cnt += len(posA) - first - 1
        second -= 1
 
    # Case with elements of A[] and B[]
    # as negative and positive respectively
    first = len(negA) - 1
    second = len(posB) - 1
 
    # Count number of pairs formed from
    # -ve and +ve integers from A[] and
    # B[] with product <= req
    for first in range(first, -1, -1):
        while (second >= 0 and negA[first] * posB[second] <= req):
            second -= 1
 
        # Update cnt
        cnt += len(posB) - second - 1
 
    # Return Answer
    return (cnt >= K)
 
 
# Function to find the Kth smallest
# number in array formed by product of
# any two elements from A[] and B[]
def kthSmallestProduct(A, B, K):
 
    posA = []
    negA = []
    posB = []
    negB = []
 
    # Loop to iterate array A[]
    for it in A:
        if (it >= 0):
            posA.append(it)
        else:
            negA.append(it)
 
        # Loop to iterate array B[]
    for it in B:
        if (it >= 0):
            posB.append(it)
        else:
            negB.append(it)
 
        # Stores the lower and upper bounds
        # of the binary search
    l = LLONG_MIN
    r = LLONG_MAX
 
    # Stores the final answer
    ans = 0
 
    # Find the kth smallest integer
    # using binary search
    while (l <= r):
 
        # Stores the mid
        mid = (l + r) // 2
 
        # If the number of elements
        # greater than mid in product
        # array are more than K
        if (check(mid, posA, posB, negA, negB, K)):
            ans = mid
            r = mid - 1
        else:
            l = mid + 1
 
        # Return answer
    return ans
 
 
# Driver Code
if __name__ == "__main__":
 
    A = [-4, -2, 0, 3]
    B = [1, 10]
    K = 7
 
    print(kthSmallestProduct(A, B, K))
 
    # This code is contributed by rakeshsahni

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
public class GFG {
 
    // Function to check if count of elements
    // greater than req in product array are
    // more than K or not
    static bool check(int req, List<int> posA, List<int> posB, List<int> negA,
            List<int> negB, int K) {
 
        // Stores the count of numbers less
        // than or equal to req
        int cnt = 0;
 
        // Case with both elements of []A and
        // []B are negative
        int first = 0;
        int second = negB.Count - 1;
 
        // Count number of pairs formed from
        // array []A and []B with both elements
        // negative and there product <= req
        while (first < negA.Count) {
            while (second >= 0 && negA[first]
                          * negB[second] <= req)
                second--;
 
            // Update cnt
            cnt += negB.Count - second - 1;
            first++;
        }
 
        // Case with both elements of []A and
        // []B are positive
        first = 0;
        second = posB.Count - 1;
 
        // Count number of pairs formed from
        // array []A and []B with both elements
        // positive and there product <= req
        while (first < posA.Count) {
            while (second >= 0 && posA[first] * posB[second] > req)
                second--;
 
            // Update cnt
            cnt += second + 1;
            first++;
        }
 
        // Case with elements of []A and []B
        // as positive and negative respectively
        first = posA.Count - 1;
        second = negB.Count - 1;
 
        // Count number of pairs formed from
        // +ve integers of []A and -ve integer
        // of array []B product <= req
        while (second >= 0) {
            while (first >= 0 && posA[first] * negB[second] <= req)
                first--;
 
            // Update cnt
            cnt += posA.Count - first - 1;
            second--;
        }
 
        // Case with elements of []A and []B
        // as negative and positive respectively
        first = negA.Count - 1;
        second = posB.Count - 1;
 
        // Count number of pairs formed from
        // -ve and +ve integers from []A and
        // []B with product <= req
        for (; first >= 0; first--) {
            while (second >= 0 && negA[first]* posB[second]<= req)
                second--;
 
            // Update cnt
            cnt += posB.Count - second - 1;
        }
 
        // Return Answer
        return (cnt >= K);
    }
 
    // Function to find the Kth smallest
    // number in array formed by product of
    // any two elements from []A and []B
    static int kthSmallestProduct(int[] A, int[] B, int K) {
        List<int> posA = new List<int>();
        List<int> negA = new List<int>();
        List<int> posB = new List<int>();
        List<int> negB = new List<int>();
        // Loop to iterate array []A
        foreach (int it in A) {
            if (it >= 0)
                posA.Add(it);
            else
                negA.Add(it);
        }
 
        // Loop to iterate array []B
        foreach (int it in B)
            if (it >= 0)
                posB.Add(it);
            else
                negB.Add(it);
 
        // Stores the lower and upper bounds
        // of the binary search
        int l = int.MinValue, r = int.MaxValue;
 
        // Stores the readonly answer
        int ans = 0;
 
        // Find the kth smallest integer
        // using binary search
        while (l <= r) {
 
            // Stores the mid
            int mid = (l + r) / 2;
 
            // If the number of elements
            // greater than mid in product
            // array are more than K
            if (check(mid, posA, posB, negA, negB, K)) {
                ans = mid;
                r = mid - 1;
            } else {
                l = mid + 1;
            }
        }
 
