Given an array A[] of integers. The task is to find the total number of ordered pairs of positive integers (X, Y) such that X occurs in A[] at least Y times and Y occurs in A at least X times.
Examples:
Input : A[] = { 1, 1, 2, 2, 3 }
Output : 4
Ordered pairs are -> { [1, 1], [1, 2], [2, 1], [2, 2] }
Input : A = { 3, 3, 2, 2, 2 }
Output : 3
Ordered pairs are -> { [3, 2], [2, 2], [2, 3] }
Approach:
- Create a hash table m[] of count of elements of array A[].
- Traverse the hash table of unique elements. Let X be current key in the hash table Y be its frequency.
- Check for each element j = (1 to Y) such that, if m[ j ] >= X increment answer by 1.
- Return the count of total ordered pairs (X, Y) of array A.
Below is the implementation of above approach:
C++
// C++ program to find number// of ordered pairs#include <bits/stdc++.h>using namespace std;// Function to find count of Ordered pairsint countOrderedPairs(int A[], int n){ // Initialize pairs to 0 int orderedPairs = 0; // Store frequencies unordered_map<int, int> m; for (int i = 0; i < n; ++i) m[A[i]]++; // Count total Ordered_pairs for (auto entry : m) { int X = entry.first; int Y = entry.second; for (int j = 1; j <= Y; j++) { if (m[j] >= X) orderedPairs++; } } return orderedPairs;}// Driver Codeint main(){ int A[] = { 1, 1, 2, 2, 3 }; int n = sizeof(A) / sizeof(A[0]); cout << countOrderedPairs(A, n); return 0;} |
Java
// Java program to find number// of ordered pairsimport java.util.HashMap;import java.util.Map;class GFG{ // Function to find count of Ordered pairs public static int countOrderedPairs(int[] A, int n) { // Initialize pairs to 0 int orderedPairs = 0; // Store frequencies HashMap<Integer, Integer> m = new HashMap<>(); for (int i = 0; i < n; i++) { if (m.get(A[i]) == null) m.put(A[i], 1); else { int a = m.get(A[i]); m.put(A[i], ++a); } } // Count total Ordered_pairs for (int entry : m.keySet()) { int X = entry; int Y = m.get(entry); for (int j = 1; j <= Y; j++) { if (m.get(j) >= X) orderedPairs++; } } return orderedPairs; } // Driver Code public static void main(String[] args) { int[] A = {1, 1, 2, 2, 3}; int n = A.length; System.out.print(countOrderedPairs(A, n)); }}// This code is contributed by// sanjeev2552 |
Python3
# Python3 program to find the # number of ordered pairs from collections import defaultdict# Function to find count of Ordered pairs def countOrderedPairs(A, n): # Initialize pairs to 0 orderedPairs = 0 # Store frequencies m = defaultdict(lambda:0) for i in range(0, n): m[A[i]] += 1 # Count total Ordered_pairs for X,Y in m.items(): for j in range(1, Y + 1): if m[j] >= X: orderedPairs += 1 return orderedPairs # Driver Code if __name__ == "__main__": A = [1, 1, 2, 2, 3] n = len(A) print(countOrderedPairs(A, n)) # This code is contributed by Rituraj Jain |
C#
// C# program to illustrate how // to create a dictionary using System; using System.Collections.Generic; class GFG{ // Function to find count of Ordered pairs public static int countOrderedPairs(int[] A, int n) { // Initialize pairs to 0 int orderedPairs = 0; // Store frequencies Dictionary<int, int> m = new Dictionary<int, int>(); for (int i = 0; i < n; i++) { if (!m.ContainsKey(A[i])) m.Add(A[i], 1); else { m[A[i]]++; } } // Count total Ordered_pairs foreach(KeyValuePair<int, int> entry in m) { int X = entry.Key; int Y = entry.Value; for (int j = 1; j <= Y; j++) { if (m[j] >= X) orderedPairs++; } } return orderedPairs; } // Driver Code public static void Main() { int[] A = {1, 1, 2, 2, 3}; int n = A.Length; Console.Write(countOrderedPairs(A, n)); }}// This code is contributed by// mohit kumar |
Javascript
<script>// JavaScript program to find number// of ordered pairs// Function to find count of Ordered pairsfunction countOrderedPairs(A, n){ // Initialize pairs to 0 let orderedPairs = 0; // Store frequencies let m = new Map(); for (let i = 0; i < n; ++i){ if(m.has(A[i])){ m.set(A[i],m.get(A[i])+1); } else m.set(A[i],1); } // Count total Ordered_pairs for (let [key,value] of m) { let X = key; let Y = value; for (let j = 1; j <= Y; j++) { if (m.get(j) >= X) orderedPairs++; } } return orderedPairs;}// Driver Codelet A =[ 1, 1, 2, 2, 3 ];let n = A.length;document.write(countOrderedPairs(A, n));// This code is contributed by shinjanpatra</script> |
Output
4
Complexity Analysis:
- Time Complexity: O(n), for traversing inside map key-value pairs
- Auxiliary Space: O(n), to create a map of size n
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!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
