Given an array arr[], the task is to find the number of quadruples of the form a[i] = a[k] and a[j] = a[l] ( 0 <= i < j < k < l <= N ) from the given array.
Examples:
Input: arr[] = {1, 2, 4, 2, 1, 5, 2}
Output: 2
Explanation: The quadruple {1, 2, 1, 2} occurs twice in the array, at indices {0, 1, 4, 6} and {0, 3, 4, 6}. Therefore, the required count is 2.Input: arr[] = {1, 2, 3, 2, 1, 3, 2}
Output: 5
Explanation: The quadruple {1, 2, 1, 2} occurs twice in the array at indices {0, 1, 4, 6} and {0, 3, 4, 6}. The quadruples {1, 3, 1, 3}, {3, 2, 3, 2} and {2, 3, 2, 3} occurs once each. Therefore, the required count is 5.
Naive Approach: The simplest approach to solve the problem is to iteratively check all combinations of 4 elements from the given array and check if it satisfies the given condition or not.
Below is the implementation of the above approach:
// C++ program of the // above approach #include <iostream> using namespace std;
const int maxN = 2002;
// Function to find the count of // the subsequence of given type int countSubsequece( int a[], int n)
{ int i, j, k, l;
// Stores the count
// of quadruples
int answer = 0;
// Generate all possible
// combinations of quadruples
for (i = 0; i < n; i++) {
for (j = i + 1; j < n; j++) {
for (k = j + 1; k < n; k++) {
for (l = k + 1; l < n; l++) {
// Check if 1st element is
// equal to 3rd element
if (a[j] == a[l] &&
// Check if 2nd element is
// equal to 4th element
a[i] == a[k]) {
answer++;
}
}
}
}
}
return answer;
} // Driver Code int main()
{ int a[7] = { 1, 2, 3, 2, 1, 3, 2 };
cout << countSubsequece(a, 7);
return 0;
} |
// Java program of the // above approach import java.util.*;
class GFG{
// Function to find the count of // the subsequence of given type static int countSubsequece( int a[], int n)
{ int i, j, k, l;
// Stores the count
// of quadruples
int answer = 0 ;
// Generate all possible
// combinations of quadruples
for (i = 0 ; i < n; i++)
{
for (j = i + 1 ; j < n; j++)
{
for (k = j + 1 ; k < n; k++)
{
for (l = k + 1 ; l < n; l++)
{
// Check if 1st element is
// equal to 3rd element
if (a[j] == a[l] &&
// Check if 2nd element is
// equal to 4th element
a[i] == a[k])
{
answer++;
}
}
}
}
}
return answer;
} // Driver code public static void main(String[] args)
{ int [] a = { 1 , 2 , 3 , 2 , 1 , 3 , 2 };
System.out.print(countSubsequece(a, 7 ));
} } // This code is contributed by code_hunt |
# Python3 program of the # above approach maxN = 2002
# Function to find the count of # the subsequence of given type def countSubsequece(a, n):
# Stores the count
# of quadruples
answer = 0
# Generate all possible
# combinations of quadruples
for i in range (n):
for j in range (i + 1 , n):
for k in range (j + 1 , n):
for l in range (k + 1 , n):
# Check if 1st element is
# equal to 3rd element
if (a[j] = = a[l] and # Check if 2nd element is
# equal to 4th element
a[i] = = a[k]):
answer + = 1
return answer
# Driver Code if __name__ = = '__main__' :
a = [ 1 , 2 , 3 , 2 , 1 , 3 , 2 ]
print (countSubsequece(a, 7 ))
# This code is contributed by bgangwar59 |
// C# program of the // above approach using System;
class GFG{
// Function to find the count of // the subsequence of given type static int countSubsequece( int [] a, int n)
{ int i, j, k, l;
// Stores the count
// of quadruples
int answer = 0;
// Generate all possible
// combinations of quadruples
for (i = 0; i < n; i++)
{
for (j = i + 1; j < n; j++)
{
for (k = j + 1; k < n; k++)
{
for (l = k + 1; l < n; l++)
{
// Check if 1st element is
// equal to 3rd element
if (a[j] == a[l] &&
// Check if 2nd element is
// equal to 4th element
a[i] == a[k])
{
answer++;
}
}
}
}
}
return answer;
} // Driver Code public static void Main()
{ int [] a = { 1, 2, 3, 2, 1, 3, 2 };
Console.WriteLine(countSubsequece(a, 7));
} } // This code is contributed by susmitakundugoaldanga |
<script> // Javascript program of the above approach
// Function to find the count of
// the subsequence of given type
function countSubsequece(a, n)
{
let i, j, k, l;
// Stores the count
// of quadruples
let answer = 0;
// Generate all possible
// combinations of quadruples
for (i = 0; i < n; i++)
{
for (j = i + 1; j < n; j++)
{
for (k = j + 1; k < n; k++)
{
for (l = k + 1; l < n; l++)
{
// Check if 1st element is
// equal to 3rd element
if (a[j] == a[l] &&
// Check if 2nd element is
// equal to 4th element
a[i] == a[k])
{
answer++;
}
}
}
}
}
return answer;
}
let a = [ 1, 2, 3, 2, 1, 3, 2 ];
document.write(countSubsequece(a, 7));
// This code is contributed by mukesh07.
</script> |
5
Time Complexity: O(N4)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, the idea is to maintain two arrays to store the count of element X on the left and right side of every index. Follow the steps below to solve the problem:
- Maintain two arrays lcount[i][j] and rcount[i][j] which stores the count of the element i in the indices less than j and rcount[i][j] stores the count of the element i in the indices greater than j.
