Given an array of N distinct integers, the task is to find the number of pairs (x, y) such that x < y.
Example:
Input: arr[] = {2, 4, 3, 1}
Output: 6
Possible pairs are (1, 2), (1, 3), (1, 4), (2, 3), (2, 4) and (3, 4).
Input: arr[] = {5, 10}
Output: 1
The only possible pair is (5, 10).
Naive approach: Find every possible pair and check whether it satisfies the given condition or not.
// C++ implementation of the approach #include <iostream> using namespace std;
// Function to return the number of // pairs (x, y) such that x < y int getPairs( int a[], int n)
{ // To store the number of valid pairs
int count = 0;
for ( int i = 0; i < n; i++)
{
for ( int j = 0; j < n; j++)
{
// If a valid pair is found
if (a[i] < a[j])
count++;
}
}
// Return the count of valid pairs
return count;
} // Driver code int main()
{ int a[] = { 2, 4, 3, 1 };
int n = sizeof (a) / sizeof (a[0]);
cout << getPairs(a, n);
return 0;
} // This code is contributed by SHUBHAMSINGH10 |
// Java implementation of the approach class GFG {
// Function to return the number of
// pairs (x, y) such that x < y
static int getPairs( int a[])
{
// To store the number of valid pairs
int count = 0 ;
for ( int i = 0 ; i < a.length; i++) {
for ( int j = 0 ; j < a.length; j++) {
// If a valid pair is found
if (a[i] < a[j])
count++;
}
}
// Return the count of valid pairs
return count;
}
// Driver code
public static void main(String[] args)
{
int a[] = { 2 , 4 , 3 , 1 };
System.out.println(getPairs(a));
}
} |
# Python3 implementation of the approach # Function to return the number of # pairs (x, y) such that x < y def getPairs(a):
# To store the number of valid pairs
count = 0
for i in range ( len (a)):
for j in range ( len (a)):
# If a valid pair is found
if (a[i] < a[j]):
count + = 1
# Return the count of valid pairs
return count
# Driver code if __name__ = = "__main__" :
a = [ 2 , 4 , 3 , 1 ]
print (getPairs(a))
# This code is contributed by ita_c |
// C# implementation of the approach using System;
class GFG
{ // Function to return the number of
// pairs (x, y) such that x < y
static int getPairs( int []a)
{
// To store the number of valid pairs
int count = 0;
for ( int i = 0; i < a.Length; i++)
{
for ( int j = 0; j < a.Length; j++)
{
// If a valid pair is found
if (a[i] < a[j])
count++;
}
}
// Return the count of valid pairs
return count;
}
// Driver code
public static void Main()
{
int []a = { 2, 4, 3, 1 };
Console.WriteLine(getPairs(a));
}
} // This code is contributed by Ryuga |
<?php // PHP implementation of the approach // Function to return the number of // pairs (x, y) such that x < y function getPairs( $a )
{ // To store the number of valid pairs
$count = 0;
for ( $i = 0; $i < sizeof( $a ); $i ++)
{
for ( $j = 0; $j < sizeof( $a ); $j ++)
{
// If a valid pair is found
if ( $a [ $i ] < $a [ $j ])
$count ++;
}
}
// Return the count of valid pairs
return $count ;
} // Driver code $a = array (2, 4, 3, 1);
echo getPairs( $a );
// This code is contributed // by Akanksha Rai ?> |
<script> // Javascript implementation of the approach // Function to return the number of
// pairs (x, y) such that x < y
function getPairs(a)
{
// To store the number of valid pairs
let count = 0;
for (let i = 0; i < a.length; i++) {
for (let j = 0; j < a.length; j++) {
// If a valid pair is found
if (a[i] < a[j])
count++;
}
}
// Return the count of valid pairs
return count;
}
// Driver code
let a=[ 2, 4, 3, 1 ];
document.write(getPairs(a));
// This code is contributed by rag2127 </script> |
6
Time Complexity: O(n2)
Auxiliary Space: O(1)
Efficient approach: For an element x. In order to find the count of valid pairs of the form (x, y1), (x, y2), …, (x, yn), we need to count the elements which are greater than x. For the smallest element, there will be n – 1 element greater than it. Similarly, the second smallest element can form n – 2 pairs and so on. Therefore, the desired count of valid pairs will be (n – 1) + (n – 2) + …. + 1 = n * (n – 1) / 2 where n is the length of the array.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to return the number of // pairs (x, y) such that x < y int getPairs( int a[])
{ // Length of the array
int n = sizeof (a[0]);
// Calculate the number valid pairs
int count = (n * (n - 1)) / 2;
// Return the count of valid pairs
return count;
} // Driver code int main()
{ int a[] = { 2, 4, 3, 1 };
cout << getPairs(a);
return 0;
} // This code is contributed // by SHUBHAMSINGH10 |
// Java implementation of the approach class GFG {
// Function to return the number of
// pairs (x, y) such that x < y
static int getPairs( int a[])
{
// Length of the array
int n = a.length;
// Calculate the number of valid pairs
int count = (n * (n - 1 )) / 2 ;
// Return the count of valid pairs
return count;
}
// Driver code
public static void main(String[] args)
{
int a[] = { 2 , 4 , 3 , 1 };
System.out.println(getPairs(a));
}
} |
# Python implementation of the approach # Function to return the number of # pairs (x, y) such that x < y def getPairs(a):
# Length of the array
n = len (a)
# Calculate the number of valid pairs
count = (n * (n - 1 )) / / 2
# Return the count of valid pairs
return count
# Driver code a = [ 2 , 4 , 3 , 1 ]
print (getPairs(a))
# This code is contributed by SHUBHAMSINGH10 |
// C# implementation of the approach using System;
class GFG
{ // Function to return the number of
// pairs (x, y) such that x < y
static int getPairs( int []a)
{
// Length of the array
int n = a.Length;
// Calculate the number of valid pairs
int count = (n * (n - 1)) / 2;
// Return the count of valid pairs
return count;
}
// Driver code
public static void Main()
{
int []a = { 2, 4, 3, 1 };
Console.Write(getPairs(a));
}
} // This code is contributed // by Akanksha Rai |
<script> // JavaScript implementation of the approach
// Function to return the number of
// pairs (x, y) such that x < y
function getPairs(a)
{
// Length of the array
let n = a.length;
// Calculate the number of valid pairs
let count = parseInt((n * (n - 1)) / 2, 10);
// Return the count of valid pairs
return count;
}
let a = [ 2, 4, 3, 1 ];
document.write(getPairs(a));
</script> |
6
Time Complexity: O(1)
Auxiliary Space: O(1)