Given an array arr[] of size N, the task is to count the number of pairs (i, j) possible from the array such that arr[j] * i = arr[i] * j, where 1 ≤ i < j ≤ N.
Examples:
Input: arr[] = {1, 3, 5, 6, 5}
Output: 2
Explanation:
Pair (1, 5) satisfies the condition, since arr[1] * 5 = arr[5] * 1.
Pair (2, 4) satisfies the condition, since arr[2] * 4 = arr[4] * 2.
Therefore, total number of pairs satisfying the given condition is 2.Input: arr[] = {2, 1, 3}
Output: 0
Naive Approach: The simplest approach to solve the problem is to generate all possible pairs from the array and check for each pair, whether the given condition satisfies or not. Increase count for the pairs for which the condition is satisfied. Finally, print the count of all such pairs.
Below is the implementation of the above idea:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to count pairs from an // array satisfying given conditions void countPairs( int arr[], int N)
{ // Stores the total
// count of pairs
int count = 0;
for ( int i = 1; i <= N; i++) {
for ( int j = i + 1; j <= N; j++) {
// Note: Indices are 1 based according to
// question
if ((arr[j - 1] * i) == (arr[i - 1] * j)) {
count++;
}
}
}
cout << count;
} // Driver Code int main()
{ // Given array
int arr[] = { 1, 3, 5, 6, 5 };
// Size of the array
int N = sizeof (arr) / sizeof (arr[0]);
// Function call to count pairs
// satisfying given conditions
countPairs(arr, N);
return 0;
} |
// Java program for the above approach import java.util.*;
class GFG{
// Function to count pairs from an
// array satisfying given conditions
static void countPairs( int [] arr, int N)
{
// Stores the total
// count of pairs
int count = 0 ;
for ( int i = 1 ; i <= N; i++) {
for ( int j = i + 1 ; j <= N; j++) {
// Note: Indices are 1 based according to
// question
if ((arr[j - 1 ] * i) == (arr[i - 1 ] * j)) {
count++;
}
}
}
System.out.print(count);
}
// Driver Code
public static void main(String args[])
{
// Given array
int [] arr = { 1 , 3 , 5 , 6 , 5 };
// Size of the array
int N = arr.length;
// Function call to count pairs
// satisfying given conditions
countPairs(arr, N);
}
} |
# Python code for the above approach # Function to count pairs from an # array satisfying given conditions def countPairs(arr, N):
# Stores the total
# count of pairs
count = 0
for i in range ( 1 , N + 1 ):
for j in range (i + 1 , N + 1 ):
# Note: Indices are 1 based according to
# question
if (arr[j - 1 ] * i) = = (arr[i - 1 ] * j):
count + = 1
print (count)
# Given array arr = [ 1 , 3 , 5 , 6 , 5 ]
# Size of the array N = len (arr)
# Function call to count pairs # satisfying given conditions countPairs(arr, N) |
// C# code to implement the approach using System;
using System.Collections.Generic;
public class Gfg {
// Function to count pairs from an
// array satisfying given conditions
static void countPairs( int []arr, int N)
{
// Stores the total
// count of pairs
int count = 0;
for ( int i = 1; i <= N; i++) {
for ( int j = i + 1; j <= N; j++) {
// Note: Indices are 1 based according to
// question
if ((arr[j - 1] * i) == (arr[i - 1] * j)) {
count++;
}
}
}
Console.WriteLine(count);
}
// Driver Code
public static void Main( string [] args)
{
// Given array
int [] arr = { 1, 3, 5, 6, 5 };
// Size of the array
int N = arr.Length;
// Function call to count pairs
// satisfying given conditions
countPairs(arr, N);
}
} |
// Javascript program for the above approach // Function to count pairs from an // array satisfying given conditions function countPairs(arr, N)
{ // Stores the total
// count of pairs
let count = 0;
for (let i = 1; i <= N; i++) {
for (let j = i + 1; j <= N; j++) {
// Note: Indices are 1 based according to
// question
if ((arr[j - 1] * i) == (arr[i - 1] * j)) {
count++;
}
}
}
document.write(count);
} // Driver Code // Given array
let arr = [1, 3, 5, 6, 5 ];
// Size of the array
let N = arr.length;
// Function call to count pairs
// satisfying given conditions
countPairs(arr, N);
|
2
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, the idea is based on the rearrangement of the given equation arr[i] * j = arr[j] * i to arr[i] / i = arr[j] / j.
Follow the steps below to solve the problem:
- Initialize a variable, say count, to store the total count of pairs satisfying the given conditions.
- Initialize an Unordered Map, say mp, to count the frequency of values arr[i] / i.
- Traverse the array arr[] and update the frequencies of arr[i]/i in the Map.
