Pair with given product | Set 1 (Find if any pair exists)

Given an array of distinct elements and a number x, find if there is a pair with a product equal to x. 

Examples :

Input : arr[] = {10, 20, 9, 40};
        int x = 400;
Output : Yes

Input : arr[] = {10, 20, 9, 40};
        int x = 190;
Output : No

Input : arr[] = {-10, 20, 9, -40};
        int x = 400;
Output : Yes

Input : arr[] = {-10, 20, 9, 40};
        int x = -400;
Output : Yes

Input : arr[] = {0, 20, 9, 40};
        int x = 0;
Output : Yes

Naive approach:  Run two loops to consider all possible pairs. For every pair, check if the product is equal to x or not.  

C++

// A simple C++ program to find if there is a pair
// with given product.
#include<bits/stdc++.h>
using namespace std;

// Returns true if there is a pair in arr[0..n-1]
// with product equal to x.
bool isProduct(int arr[], int n, int x)
{
    // Consider all possible pairs and check for
    // every pair.
    for (int i=0; i<n-1; i++)
       for (int j=i+1; j<n; j++)
          if (arr[i] * arr[j] == x)
              return true;

    return false;
}

// Driver code
int main()
{
    int arr[] = {10, 20, 9, 40};
    int x = 400;
    int n = sizeof(arr)/sizeof(arr[0]);
    isProduct(arr, n, x)? cout << "Yes\n"
                        : cout << "No\n";
    x = 190;
    isProduct(arr, n, x)? cout << "Yes\n"
                        : cout << "No\n";
    return 0;
}

Java

// Java program to find if there is a pair
// with given product.
class GFG
{
    // Returns true if there is a pair in
    // arr[0..n-1] with product equal to x.
    boolean isProduct(int arr[], int n, int x)
    {
        for (int i=0; i<n-1; i++)
            for (int j=i+1; j<n; j++)
                if (arr[i]*arr[j] == x)
                    return true;
        return false;
    }

    // Driver code
    public static void main(String[] args)
    {
        GFG g = new GFG();
        int arr[] = {10, 20, 9, 40};
        int x = 400;
        int n = arr.length;
        if (g.isProduct(arr, n, x))
            System.out.println("Yes");
        else
            System.out.println("No");

        x = 190;
        if (g.isProduct(arr, n, x))
            System.out.println("Yes");
        else
            System.out.println("No");

    }
}
// This code is contributed by Kamal Rawal

Python3

# Python3 program to find if there
# is a pair with given product.

# Returns true if there is a
# pair in arr[0..n-1] with
# product equal to x
def isProduct(arr, n, x):
    for i in arr:
        for j in arr:
            if i * j == x:
                return True
    return False


# Driver code
arr = [10, 20, 9, 40]
x = 400
n = len(arr)
if(isProduct(arr,n, x) == True):
    print ("Yes")

else:
    print("No")

x = 900
if(isProduct(arr, n, x)):
    print("Yes")

else:
    print("No")

# This code is contributed
# by prerna saini

C#

// C# program to find
// if there is a pair
// with given product.
using System;

class GFG
{

// Returns true if there
// is a pair in arr[0..n-1]
// with product equal to x.
static bool isProduct(int []arr,
                      int n, int x)
{
    for (int i = 0; i < n - 1; i++)
        for (int j = i + 1; j < n; j++)
            if (arr[i] * arr[j] == x)
                return true;
    return false;
}

// Driver Code
static void Main()
{
    int []arr = {10, 20, 9, 40};
    int x = 400;
    int n = arr.Length;
    if (isProduct(arr, n, x))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");

    x = 190;
    if (isProduct(arr, n, x))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}

// This code is contributed
// by Sam007

PHP

<?php
// A simple php program to
// find if there is a pair
// with given product.

