AuxilGiven an array of n elements. The task is to count the total number of indices (i, j) such that arr[i] = arr[j] and i < j
Examples :
Input : arr[] = {1, 1, 2} Output : 1 As arr[0] = arr[1], the pair of indices is (0, 1) Input : arr[] = {1, 1, 1} Output : 3 As arr[0] = arr[1], the pair of indices is (0, 1), (0, 2) and (1, 2) Input : arr[] = {1, 2, 3} Output : 0
Method 1 (Brute Force): For each index i, find element after it with same value as arr[i]. Below is the implementation of this approach:
Implementation:
// C++ program to count of pairs with equal // elements in an array. #include<bits/stdc++.h> using namespace std;
// Return the number of pairs with equal // values. int countPairs( int arr[], int n)
{ int ans = 0;
// for each index i and j
for ( int i = 0; i < n; i++)
for ( int j = i+1; j < n; j++)
// finding the index with same
// value but different index.
if (arr[i] == arr[j])
ans++;
return ans;
} // Driven Program int main()
{ int arr[] = { 1, 1, 2 };
int n = sizeof (arr)/ sizeof (arr[0]);
cout << countPairs(arr, n) << endl;
return 0;
} |
// Java program to count of pairs with equal // elements in an array. class GFG {
// Return the number of pairs with equal
// values.
static int countPairs( int arr[], int n)
{
int ans = 0 ;
// for each index i and j
for ( int i = 0 ; i < n; i++)
for ( int j = i+ 1 ; j < n; j++)
// finding the index with same
// value but different index.
if (arr[i] == arr[j])
ans++;
return ans;
}
//driver code
public static void main (String[] args)
{
int arr[] = { 1 , 1 , 2 };
int n = arr.length;
System.out.println(countPairs(arr, n));
}
} // This code is contributed by Anant Agarwal. |
# Python3 program to # count of pairs with equal # elements in an array. # Return the number of # pairs with equal values. def countPairs(arr, n):
ans = 0
# for each index i and j
for i in range ( 0 , n):
for j in range (i + 1 , n):
# finding the index
# with same value but
# different index.
if (arr[i] = = arr[j]):
ans + = 1
return ans
# Driven Code arr = [ 1 , 1 , 2 ]
n = len (arr)
print (countPairs(arr, n))
# This code is contributed # by Smitha |
// C# program to count of pairs with equal // elements in an array. using System;
class GFG {
// Return the number of pairs with equal
// values.
static int countPairs( int []arr, int n)
{
int ans = 0;
// for each index i and j
for ( int i = 0; i < n; i++)
for ( int j = i+1; j < n; j++)
// finding the index with same
// value but different index.
if (arr[i] == arr[j])
ans++;
return ans;
}
// Driver code
public static void Main ()
{
int []arr = { 1, 1, 2 };
int n = arr.Length;
Console.WriteLine(countPairs(arr, n));
}
} // This code is contributed by anuj_67. |
<?php // PHP program to count of // pairs with equal elements // in an array. // Return the number of pairs // with equal values. function countPairs( $arr , $n )
{ $ans = 0;
// for each index i and j
for ( $i = 0; $i < $n ; $i ++)
for ( $j = $i + 1; $j < $n ; $j ++)
// finding the index with same
// value but different index.
if ( $arr [ $i ] == $arr [ $j ])
$ans ++;
return $ans ;
} // Driven Code $arr = array ( 1, 1, 2 );
$n = count ( $arr );
echo countPairs( $arr , $n ) ;
// This code is contributed by anuj_67. ?> |
<script> // Javascript program to count of pairs with equal // elements in an array. // Return the number of pairs with equal
// values.
function countPairs(arr, n)
{
let ans = 0;
// for each index i and j
for (let i = 0; i < n; i++)
for (let j = i+1; j < n; j++)
// finding the index with same
// value but different index.
if (arr[i] == arr[j])
ans++;
return ans;
}
// Driver code let arr = [ 1, 1, 2 ];
let n = arr.length;
document.write(countPairs(arr, n));
// This code is contributed by susmitakundugoaldanga.
</script> |
1
Time Complexity : O(n2)
Auxiliary Space: O(1)
Method 2 (Efficient approach):
The idea is to count the frequency of each number and then find the number of pairs with equal elements. Suppose, a number x appears k times at index i1, i2,….,ik. Then pick any two indexes ix and iy which will be counted as 1 pair. Similarly, iy and ix can also be pair. So, choose nC2 is the number of pairs such that arr[i] = arr[j] = x.
