Given an array a[] of non-negative integers. The task is to count the number of pairs (i, j) in the array such that LCM(a[i], a[j]) = HCF(a[i], a[j]).
Note: The pair (i, j) and (j, i) are considered same and i should not be equal to j.
Examples:
Input : a[] = {3, 4, 3, 4, 5}
Output : 2
Pairs are (3, 3) and (4, 4)
Input : a[] = {1, 1, 1}
Output : 3
Naive approach: Generate all possible pairs and count pairs with equal HCF and LCM.
C++
// Naive C++ program to count number of pairs// such that their hcf and lcm are equal#include <bits/stdc++.h>using namespace std;// Function to return HCF of two numbersint gcd(int a, int b){ if (a == 0) return b; return gcd(b % a, a);}// Function to return LCM of two numbersint lcm(int a, int b){ return (a * b) / gcd(a, b);}// Returns the number of valid pairsint countPairs(int arr[], int n){ int ans = 0; // initializing answer // Traversing the array. For each array // element, checking if it // follow the condition for (int i = 0; i < n; i++) for (int j = i + 1; j < n; j++) if (lcm(arr[i], arr[j]) == gcd(arr[i], arr[j])) ans++; return ans;}// Driver functionint main(){ int arr[] = { 1, 1, 1 }; int n = sizeof(arr) / sizeof(arr[0]); cout << countPairs(arr, n); return 0;} |
Java
// Naive Java program to count number of pairs// such that their hcf and lcm are equalimport java.util.*;class GFG{ // Function to return HCF of two numbersstatic int gcd(int a, int b){ if (a == 0) return b; return gcd(b % a, a);} // Function to return LCM of two numbersstatic int lcm(int a, int b){ return (a * b) / gcd(a, b);} // Returns the number of valid pairsstatic int countPairs(int arr[], int n){ int ans = 0; // initializing answer // Traversing the array. For each array // element, checking if it // follow the condition for (int i = 0; i < n; i++) for (int j = i + 1; j < n; j++) if (lcm(arr[i], arr[j]) == gcd(arr[i], arr[j])) ans++; return ans;} // Driver functionpublic static void main(String[] args){ int arr[] = { 1, 1, 1 }; int n = arr.length; System.out.print(countPairs(arr, n));}}// This code is contributed by Rajput-Ji |
Python3
# Naive Python program to count number of pairs# such that their hcf and lcm are equal# Function to return HCF of two numbersdef gcd(a, b): if (a == 0): return b; return gcd(b % a, a);# Function to return LCM of two numbersdef lcm(a, b): return (a * b) / gcd(a, b);# Returns the number of valid pairsdef countPairs(arr, n): ans = 0; # initializing answer # Traversing the array. For each array # element, checking if it # follow the condition for i in range(n): for j in range(i+1,n): if (lcm(arr[i], arr[j]) == gcd(arr[i], arr[j])): ans+=1; return ans;# Driver functionif __name__ == '__main__': arr = [ 1, 1, 1 ]; n = len(arr); print(countPairs(arr, n));# This code is contributed by 29AjayKumar |
C#
// Naive C# program to count number of pairs// such that their hcf and lcm are equalusing System;class GFG{ // Function to return HCF of two numbersstatic int gcd(int a, int b){ if (a == 0) return b; return gcd(b % a, a);} // Function to return LCM of two numbersstatic int lcm(int a, int b){ return (a * b) / gcd(a, b);} // Returns the number of valid pairsstatic int countPairs(int []arr, int n){ int ans = 0; // initializing answer // Traversing the array. For each array // element, checking if it // follow the condition for (int i = 0; i < n; i++) for (int j = i + 1; j < n; j++) if (lcm(arr[i], arr[j]) == gcd(arr[i], arr[j])) ans++; return ans;} // Driver functionpublic static void Main(String[] args){ int []arr = { 1, 1, 1 }; int n = arr.Length; Console.Write(countPairs(arr, n));}}// This code is contributed by 29AjayKumar |
Javascript
<script>// Naive Javascript program to count number of pairs// such that their hcf and lcm are equal// Function to return HCF of two numbersfunction gcd(a, b){ if (a == 0) return b; return gcd(b % a, a);}// Function to return LCM of two numbersfunction lcm(a, b){ return (a * b) / gcd(a, b);}// Returns the number of valid pairsfunction countPairs(arr, n){ var ans = 0; // initializing answer // Traversing the array. For each array // element, checking if it // follow the condition for (var i = 0; i < n; i++) for (var j = i + 1; j < n; j++) if (lcm(arr[i], arr[j]) == gcd(arr[i], arr[j])) ans++; return ans;}// Driver functionvar arr = [ 1, 1, 1 ];var n = arr.length;document.write(countPairs(arr, n));// This code is contributed by rutvik_56.</script> |
Output:
3
Time Complexity: O(n2 * log(max(a, b)))
Auxiliary Space: O(log(max(a, b)))
Efficient Approach: If we observe carefully, it can be proved that the HCF and LCM of two numbers can be equal only when the numbers are also equal.
