Given an array of distinct integers, the task is to find two pairs (a, b) and (c, d) such that ab = cd, where a, b, c and d are distinct elements.
Examples:
Input : arr[] = {3, 4, 7, 1, 2, 9, 8}
Output : 4 2 and 1 8
Product of 4 and 2 is 8 and
also product of 1 and 8 is 8 .
Input : arr[] = {1, 6, 3, 9, 2, 10};
Output : 6 3 and 9 2
A Simple Solution is to run four loops to generate all possible quadruples of array element. For every quadruple (a, b, c, d), check if a*b = c*d. Time complexity of this solution is O(n4).
An Efficient Solution of this problem is to use hashing. We use product as key and pair as value in hash table.
1. For i=0 to n-1
2. For j=i+1 to n-1
a) Find prod = arr[i]*arr[j]
b) If prod is not available in hash then make
H[prod] = make_pair(i, j) // H is hash table
c) If product is also available in hash
then print previous and current elements
of array
C++
// C++ program to find four elements a, b, c // and d in array such that ab = cd #include<bits/stdc++.h> using namespace std; // Function to find out four elements in array // whose product is ab = cd void findPairs(int arr[], int n) { bool found = false; unordered_map<int, pair < int, int > > H; for (int i=0; i<n; i++) { for (int j=i+1; j<n; j++) { // If product of pair is not in hash table, // then store it int prod = arr[i]*arr[j]; if (H.find(prod) == H.end()) H[prod] = make_pair(i,j); // If product of pair is also available in // then print current and previous pair else { pair<int,int> pp = H[prod]; cout << arr[pp.first] << " " << arr[pp.second] << " and " << arr[i]<<" "<<arr[j]<<endl; found = true; } } } // If no pair find then print not found if (found == false) cout << "No pairs Found" << endl; } //Driven code int main() { int arr[] = {1, 2, 3, 4, 5, 6, 7, 8}; int n = sizeof(arr)/sizeof(int); findPairs(arr, n); return 0; } |
Java
// Java program to find four elements a, b, c // and d in array such that ab = cd import java.io.*; import java.util.*; class GFG { public static class pair { int first,second; pair(int f, int s) { first = f; second = s; } }; // Function to find out four elements // in array whose product is ab = cd public static void findPairs(int arr[], int n) { boolean found = false; HashMap<Integer, pair> hp = new HashMap<Integer, pair>(); for(int i = 0; i < n; i++) { for(int j = i + 1; j < n; j++) { // If product of pair is not in // hash table, then store it int prod = arr[i] * arr[j]; if(!hp.containsKey(prod)) hp.put(prod, new pair(i,j)); // If product of pair is also // available in then print // current and previous pair else { pair p = hp.get(prod); System.out.println(arr[p.first] + " " + arr[p.second] + " " + "and" + " " + arr[i] + " " + arr[j]); found = true; } } } // If no pair find then print not found if(found == false) System.out.println("No pairs Found"); } // Driver code public static void main (String[] args) { int arr[] = {1, 2, 3, 4, 5, 6, 7, 8}; int n = arr.length; findPairs(arr, n); } } // This code is contributed by akash1295. |
Python3
# Python3 program to find four elements # a, b, c and d in array such that ab = cd # Function to find out four elements in array # whose product is ab = cd def findPairs(arr, n): found = False H = dict() for i in range(n): for j in range(i + 1, n): # If product of pair is not in hash table, # then store it prod = arr[i] * arr[j] if (prod not in H.keys()): H[prod] = [i, j] # If product of pair is also available in # then prcurrent and previous pair else: pp = H[prod] print(arr[pp[0]], arr[pp[1]], "and", arr[i], arr[j]) found = True # If no pair find then prnot found if (found == False): print("No pairs Found") # Driver code arr = [1, 2, 3, 4, 5, 6, 7, 8] n = len(arr) findPairs(arr, n) # This code is contributed # by mohit kumar |
C#
// C# program to find four elements a, b, c // and d in array such that ab = cd using System; using System.Collections.Generic; class GFG { public class pair { public int first,second; public pair(int f, int s) { first = f; second = s; } }; // Function to find out four elements // in array whose product is ab = cd public static void findPairs(int[] arr, int n) { bool found = false; Dictionary<int, pair> hp = new Dictionary<int, pair>(); for(int i = 0; i < n; i++) { for(int j = i + 1; j < n; j++) { // If product of pair is not in // hash table, then store it int prod = arr[i] * arr[j]; if(!hp.ContainsKey(prod)) hp.Add(prod, new pair(i,j)); // If product of pair is also // available in then print // current and previous pair else { pair p = hp[prod]; Console.WriteLine(arr[p.first] + " " + arr[p.second] + " " + "and" + " " + arr[i] + " " + arr[j]); found = true; } } } // If no pair find then print not found if(found == false) Console.WriteLine("No pairs Found"); } // Driver code public static void Main (String[] args) { int []arr = {1, 2, 3, 4, 5, 6, 7, 8}; int n = arr.Length; findPairs(arr, n); } } /* This code contributed by PrinciRaj1992 */ |
Output:
1 6 and 2 3 1 8 and 2 4 2 6 and 3 4 3 8 and 4 6
Time Complexity : O(n2) assuming hash search and insert operations take O(1) time.
Related Article:
Find four elements a, b, c and d in an array such that a+b = c+d
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