Below is the implementation of this approach:
// C++ program to count of index pairs with // equal elements in an array. #include<bits/stdc++.h> using namespace std;
// Return the number of pairs with equal // values. int countPairs( int arr[], int n)
{ unordered_map< int , int > mp;
// Finding frequency of each number.
for ( int i = 0; i < n; i++)
mp[arr[i]]++;
// Calculating pairs of each value.
int ans = 0;
for ( auto it=mp.begin(); it!=mp.end(); it++)
{
int count = it->second;
ans += (count * (count - 1))/2;
}
return ans;
} // Driven Program int main()
{ int arr[] = {1, 1, 2};
int n = sizeof (arr)/ sizeof (arr[0]);
cout << countPairs(arr, n) << endl;
return 0;
} |
// Java program to count of index pairs with // equal elements in an array. import java.util.*;
class GFG {
public static int countPairs( int arr[], int n)
{
//A method to return number of pairs with
// equal values
HashMap<Integer,Integer> hm = new HashMap<>();
// Finding frequency of each number.
for ( int i = 0 ; i < n; i++)
{
if (hm.containsKey(arr[i]))
hm.put(arr[i],hm.get(arr[i]) + 1 );
else
hm.put(arr[i], 1 );
}
int ans= 0 ;
// Calculating count of pairs with equal values
for (Map.Entry<Integer,Integer> it : hm.entrySet())
{
int count = it.getValue();
ans += (count * (count - 1 )) / 2 ;
}
return ans;
}
// Driver code
public static void main(String[] args)
{
int arr[] = new int []{ 1 , 2 , 3 , 1 };
System.out.println(countPairs(arr,arr.length));
}
} // This Code is Contributed // by Adarsh_Verma |
# Python3 program to count of index pairs # with equal elements in an array. import math as mt
# Return the number of pairs with # equal values. def countPairs(arr, n):
mp = dict ()
# Finding frequency of each number.
for i in range (n):
if arr[i] in mp.keys():
mp[arr[i]] + = 1
else :
mp[arr[i]] = 1
# Calculating pairs of each value.
ans = 0
for it in mp:
count = mp[it]
ans + = (count * (count - 1 )) / / 2
return ans
# Driver Code arr = [ 1 , 1 , 2 ]
n = len (arr)
print (countPairs(arr, n))
# This code is contributed by mohit kumar 29 |
// C# program to count of index pairs with // equal elements in an array. using System;
using System.Collections.Generic;
class GFG
{ // Return the number of pairs with
// equal values.
public static int countPairs( int []arr, int n)
{
// A method to return number of pairs
// with equal values
Dictionary< int ,
int > hm = new Dictionary< int ,
int >();
// Finding frequency of each number.
for ( int i = 0; i < n; i++)
{
if (hm.ContainsKey(arr[i]))
{
int a = hm[arr[i]];
hm.Remove(arr[i]);
hm.Add(arr[i], a + 1);
}
else
hm.Add(arr[i], 1);
}
int ans = 0;
// Calculating count of pairs with
// equal values
foreach ( var it in hm)
{
int count = it.Value;
ans += (count * (count - 1)) / 2;
}
return ans;
}
// Driver code
public static void Main()
{
int []arr = new int []{1, 2, 3, 1};
Console.WriteLine(countPairs(arr,arr.Length));
}
} // This code is contributed by 29AjayKumar |
<script> // Javascript program to count of index pairs with // equal elements in an array. function countPairs(arr,n)
{
//A method to return number of pairs with
// equal values
let hm = new Map();
// Finding frequency of each number.
for (let i = 0; i < n; i++)
{
if (hm.has(arr[i]))
hm.set(arr[i],hm.get(arr[i]) + 1);
else
hm.set(arr[i], 1);
}
let ans=0;
// Calculating count of pairs with equal values
for (let [key, value] of hm.entries())
{
let count = value;
ans += (count * (count - 1)) / 2;
}
return ans;
}
// Driver code
let arr=[1, 2, 3, 1];
document.write(countPairs(arr,arr.length));
// This code is contributed by patel2127 </script> |
1
Time Complexity : O(n)
Auxiliary Space: O(n)