// Returns true if there
// is a pair in arr[0..n-1]
// with product equal to x.
function isProduct($arr, $n, $x)
{
    // Consider all possible
    // pairs and check for
    // every pair.
    for ($i = 0;
         $i < $n - 1; $i++)
    for ($j = $i + 1;
         $i < $n; $i++)
        if ($arr[$i] *
            $arr[$j] == $x)
            return true;

    return false;
}

// Driver code
$arr = array(10, 20, 9, 40);
$x = 400;
$n = count($arr);
if(isProduct($arr, $n, $x))
echo "Yes\n";
else
echo "No\n";

$x = 190;
if(isProduct($arr, $n, $x))
echo "Yes\n";
else
echo "No\n";

// This code is contributed
// by Sam007
?>

Javascript

<script>

// A simple Javascript program to find if there is a pair
// with given product.

// Returns true if there is a pair in arr[0..n-1]
// with product equal to x.
function isProduct(arr, n, x)
{
    // Consider all possible pairs and check for
    // every pair.
    for (var i=0; i<n-1; i++)
    for (var j=i+1; i<n; i++)
        if (arr[i] * arr[j] == x)
            return true;

    return false;
}

// Driver code
var arr = [10, 20, 9, 40];
var x = 400;
var n = arr.length;
isProduct(arr, n, x)? document.write("Yes<br>")
                    : document.write("No<br>");
x = 190;
isProduct(arr, n, x)? document.write("Yes")
                    : document.write("No");

</script>

Output

Yes
No

Time Complexity: O(n²)
Auxiliary Space: O(1)

Better Solution: We sort the given array. After sorting, we traverse the array and for every element arr[i], we do binary search for x/arr[i] in the subarray on the right of arr[i], i.e., in subarray arr[i+1..n-1]

C++

// A simple C++ program to find if there is a pair
// with given product.
#include<bits/stdc++.h>
using namespace std;

//Function to check if x is present in the array or not
// such that arr[i] + x == given sum
bool binarysearch(int arr[], int n, int i,int x)
{
    int l = 0, r = n - 1;

    while (l <= r) {
        int mid = (l + r) / 2;

        // Checking if the middle element is equal to x
        if (arr[mid]*arr[i] == x)
        {
           if(i!=mid)//if position is no same
           {     return true; }
          else{ //if position is  same
            if(mid>0 && arr[mid-1]==arr[mid])
            {  return true;  }//if exist adjacent element
            else if(mid<n-1 && arr[mid+1]==arr[mid])
              {  return true;  }//if exist adjacent element
            else {   return false;  }
          }
        }
        else if (arr[mid]*arr[i] < x)
        {
            l = mid + 1;
        }
        else {
            r = mid - 1;
        }
    }
    // return true , if element x is present in the array
    // else false
    return false;
}
// Returns true if there is a pair in arr[0..n-1]
// with product equal to x.
bool isProduct(int arr[], int n, int x)
{   sort( arr , arr+n);//sorting array for Binary search

    // Consider all possible pairs and check for
    // every pair.
    for (int i=0; i<n; i++)
    {  //Using binary search to check if there exist a
      //index in arr such that arr[i]*arr[index]==given sum
      if(binarysearch( arr, n, i , x))
         {   return true;  }// Return true if pair found
    }
    return false;
}

// Driver code
int main()
{
    int arr[] = {10, 20, 9, 40};
    int n = sizeof(arr)/sizeof(arr[0]);

    int x = 400;
    // Function call test case 1
    if(isProduct(arr, n, x))
    { cout<<"Yes"<<endl;   }
    else { cout << "No"<<endl; }

    x = 190;
    // Function call test case 2
    if(isProduct(arr, n, x))
    { cout<<"Yes"<<endl;   }
    else { cout << "No"<<endl; }

    return 0;
}

Java

// A simple Java program to find if there is a pair
// with given product.
import java.util.*;

public class Gfg {

    //Function to check if x is present in the array or not
    // such that arr[i] + x == given sum
    public static boolean binarysearch(int[] arr, int n, int i,int x)
    {
        int l = 0, r = n - 1;

        while (l <= r) {
            int mid = (l + r) / 2;