        // Return answer
        return ans;
    }
 
    // Driver Code
    public static void Main(String[] args) {
        int[] A = { -4, -2, 0, 3 };
        int[] B = { 1, 10 };
        int K = 7;
 
        Console.Write(kthSmallestProduct(A, B, K));
 
    }
}
 
// This code is contributed by gauravrajput1 

Javascript

<script>
// Javascript program for the above approach
 
// Function to check if count of elements
// greater than req in product array are
// more than K or not
function check(req, posA, posB, negA, negB, K)
{
 
  // Stores the count of numbers less
  // than or equal to req
  let cnt = 0;
 
  // Case with both elements of A[] and
  // B[] are negative
  let first = 0;
  let second = negB.length - 1;
 
  // Count number of pairs formed from
  // array A[] and B[] with both elements
  // negative and there product <= req
  while (first < negA.length) {
    while (second >= 0 && negA[first] * negB[second] <= req) second--;
 
    // Update cnt
    cnt += negB.length - second - 1;
    first++;
  }
 
  // Case with both elements of A[] and
  // B[] are positive
  first = 0;
  second = posB.length - 1;
 
  // Count number of pairs formed from
  // array A[] and B[] with both elements
  // positive and there product <= req
  while (first < posA.length) {
    while (second >= 0 && posA[first] * posB[second] > req) second--;
 
    // Update cnt
    cnt += second + 1;
    first++;
  }
 
  // Case with elements of A[] and B[]
  // as positive and negative respectively
  first = posA.length - 1;
  second = negB.length - 1;
 
  // Count number of pairs formed from
  // +ve integers of A[] and -ve integer
  // of array B[] product <= req
  while (second >= 0) {
    while (first >= 0 && posA[first] * negB[second] <= req) first--;
 
    // Update cnt
    cnt += posA.length - first - 1;
    second--;
  }
 
  // Case with elements of A[] and B[]
  // as negative and positive respectively
  first = negA.length - 1;
  second = posB.length - 1;
 
  // Count number of pairs formed from
  // -ve and +ve integers from A[] and
  // B[] with product <= req
  for (; first >= 0; first--) {
    while (second >= 0 && negA[first] * posB[second] <= req) second--;
 
    // Update cnt
    cnt += posB.length - second - 1;
  }
 
  // Return Answer
  return cnt >= K;
}
 
// Function to find the Kth smallest
// number in array formed by product of
// any two elements from A[] and B[]
function kthSmallestProduct(A, B, K) {
  let posA = [],
    negA = [],
    posB = [],
    negB = [];
 
  // Loop to iterate array A[]
  for (it of A) {
    if (it >= 0) posA.push(it);
    else negA.push(it);
  }
 
  // Loop to iterate array B[]
  for (it of B)
    if (it >= 0) posB.push(it);
    else negB.push(it);
 
  // Stores the lower and upper bounds
  // of the binary search
  let l = Number.MIN_SAFE_INTEGER,
    r = Number.MAX_SAFE_INTEGER;
 
  // Stores the final answer
  let ans;
 
  // Find the kth smallest integer
  // using binary search
  while (l <= r) 
  {
   
    // Stores the mid
    let mid = (l + r) / 2;
 
    // If the number of elements
    // greater than mid in product
    // array are more than K
    if (check(mid, posA, posB, negA, negB, K)) {
      ans = mid;
      r = mid - 1;
    } else {
      l = mid + 1;
    }
  }
 
  // Return answer
  return ans;
}
 
// Driver Code
let A = [-4, -2, 0, 3];
let B = [1, 10];
let K = 7;
 
document.write(kthSmallestProduct(A, B, K));
 
// This code is contributed by gfgking.
</script>

Output:

Time Complexity: O((N+M)*log 2⁶⁴) or O((N+M)*64)
Auxiliary Space: O(N+M)

Recommended

Solve DSA problems on GfG Practice.

Solve Problems

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!

Kth smallest number in array formed by product of any two elements from two arrays

C++

Java

Python3

C#

Javascript

From Word2Vec to LLM2Vec: How to Choose the Right Embedding Model for RAG

How to Debug Slow Search Requests in Milvus

When Context Engineering Is Done Right, Hallucinations Can Be the Spark of AI Creativity

LEAVE A REPLY Cancel reply

Most Popular

Pixel phones on Android 16 QPR1 Beta 3.1 are experiencing Google Wallet errors

Google’s underrated AI shopping tool just made online shoe hunts surprisingly easy

Google Home chief says he wants to earn back love and trust

UFS 5.0 is on the way, but most Android users probably won’t notice

EDITOR PICKS

Pixel phones on Android 16 QPR1 Beta 3.1 are experiencing Google Wallet errors

Google’s underrated AI shopping tool just made online shoe hunts surprisingly easy

Google Home chief says he wants to earn back love and trust

POPULAR POSTS

Pixel phones on Android 16 QPR1 Beta 3.1 are experiencing Google Wallet errors

Google’s underrated AI shopping tool just made online shoe hunts surprisingly easy

Google Home chief says he wants to earn back love and trust

POPULAR CATEGORY

ABOUT US

FOLLOW US