Given an array arr[], the task is to print the count of unique pairs (arr[i], arr[j]) such that i < j.
Examples:
Input: arr[] = {1, 2, 1, 4, 5, 2}
Output: 11
The possible pairs are (1, 2), (1, 1), (1, 4), (1, 5), (2, 1), (2, 4), (2, 5), (2, 2), (4, 5), (4, 2), (5, 2)Input: arr[] = {1, 2, 3, 4}
Output: 6
The possible pairs are (1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)
Naive approach: The easiest way is to iterate through every possible pair and if it satisfies the condition, then add it to a set. Then, we can return the size of the set as our answer.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <iostream> #include <set> using namespace std; // Function to return the count // of unique pairs in the array int getPairs( int arr[], int n) { // Set to store unique pairs set<pair< int , int >> h; for ( int i = 0; i < (n - 1); i++) { for ( int j = i + 1; j < n; j++) { // Create pair of (arr[i], arr[j]) // and add it to the hashset h.insert(make_pair(arr[i], arr[j])); } } // Return the size of the HashSet return h.size(); } // Driver code int main() { int arr[] = { 1, 2, 2, 4, 2, 5, 3, 5 }; int n = sizeof (arr) / sizeof (arr[0]); printf ( "%d" , getPairs(arr, n)) ; return 0; } // This code is contributed by SHUBHAMSINGH10 |
Java
// Java implementation of the approach import java.util.HashSet; import javafx.util.Pair; class GFG { // Function to return the count // of unique pairs in the array static int getPairs( int arr[], int n) { // HashSet to store unique pairs HashSet<Pair> h = new HashSet<Pair>(); for ( int i = 0 ; i < n - 1 ; i++) { for ( int j = i + 1 ; j < n; j++) { // Create pair of (a[i], a[j]) // and add it to the hashset Pair<Integer, Integer> p = new Pair<>(arr[i], arr[j]); h.add(p); } } // Return the size of the HashSet return h.size(); } // Driver code public static void main(String[] args) { int arr[] = { 1 , 2 , 2 , 4 , 2 , 5 , 3 , 5 }; int n = arr.length; System.out.println(getPairs(arr, n)); } } |
Python3
# Python3 implementation of the approach # Function to return the count # of unique pairs in the array def getPairs(arr, n) : # Set to store unique pairs h = set () for i in range (n - 1 ) : for j in range (i + 1 , n) : # Create pair of (a[i], a[j]) # and add it to the hashset h.add((arr[i], arr[j])); # Return the size of the HashSet return len (h); # Driver code if __name__ = = "__main__" : arr = [ 1 , 2 , 2 , 4 , 2 , 5 , 3 , 5 ] n = len (arr) print (getPairs(arr, n)) # This code is contributed by Ryuga |
C#
// C# implementation of the approach using System; using System.Collections.Generic; class GFG{ // Function to return the count // of unique pairs in the array static int getPairs( int []arr, int n) { // HashSet to store unique pairs HashSet<Tuple< int , int >> h = new HashSet<Tuple< int , int >>(); for ( int i = 0; i < n - 1; i++) { for ( int j = i + 1; j < n; j++) { // Create pair of (a[i], a[j]) // and add it to the hashset Tuple< int , int > p = new Tuple< int , int >(arr[i], arr[j]); h.Add(p); } } // Return the size of the HashSet return h.Count; } // Driver code public static void Main(String[] args) { int []arr = { 1, 2, 2, 4, 2, 5, 3, 5 }; int n = arr.Length; Console.WriteLine(getPairs(arr, n)); } } // This code is contributed by Amit Katiyar |
14
Time Complexity: O(n2)
Note: Please use an offline IDE to compile the above code. Online compilers might not support JavaFX.