            // Checking if the middle element is equal to x
            if (arr[mid]*arr[i] == x)
            {
               if(i!=mid)//if position is no same
               {     return true; }
              else{ //if position is  same
                if(mid>0 && arr[mid-1]==arr[mid])
                {  return true;  }//if exist adjacent element
                else if(mid<n-1 && arr[mid+1]==arr[mid])
                  {  return true;  }//if exist adjacent element
                else {   return false;  }
              }
            }
            else if (arr[mid]*arr[i] < x)
            {
                l = mid + 1;
            }
            else {
                r = mid - 1;
            }
        }
        // return true , if element x is present in the array
        // else false
        return false;
    }

    // Returns true if there is a pair in arr[0..n-1]
    // with product equal to x.
    public static boolean isProduct(int[] arr, int n, int x)
    {   Arrays.sort(arr);//sorting array for Binary search

        // Consider all possible pairs and check for
        // every pair.
        for (int i=0; i<n; i++)
        {  //Using binary search to check if there exist a
          //index in arr such that arr[i]*arr[index]==given sum
          if(binarysearch( arr, n, i , x))
             {   return true;  }// Return true if pair found
        }
        return false;
    }

    // Driver code
    public static void main(String[] args)
    {
        int[] arr = {10, 20, 9, 40};
        int n = arr.length;

        int x = 400;
        // Function call test case 1
        if(isProduct(arr, n, x))
        { System.out.println("Yes");   }
        else { System.out.println("No"); }

        x = 190;
        // Function call test case 2
        if(isProduct(arr, n, x))
        { System.out.println("Yes");   }
        else { System.out.println("No"); }

    }
}

Python3

# Python program to find if there is a pair
# with given product.

# Function to check if x is present in the array or not
# such that arr[i] * x == given product
def binarysearch(arr, n, i, x):
    l = 0
    r = n - 1

    while l <= r:
        mid = (l + r) // 2

        # Checking if the middle element is equal to x
        if arr[mid] * arr[i] == x:
            # if position is not same
            if i != mid:
                return True
            else:
                # if position is same
                if mid > 0 and arr[mid-1] == arr[mid]:
                    # if exist adjacent element
                    return True
                elif mid < n-1 and arr[mid+1] == arr[mid]:
                    # if exist adjacent element
                    return True
                else:
                    return False
        elif arr[mid] * arr[i] < x:
            l = mid + 1
        else:
            r = mid - 1

    # return True, if element x is present in the array, else False
    return False

# Returns True if there is a pair in arr[0..n-1]
# with product equal to x.
def isProduct(arr, n, x):
    # Sorting array for Binary search
    arr.sort()

    # Consider all possible pairs and check for every pair
    for i in range(n):
        # Using binary search to check if there exists an
        # index in arr such that arr[i] * arr[index] == given product
        if binarysearch(arr, n, i, x):
            return True # Return True if pair found
    return False

# Driver code
arr = [10, 20, 9, 40]
n = len(arr)

x = 400
# Function call test case 1
if isProduct(arr, n, x):
    print("Yes")
else:
    print("No")

x = 190
# Function call test case 2
if isProduct(arr, n, x):
    print("Yes")
else:
    print("No")

C#

using System;

class Program
{
  // Function to check if x is present in the array or not
  // such that arr[i] + x == given sum
  static bool binarysearch(int[] arr, int n, int i, int x)
  {
    int l = 0, r = n - 1;

    while (l <= r)
    {
      int mid = (l + r) / 2;

      // Checking if the middle element is equal to x
      if (arr[mid] * arr[i] == x)
      {
        if (i != mid) //if position is not the same
        {
          return true;
        }
        else
        { //if position is same
          if (mid > 0 && arr[mid - 1] == arr[mid])
          {
            return true;
          } //if exist adjacent element
          else if (mid < n - 1 && arr[mid + 1] == arr[mid])
          {
            return true;
          } //if exist adjacent element
          else
          {
            return false;
          }
        }
      }
      else if (arr[mid] * arr[i] < x)
      {
        l = mid + 1;
      }
      else
      {
        r = mid - 1;
      }
    }
    // return true , if element x is present in the array
    // else false
    return false;
  }

  // Returns true if there is a pair in arr[0..n-1]
  // with product equal to x.
  static bool isProduct(int[] arr, int n, int x)
  {
    Array.Sort(arr); //sorting array for Binary search