Proof:
Let, HCF(a[i], a[j]) = LCM(a[i], a[j]) = K Since HCF(a[i], a[j]) = k, a[i] = k*n1, a[j] = k*n2, for some natural numbers n1, n2 We know that, HCF × LCM = Product of the two numbers Therefore, k*k = k*n1 × k*n2 or, n1*n2 = 1 Implies, n1 = n2 = 1, since n1, n2 are natural numbers. Therefore, a[i] = k*n1 = k, a[j] = k*n2 = k i.e. the numbers must be equal.
So we have to count pairs with same elements in the pair.
It is observed that only the pairs of the form (arr[i], arr[j]) where arr[i] = arr[j] will satisfy the given condition. So, the problem now gets reduced to finding the number of pairs (arr[i], arr[j]) such that arr[i] = arr[j].
Below is the implementation of the above approach:
C++
// C++ program to count number of pairs// such that their hcf and lcm are equal#include <bits/stdc++.h>using namespace std;// Function to count number of pairs// such that their hcf and lcm are equalint countPairs(int a[], int n){ // Store frequencies of array elements unordered_map<int, int> frequency; for (int i = 0; i < n; i++) { frequency[a[i]]++; } int count = 0; // Count of pairs (arr[i], arr[j]) // where arr[i] = arr[j] for (auto x : frequency) { int f = x.second; count += f * (f - 1) / 2; } // Count of pairs (arr[i], arr[j]) where // arr[i] = arr[j], return count;}// Driver functionint main(){ int arr[] = { 1, 1, 1 }; int n = sizeof(arr) / sizeof(arr[0]); cout << countPairs(arr, n); return 0;} |
Java
// Java program to count number of pairs// such that their hcf and lcm are equalimport java.util.*;class GFG{ // Function to count number of pairs// such that their hcf and lcm are equalstatic int countPairs(int arr[], int n){ // Store frequencies of array elements HashMap<Integer,Integer> frequency = new HashMap<Integer,Integer>(); for (int i = 0; i < n; i++) { if(frequency.containsKey(arr[i])){ frequency.put(arr[i], frequency.get(arr[i])+1); }else{ frequency.put(arr[i], 1); } } int count = 0; // Count of pairs (arr[i], arr[j]) // where arr[i] = arr[j] for (Map.Entry<Integer,Integer> x : frequency.entrySet()) { int f = x.getValue(); count += f * (f - 1) / 2; } // Count of pairs (arr[i], arr[j]) where // arr[i] = arr[j], return count;} // Driver functionpublic static void main(String[] args){ int arr[] = { 1, 1, 1 }; int n = arr.length; System.out.print(countPairs(arr, n));}}// This code contributed by sapnasingh4991 |
C#
// C# program to count number of pairs// such that their hcf and lcm are equalusing System;using System.Collections.Generic;class GFG{// Function to count number of pairs// such that their hcf and lcm are equalstatic int countPairs(int []a, int n){ // Store frequencies of array elements Dictionary<int, int> frequency = new Dictionary<int,int>(); for (int i = 0; i < n; i++) { if (frequency.ContainsKey(a[i])) { var val = frequency[a[i]]; frequency.Remove(a[i]); frequency.Add(a[i], val + 1); } else { frequency.Add(a[i], 1); } } int count = 0; // Count of pairs (arr[i], arr[j]) // where arr[i] = arr[j] foreach(KeyValuePair<int, int> entry in frequency) { int f = entry.Value; count += f * (f - 1) / 2; } // Count of pairs (arr[i], arr[j]) where // arr[i] = arr[j], return count;}// Driver functionpublic static void Main(String[] args){ int []arr = { 1, 1, 1 }; int n = arr.Length; Console.Write(countPairs(arr, n));}}// This code is contributed by shivanisinghss2110 |
Python3
# Python 3 program to count number of pairs# such that their hcf and lcm are equalfrom collections import defaultdict # Function to count number of pairs# such that their hcf and lcm are equaldef countPairs(a, n): # Store frequencies of array elements frequency = defaultdict(int) for i in range(n) : frequency[a[i]] += 1 count = 0 # Count of pairs (arr[i], arr[j]) # where arr[i] = arr[j] for x in frequency.keys(): f = frequency[x] count += f * (f - 1) // 2 # Count of pairs (arr[i], arr[j]) where # arr[i] = arr[j], return count # Driver functionif __name__ == "__main__": arr = [ 1, 1, 1 ] n = len(arr) print(countPairs(arr, n))# This code is contributed by chitranayal |
Javascript
<script>// JavaScript program to count number of pairs// such that their hcf and lcm are equal// Function to count number of pairs// such that their hcf and lcm are equalfunction countPairs(arr,n){ // Store frequencies of array elements let frequency = new Map(); for (let i = 0; i < n; i++) { if(frequency.has(arr[i])){ frequency.set(arr[i], frequency.get(arr[i])+1); }else{ frequency.set(arr[i], 1); } } let count = 0; // Count of pairs (arr[i], arr[j]) // where arr[i] = arr[j] for (let [key, value] of frequency.entries()) { let f = value; count += f * (f - 1) / 2; } // Count of pairs (arr[i], arr[j]) where // arr[i] = arr[j], return count;}// Driver functionlet arr=[1, 1, 1 ];let n = arr.length;document.write(countPairs(arr, n));// This code is contributed by avanitrachhadiya2155</script> |
Output:
3
Time Complexity: O(n)
Auxiliary Space: O(n)
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