- Print the count as the answer.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to count pairs from an // array satisfying given conditions void countPairs( int arr[], int N)
{ // Stores the total
// count of pairs
int count = 0;
// Stores count of a[i] / i
unordered_map< double , int > mp;
// Traverse the array
for ( int i = 0; i < N; i++) {
double val = 1.0 * arr[i];
double idx = 1.0 * (i + 1);
// Updating count
count += mp[val / idx];
// Update frequency
// in the Map
mp[val / idx]++;
}
// Print count of pairs
cout << count;
} // Driver Code int main()
{ // Given array
int arr[] = { 1, 3, 5, 6, 5 };
// Size of the array
int N = sizeof (arr) / sizeof (arr[0]);
// Function call to count pairs
// satisfying given conditions
countPairs(arr, N);
return 0;
} |
// Java program for the above approach import java.util.*;
class GFG{
// Function to count pairs from an // array satisfying given conditions static void countPairs( int []arr, int N)
{ // Stores the total
// count of pairs
int count = 0 ;
// Stores count of a[i]/i
Map<Double, Integer> mp
= new HashMap<Double, Integer>();
// Traverse the array
for ( int i = 0 ; i < N; i++)
{
Double val = 1.0 * arr[i];
Double idx = 1.0 * (i + 1 );
// Updating count
if (mp.containsKey(val / idx))
count += mp.get(val/idx);
// Update frequency
// in the Map
if (mp.containsKey(val / idx))
mp.put(val / idx, mp.getOrDefault(val / idx, 0 ) + 1 );
else
mp.put(val/idx, 1 );
}
// Print count of pairs
System.out.print(count);
} // Driver Code public static void main(String args[])
{ // Given array
int []arr = { 1 , 3 , 5 , 6 , 5 };
// Size of the array
int N = arr.length;
// Function call to count pairs
// satisfying given conditions
countPairs(arr, N);
} } // This code is contributed by ipg2016107. |
# Python3 program for the above approach from collections import defaultdict
# Function to count pairs from an # array satisfying given conditions def countPairs(arr, N):
# Stores the total
# count of pairs
count = 0
# Stores count of a[i] / i
mp = defaultdict( int )
# Traverse the array
for i in range (N):
val = 1.0 * arr[i]
idx = 1.0 * (i + 1 )
# Updating count
count + = mp[val / idx]
# Update frequency
# in the Map
mp[val / idx] + = 1
# Print count of pairs
print (count)
# Driver Code if __name__ = = "__main__" :
# Given array
arr = [ 1 , 3 , 5 , 6 , 5 ]
# Size of the array
N = len (arr)
# Function call to count pairs
# satisfying given conditions
countPairs(arr, N)
# This code is contributed by ukasp |
// C# program for the above approach using System;
using System.Collections.Generic;
class GFG{
// Function to count pairs from an // array satisfying given conditions static void countPairs( int []arr, int N)
{ // Stores the total
// count of pairs
int count = 0;
// Stores count of a[i]/i
Dictionary< double ,
int > mp = new Dictionary< double ,
int >();
// Traverse the array
for ( int i = 0; i < N; i++)
{
double val = 1.0 * arr[i];
double idx = 1.0 * (i + 1);
// Updating count
if (mp.ContainsKey(val / idx))
count += mp[val/idx];
// Update frequency
// in the Map
if (mp.ContainsKey(val / idx))
mp[val / idx]++;
else
mp[val/idx] = 1;
}
// Print count of pairs
Console.WriteLine(count);
} // Driver Code public static void Main()
{ // Given array
int []arr = { 1, 3, 5, 6, 5 };
// Size of the array
int N = arr.Length;
// Function call to count pairs
// satisfying given conditions
countPairs(arr, N);
} } // This code is contributed by SURENDRA_GANGWAR |
<script> // Javascript program for the above approach // Function to count pairs from an // array satisfying given conditions function countPairs(arr, N)
{ // Stores the total
// count of pairs
var count = 0;
// Stores count of a[i] / i
var mp = new Map();
// Traverse the array
for ( var i = 0; i < N; i++) {
var val = 1.0 * arr[i];
var idx = 1.0 * (i + 1);
// Updating count
count += mp.has(val/idx)?mp.get(val/idx):0
// Update frequency
// in the Map
if (mp.has(val/idx))
mp.set(val/idx, mp.get(val/idx)+1)
else
mp.set(val/idx, 1)
}
// Print count of pairs
document.write( count);
} // Driver Code // Given array var arr = [1, 3, 5, 6, 5];
// Size of the array var N = arr.length;
// Function call to count pairs // satisfying given conditions countPairs(arr, N); // This code is contributed by itsok. </script> |
2
Time Complexity: O(N)
Auxiliary Space: O(N)