    // Consider all possible pairs and check for
    // every pair.
    for (int i = 0; i < n; i++)
    {
      //Using binary search to check if there exist a
      //index in arr such that arr[i]*arr[index]==given sum
      if (binarysearch(arr, n, i, x))
      {
        return true; // Return true if pair found
      }
    }
    return false;
  }

  // Driver code
  static void Main()
  {
    int[] arr = { 10, 20, 9, 40 };
    int n = arr.Length;

    int x = 400;
    // Function call test case 1
    if (isProduct(arr, n, x))
    {
      Console.WriteLine("Yes");
    }
    else
    {
      Console.WriteLine("No");
    }

    x = 190;
    // Function call test case 2
    if (isProduct(arr, n, x))
    {
      Console.WriteLine("Yes");
    }
    else
    {
      Console.WriteLine("No");
    }
  }
}

Javascript

// Javascript equivalent code
function binarysearch(arr, n, i, x) {
  let l = 0;
  let r = n - 1;

  while (l <= r) {
    let mid = Math.floor((l + r) / 2);

    // Checking if the middle element is equal to x
    if (arr[mid] * arr[i] == x) {
      // if position is not same
      if (i != mid) {
        return true;
      } else {
        // if position is same
        if (mid > 0 && arr[mid - 1] == arr[mid]) {
          // if exist adjacent element
          return true;
        } else if (mid < n - 1 && arr[mid + 1] == arr[mid]) {
          // if exist adjacent element
          return true;
        } else {
          return false;
        }
      }
    } else if (arr[mid] * arr[i] < x) {
      l = mid + 1;
    } else {
      r = mid - 1;
    }
  }

  // return True, if element x is present in the array, else False
  return false;
}

// Returns True if there is a pair in arr[0..n-1]
// with product equal to x.
function isProduct(arr, n, x) {
  // Sorting array for Binary search
  arr.sort();

  // Consider all possible pairs and check for every pair
  for (let i = 0; i < n; i++) {
    // Using binary search to check if there exists an
    // index in arr such that arr[i] * arr[index] == given product
    if (binarysearch(arr, n, i, x)) {
      return true; // Return True if pair found
    }
  }
  return false;
}

// Driver code
let arr = [10, 20, 9, 40];
let n = arr.length;

let x = 400;
// Function call test case 1
if (isProduct(arr, n, x)) {
  console.log("Yes");
} else {
  console.log("No");
}

x = 190;
// Function call test case 2
if (isProduct(arr, n, x)) {
  console.log("Yes");
} else {
  console.log("No");
}

Output

Yes
No

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

Another Approach ( Using two pointer technique):

This approach to solve the problem is to sort the array in ascending order and then use Two pointer approach ( l = 0, r = arr.size()-1) to traverse that sorted array. If product of arr[l] and arr[r] is equal to x, then return true. If product is less than k then increase l else decrease r.

Algorithm:

1.  Define a function isProduct that takes an integer array arr, an integer n, and an integer x as inputs and returns a boolean value. The        function first sorts the array arr in ascending order. It then initializes two indices l and r to the beginning and end of the array,                      respectively. While l < r, the function calculates the product of the elements at indices l and r, compares it with x, and adjusts l and r    accordingly. If the product equals x, it returns true. Otherwise, if the product is less than x, it increments l to increase the value of the        product, and if the product is greater than x, it decrements r to decrease the value of the product. If no such pair exists in the array, the  function returns false.
2.  In the main function:
      a. Create an integer array according to the input specification.
      b. Initialize an integer n as the size of the array.
      c. Initialize an integer x according to the input specification.
      d. Call the isProduct function with the array, n, and x as inputs.
      e. Print “Yes” if the function returns true, and “No” otherwise.

