Given an array arr[] consisting of N integers, the task is to count the number of distinct pairs of array elements having the product of indices equals the product of elements at that indices. The pairs (x, y) and (y, x) are considered as the same pairs.
Examples:
Input: arr[] = {1, 0, 3, 2, 6}
Output: 3
Explanation: All possible pairs satisfying the given criteria are:
- (0, 1): Product of indices = 1 * 0 = 0. Product of elements = 1 * 0 = 0.
- (2, 3): Product of indices = 2 * 3 = 6. Product of elements = 3 * 2 = 6.
- (3, 4): Product of indices = 3 * 4 = 12. Product of elements = 2 * 6 = 12.
Therefore, the total count of pairs is 3.
Input: arr[] = {4, -1, 2, 6, 2, 10}
Output: 2
Approach: The given problem can be solved by generating all possible distinct pairs of elements (arr[i], arr[j]) from the given array and if the product of i and j is equal to the product of array elements arr[i] and arr[j], then increment the count of such pairs. After checking for all the distinct pairs, print the total count as the result.
Below is the implementation of the above approach:
// C++ program for the above approach #include <iostream> using namespace std;
// Function to count the number of pairs // having product of indices equal to the // product of elements at that indices int CountPairs( int arr[], int n)
{ // Stores the count of valid pairs
int count = 0;
// Generate all possible pairs
for ( int i = 0; i < n - 1; i++) {
for ( int j = i + 1; j < n; j++) {
// If the condition is satisfied
if ((i * j) == (arr[i] * arr[j]))
// Increment the count
count++;
}
}
// Return the total count
return count;
} // Driver Code int main()
{ int arr[] = { 1, 0, 3, 2, 6 };
int N = sizeof (arr) / sizeof (arr[0]);
cout << CountPairs(arr, N);
return 0;
} |
// Java program for the above approach import java.util.*;
class GFG{
// Function to count the number of pairs // having product of indices equal to the // product of elements at that indices static int CountPairs( int [] arr, int n)
{ // Stores the count of valid pairs
int count = 0 ;
// Generate all possible pairs
for ( int i = 0 ; i < n - 1 ; i++)
{
for ( int j = i + 1 ; j < n; j++)
{
// If the condition is satisfied
if ((i * j) == (arr[i] * arr[j]))
// Increment the count
count++;
}
}
// Return the total count
return count;
} // Driver Code public static void main (String[] args)
{ int [] arr = { 1 , 0 , 3 , 2 , 6 };
int N = arr.length;
System.out.println(CountPairs(arr, N));
} } // This code is contributed by sanjoy_62 |
# Python3 program for the above approach # Function to count the number of pairs # having product of indices equal to the # product of elements at that indices def CountPairs(arr, n):
# Stores the count of valid pairs
count = 0
# Generate all possible pairs
for i in range (n - 1 ):
for j in range (i + 1 , n):
# If the condition is satisfied
if ((i * j) = = (arr[i] * arr[j])):
# Increment the count
count + = 1
# Return the total count
return count
# Driver Code if __name__ = = "__main__" :
arr = [ 1 , 0 , 3 , 2 , 6 ]
N = len (arr)
print (CountPairs(arr, N))
# This code is contributed by AnkThon |
// C# program for the above approach using System;
class GFG{
// Function to count the number of pairs // having product of indices equal to the // product of elements at that indices static int CountPairs( int [] arr, int n)
{ // Stores the count of valid pairs
int count = 0;
// Generate all possible pairs
for ( int i = 0; i < n - 1; i++)
{
for ( int j = i + 1; j < n; j++)
{
// If the condition is satisfied
if ((i * j) == (arr[i] * arr[j]))
// Increment the count
count++;
}
}
// Return the total count
return count;
} // Driver Code public static void Main()
{ int [] arr = { 1, 0, 3, 2, 6 };
int N = arr.Length;
Console.Write(CountPairs(arr, N));
} } // This code is contributed by ukasp |
<script> // javascript program for the above approach // Function to count the number of pairs
// having product of indices equal to the
// product of elements at that indices
function CountPairs(arr , n) {
// Stores the count of valid pairs
var count = 0;
// Generate all possible pairs
for (i = 0; i < n - 1; i++) {
for (j = i + 1; j < n; j++) {
// If the condition is satisfied
if ((i * j) == (arr[i] * arr[j]))
// Increment the count
count++;
}
}
// Return the total count
return count;
}
// Driver Code
var arr = [ 1, 0, 3, 2, 6 ];
var N = arr.length;
document.write(CountPairs(arr, N));
// This code contributed by Rajput-Ji </script> |
3
Time Complexity: O(N2)
Auxiliary Space: O(1)