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 ( O(n2) ) is to run two loops to consider all possible pairs. For every pair, check if 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; i<n; i++) 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 << "Yesn" : cout << "Non"; x = 190; isProduct(arr, n, x)? cout << "Yesn" : cout << "Non"; 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 = 400n = len(arr) if(isProduct(arr,n, x) == True): print ("Yes") else: print("No") x = 900if(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"; elseecho "No\n"; $x = 190; if(isProduct($arr, $n, $x)) echo "Yes\n"; elseecho "No\n"; // This code is contributed // by Sam007 ?> |
Output :
Yes No
Better Solution (O(n Log n) : 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 subarry on right of arr[i], i.e., in subarray arr[i+1..n-1]
Efficient Solution ( O(n) ): We can improve time complexity to O(n) using hashing. Below are steps.
- Create an empty hash table
- Traverse array elements and do 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 table, return true.
- Insert arr[i] into the hash table.
- Return false
Below is the implementation of above idea.
C++
// 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 << "Non"; x = 190; isProduct(arr, n, x)? cout << "Yes\n" : cout << "Non"; 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 |
Output :
Yes No
In the next set, we will be discussing approaches to print all pairs with product equal to 0.
This article is contributed by Shubham Goyal. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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