Below is the implementation of the above idea:

C++

// C++ code for the approach

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

// Returns true if there is a pair in arr[0..n-1]
// with product equal to x.
bool isProduct(int arr[], int n, int x) {
    // sort the array arr
    sort(arr, arr + n);

    int l = 0, r = n - 1;

    // traverse the array inorder
    // using two pointer l and r
    while (l < r) {
        int prod = arr[l] * arr[r];

        // if product of element
        // at the two pinters is k
        // return this as res
        if (prod == x) {
            return true;
        }
        // if prod is less then
        // increase l as we have to
        // increase element value
        else if (prod < x)
            l++;
        // if prod is greater then
        // decrease r as we have to
        // decrease element value
        else
            r--;
    }

    return false;
}

// Driver code
int main() {
    int arr[] = { 10, 20, 9, 40 };
    int n = sizeof(arr) / sizeof(arr[0]);

    int x = 400;
    // Function call test case 1
    if (isProduct(arr, n, x)) {
        cout << "Yes" << endl;
    }
    else {
        cout << "No" << endl;
    }

    x = 190;
    // Function call test case 2
    if (isProduct(arr, n, x)) {
        cout << "Yes" << endl;
    }
    else {
        cout << "No" << endl;
    }

    return 0;
}

// This code is contributed by Chandramani Kumar

Output

Yes
No

Time Complexity: O(N * logN) where N is size of input array. This is because sort has been called which takes N*logN time.

Space Complexity: O(1) as no extra space has been taken.

Efficient Solution: We can improve time complexity to O(n) using hashing. Below are the steps.  

1. Create an empty hash table
2. Traverse array elements and do the following for every element arr[i]. 
   • If arr[i] is 0 and x is also 0, return true, else ignore arr[i].
   • If x % arr[i] is 0 and x/arr[i] exists in the table, it returns true.
   • Insert arr[i] into the hash table.
3. Return false

Below is the implementation of the above idea.  

C++14

// C++ program to find if there is a pair
// with given product.
#include<bits/stdc++.h>
using namespace std;

// Returns true if there is a pair in arr[0..n-1]
// with product equal to x.
bool isProduct(int arr[], int n, int x)
{
    if (n < 2)
        return false;

    // Create an empty set and insert first
    // element into it
    unordered_set<int> s;

    // Traverse remaining elements
    for (int i=0; i<n; i++)
    {
        // 0 case must be handles explicitly as
        // x % 0 is undefined behaviour in C++
        if (arr[i] == 0)
        {
           if (x == 0)
               return true;
           else
               continue;
        }

        // x/arr[i] exists in hash, then we
        // found a pair
        if (x%arr[i] == 0)
        {
            if (s.find(x/arr[i]) != s.end())
               return true;

            // Insert arr[i]
            s.insert(arr[i]);
        }
    }
    return false;
}

// Driver code
int main()
{
    int arr[] = {10, 20, 9, 40};
    int x = 400;

    int n = sizeof(arr)/sizeof(arr[0]);
    isProduct(arr, n, x)? cout << "Yes\n"
                       : cout << "No";

    x = 190;
    isProduct(arr, n, x)? cout << "Yes\n"
                        : cout << "No";

    return 0;
}

Java

// Java program if there exists a pair for given product
import java.util.HashSet;

class GFG
{
    // Returns true if there is a pair in arr[0..n-1]
    // with product equal to x.
    static boolean isProduct(int arr[], int n, int x)
    {
        // Create an empty set and insert first
        // element into it
        HashSet<Integer> hset = new HashSet<>();

        if(n < 2)
            return false;

        // Traverse remaining elements
        for(int i = 0; i < n; i++)
        {
            // 0 case must be handles explicitly as
            // x % 0 is undefined
            if(arr[i] == 0)
            {
                if(x == 0)
                    return true;
                else
                    continue;
            }

            // x/arr[i] exists in hash, then we
            // found a pair
            if(x % arr[i] == 0)
            {
                if(hset.contains(x / arr[i]))
                    return true;

            // Insert arr[i]
            hset.add(arr[i]);
            }
        }
        return false;
    }

    // driver code
    public static void main(String[] args)
    {
        int arr[] = {10, 20, 9, 40};
        int x = 400;
        int n = arr.length;

        if(isProduct(arr, arr.length, x))
        System.out.println("Yes");
        else
        System.out.println("No");

        x = 190;

        if(isProduct(arr, arr.length, x))
        System.out.println("Yes");
        else
        System.out.println("No");
    }
}