Efficient Approach: Every element arr[i] can form a pair with the element arr[j] if i < j. But (arr[i], arr[j]) should be unique therefore for every unique arr[i], possible pairs will be equal to the number of distinct numbers in the sub-array arr[i + 1], arr[i + 2], …, arr[n – 1]. So for every arr[i], we will find the unique elements from right to left. For this task, it is easy to keep track of the elements visited by using a Hash Tablle. In this way, we will have unique arr[i] for every unique arr[j]. Now, we will sum these values for every unique arr[i] which is the desired count of pairs.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include<bits/stdc++.h> using namespace std; // Function to return the count // of unique pairs in the array int getPairs( int a[], int n) { set< int > visited1; // un[i] stores number of unique elements // from un[i + 1] to un[n - 1] int un[n] ; // Last element will have no unique elements // after it un[n - 1] = 0; // To count unique elements after every a[i] int count = 0; // auto pos = s.find(3); // prints the set elements // cout << "The set elements after 3 are: "; // for (auto it = pos; it != s.end(); it++) // cout << *it << " " for ( int i = n - 1; i > 0; i--) { // If current element has already been used // i.e. not unique auto pos = visited1.find(a[i]); if (pos != visited1.end()) un[i - 1] = count; else un[i - 1] = ++count; // Set to true if a[i] is visited visited1.insert(a[i]); } set< int >visited2; // To know which a[i] is already visited int answer = 0; for ( int i = 0; i < n - 1; i++) { // If visited, then the pair would // not be unique auto pos = visited2.find(a[i]); if (pos != visited2.end()) continue ; // Calculating total unqiue pairs answer += un[i]; // Set to true if a[i] is visited visited2.insert(a[i]); } return answer; } // Driver code int main() { int a[] = { 1, 2, 2, 4, 2, 5, 3, 5 }; int n = sizeof (a)/ sizeof (a[0]); // Print the count of unique pairs cout<<(getPairs(a, n)); } // This code is contributed by Rajput-Ji |
Java
// Java implementation of the approach import java.util.HashSet; public class GFG { // Function to return the count // of unique pairs in the array static int getPairs( int a[], int n) { HashSet<Integer> visited1 = new HashSet<Integer>(); // un[i] stores number of unique elements // from un[i + 1] to un[n - 1] int un[] = new int [n]; // Last element will have no unique elements // after it un[n - 1 ] = 0 ; // To count unique elements after every a[i] int count = 0 ; for ( int i = n - 1 ; i > 0 ; i--) { // If current element has already been used // i.e. not unique if (visited1.contains(a[i])) un[i - 1 ] = count; else un[i - 1 ] = ++count; // Set to true if a[i] is visited visited1.add(a[i]); } HashSet<Integer> visited2 = new HashSet<Integer>(); // To know which a[i] is already visited int answer = 0 ; for ( int i = 0 ; i < n - 1 ; i++) { // If visited, then the pair would // not be unique if (visited2.contains(a[i])) continue ; // Calculating total unqiue pairs answer += un[i]; // Set to true if a[i] is visited visited2.add(a[i]); } return answer; } // Driver code public static void main(String[] args) { int a[] = { 1 , 2 , 2 , 4 , 2 , 5 , 3 , 5 }; int n = a.length; // Print the count of unique pairs System.out.println(getPairs(a, n)); } } |
Python3
# Python3 implementation of the approach # Function to return the count # of unique pairs in the array def getPairs(a, n): visited1 = set () # un[i] stores number of unique elements # from un[i + 1] to un[n - 1] un = [ 0 ] * n # Last element will have no unique elements # after it un[n - 1 ] = 0 # To count unique elements after every a[i] count = 0 for i in range (n - 1 , - 1 , - 1 ): # If current element has already been used # i.e. not unique if (a[i] in visited1): un[i - 1 ] = count else : count + = 1 un[i - 1 ] = count # Set to true if a[i] is visited visited1.add(a[i]) visited2 = set () # To know which a[i] is already visited answer = 0 for i in range (n - 1 ): # If visited, then the pair would # not be unique if (a[i] in visited2): continue # Calculating total unqiue pairs answer + = un[i] # Set to true if a[i] is visited visited2.add(a[i]) return answer # Driver code a = [ 1 , 2 , 2 , 4 , 2 , 5 , 3 , 5 ] n = len (a) # Print the count of unique pairs print (getPairs(a, n)) # This code is contributed by SHUBHAMSINGH10 |
C#
// C# implementation of the approach using System; using System.Collections.Generic; class GFG { // Function to return the count // of unique pairs in the array static int getPairs( int []a, int n) { HashSet< int > visited1 = new HashSet< int >(); // un[i] stores number of unique elements // from un[i + 1] to un[n - 1] int []un = new int [n]; // Last element will have no unique elements // after it un[n - 1] = 0; // To count unique elements after every a[i] int count = 0; for ( int i = n - 1; i > 0; i--) { // If current element has already been used // i.e. not unique if (visited1.Contains(a[i])) un[i - 1] = count; else un[i - 1] = ++count; // Set to true if a[i] is visited visited1.Add(a[i]); } HashSet< int > visited2 = new HashSet< int >(); // To know which a[i] is already visited int answer = 0; for ( int i = 0; i < n - 1; i++) { // If visited, then the pair would // not be unique if (visited2.Contains(a[i])) continue ; // Calculating total unqiue pairs answer += un[i]; // Set to true if a[i] is visited visited2.Add(a[i]); } return answer; } // Driver code public static void Main(String[] args) { int []a = { 1, 2, 2, 4, 2, 5, 3, 5 }; int n = a.Length; // Print the count of unique pairs Console.WriteLine(getPairs(a, n)); } } /* This code contributed by PrinciRaj1992 */ |
14
Time Complexity: O(n)
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