// This code is contributed by Kamal Rawal

Python3

# Python3 program to find if there
# is a pair with the given product.

# Returns true if there is a pair in
# arr[0..n-1] with product equal to x.
def isProduct(arr, n, x):

    if n < 2:
        return False

    # Create an empty set and insert
    # first element into it
    s = set()

    # Traverse remaining elements
    for i in range(0, n):

        # 0 case must be handles explicitly as
        # x % 0 is undefined behaviour in C++
        if arr[i] == 0:

            if x == 0:
                return True
            else:
                continue

        # x/arr[i] exists in hash, then
        # we found a pair
        if x % arr[i] == 0:

            if x // arr[i] in s:
                return True

            # Insert arr[i]
            s.add(arr[i])

    return False

# Driver code
if __name__ == "__main__":

    arr = [10, 20, 9, 40]
    x = 400

    n = len(arr)
    if isProduct(arr, n, x):
        print("Yes")
    else:
        print("No")

    x = 190
    if isProduct(arr, n, x):
        print("Yes")
    else:
        print("No")

# This code is contributed by
# Rituraj Jain

C#

// C# program if there exists a
// pair for given product
using System;
using System.Collections.Generic;

class GFG
{
// Returns true if there is a pair
// in arr[0..n-1] with product equal to x.
public static bool isProduct(int[] arr,
                             int n, int x)
{
    // Create an empty set and insert
    // first element into it
    HashSet<int> hset = new HashSet<int>();

    if (n < 2)
    {
        return false;
    }

    // Traverse remaining elements
    for (int i = 0; i < n; i++)
    {
        // 0 case must be handles explicitly
        // as x % 0 is undefined
        if (arr[i] == 0)
        {
            if (x == 0)
            {
                return true;
            }
            else
            {
                continue;
            }
        }

        // x/arr[i] exists in hash, then
        // we found a pair
        if (x % arr[i] == 0)
        {
            if (hset.Contains(x / arr[i]))
            {
                return true;
            }

        // Insert arr[i]
        hset.Add(arr[i]);
        }
    }
    return false;
}

// Driver Code
public static void Main(string[] args)
{
    int[] arr = new int[] {10, 20, 9, 40};
    int x = 400;
    int n = arr.Length;

    if (isProduct(arr, arr.Length, x))
    {
        Console.WriteLine("Yes");
    }
    else
    {
        Console.WriteLine("No");
    }

    x = 190;

    if (isProduct(arr, arr.Length, x))
    {
        Console.WriteLine("Yes");
    }
    else
    {
        Console.WriteLine("No");
    }
}
}

// This code is contributed by Shrikant13

Javascript

<script>

// Javascript program if there exists a pair for given product

// Returns true if there is a pair in arr[0..n-1]
    // with product equal to x.
    function isProduct(arr, n, x)
    {
        // Create an empty set and insert first
        // element into it
        let hset = new Set();

        if(n < 2)
            return false;

        // Traverse remaining elements
        for(let i = 0; i < n; i++)
        {
            // 0 case must be handles explicitly as
            // x % 0 is undefined
            if(arr[i] == 0)
            {
                if(x == 0)
                    return true;
                else
                    continue;
            }

            // x/arr[i] exists in hash, then we
            // found a pair
            if(x % arr[i] == 0)
            {
                if(hset.has(x / arr[i]))
                    return true;

            // Insert arr[i]
            hset.add(arr[i]);
            }
        }
        return false;
    }


// Driver program

         let arr = [10, 20, 9, 40];
        let x = 400;
        let n = arr.length;

        if(isProduct(arr, arr.length, x))
            document.write("Yes" + "<br/>");
        else
            document.write("No" + "<br/>");

        x = 190;

        if(isProduct(arr, arr.length, x))
            document.write("Yes" + "<br/>");
        else
            document.write("No" + "<br/>");

</script>

Output

Yes
No

Time Complexity : O(n²)
Auxiliary Space: O(n)

In the next set, we will be discussing approaches to print all pairs with products equal to 0.
This article is contributed by Aarti_Rathi and Shubham Goyal. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above