- Iterate over the nested loop from 1 to N and find all the subsequence of type XYXY
answer += lcount[a[i]][j-1] * rcount[a[j]][i-1]
Below is the implementation of the above approach:
// C++ program of the // above approach #include <cstring> #include <iostream> using namespace std;
const int maxN = 2002;
// lcount[i][j]: Stores the count of // i on left of index j int lcount[maxN][maxN];
// rcount[i][j]: Stores the count of // i on right of index j int rcount[maxN][maxN];
// Function to count unique elements // on left and right of any index void fill_counts( int a[], int n)
{ int i, j;
// Find the maximum array element
int maxA = a[0];
for (i = 0; i < n; i++) {
if (a[i] > maxA) {
maxA = a[i];
}
}
memset (lcount, 0, sizeof (lcount));
memset (rcount, 0, sizeof (rcount));
for (i = 0; i < n; i++) {
lcount[a[i]][i] = 1;
rcount[a[i]][i] = 1;
}
for (i = 0; i <= maxA; i++) {
// Calculate prefix sum of
// counts of each value
for (j = 0; j < n; j++) {
lcount[i][j] = lcount[i][j - 1]
+ lcount[i][j];
}
// Calculate suffix sum of
// counts of each value
for (j = n - 2; j >= 0; j--) {
rcount[i][j] = rcount[i][j + 1]
+ rcount[i][j];
}
}
} // Function to count quadruples // of the required type int countSubsequence( int a[], int n)
{ int i, j;
fill_counts(a, n);
int answer = 0;
for (i = 1; i < n; i++) {
for (j = i + 1; j < n - 1; j++) {
answer += lcount[a[j]][i - 1]
* rcount[a[i]][j + 1];
}
}
return answer;
} // Driver Code int main()
{ int a[7] = { 1, 2, 3, 2, 1, 3, 2 };
cout << countSubsequence(a, 7);
return 0;
} |
// Java program of the // above approach import java.util.*;
class GFG{
static int maxN = 2002 ;
// lcount[i][j]: Stores the // count of i on left of index j static int [][]lcount =
new int [maxN][maxN];
// rcount[i][j]: Stores the // count of i on right of index j static int [][]rcount =
new int [maxN][maxN];
// Function to count unique // elements on left and right // of any index static void fill_counts( int a[],
int n)
{ int i, j;
// Find the maximum
// array element
int maxA = a[ 0 ];
for (i = 0 ; i < n; i++)
{
if (a[i] > maxA)
{
maxA = a[i];
}
}
for (i = 0 ; i < n; i++)
{
lcount[a[i]][i] = 1 ;
rcount[a[i]][i] = 1 ;
}
for (i = 0 ; i <= maxA; i++)
{
// Calculate prefix sum of
// counts of each value
for (j = 1 ; j < n; j++)
{
lcount[i][j] = lcount[i][j - 1 ] +
lcount[i][j];
}
// Calculate suffix sum of
// counts of each value
for (j = n - 2 ; j >= 0 ; j--)
{
rcount[i][j] = rcount[i][j + 1 ] +
rcount[i][j];
}
}
} // Function to count quadruples // of the required type static int countSubsequence( int a[],
int n)
{ int i, j;
fill_counts(a, n);
int answer = 0 ;
for (i = 1 ; i < n; i++)
{
for (j = i + 1 ; j < n - 1 ; j++)
{
answer += lcount[a[j]][i - 1 ] *
rcount[a[i]][j + 1 ];
}
}
return answer;
} // Driver Code public static void main(String[] args)
{ int a[] = { 1 , 2 , 3 , 2 , 1 , 3 , 2 };
System.out.print(
countSubsequence(a, a.length));
} } // This code is contributed by shikhasingrajput |
# Python3 program of the # above approach maxN = 2002
# lcount[i][j]: Stores the count of # i on left of index j lcount = [[ 0 for i in range (maxN)]
for j in range (maxN)]
# rcount[i][j]: Stores the count of # i on right of index j rcount = [[ 0 for i in range (maxN)]
for j in range (maxN)]
# Function to count unique elements # on left and right of any index def fill_counts(a, n):
# Find the maximum array element
maxA = a[ 0 ]
for i in range (n):
if (a[i] > maxA):
maxA = a[i]
for i in range (n):
lcount[a[i]][i] = 1
rcount[a[i]][i] = 1
for i in range (maxA + 1 ):
# Calculate prefix sum of
# counts of each value
for j in range (n) :
lcount[i][j] = (lcount[i][j - 1 ] + lcount[i][j])
# Calculate suffix sum of
# counts of each value
for j in range (n - 2 , - 1 , - 1 ):
rcount[i][j] = (rcount[i][j + 1 ] + rcount[i][j])
# Function to count quadruples # of the required type def countSubsequence(a, n):
fill_counts(a, n)
answer = 0
for i in range ( 1 , n):
for j in range (i + 1 , n - 1 ):
answer + = (lcount[a[j]][i - 1 ] *
rcount[a[i]][j + 1 ])
return answer
# Driver Code a = [ 1 , 2 , 3 , 2 , 1 , 3 , 2 ]
print (countSubsequence(a, 7 ))
# This code is contributed by divyesh072019 |
// C# program of the // above approach using System;
class GFG{
static int maxN = 2002;
// lcount[i,j]: Stores the // count of i on left of index j static int [,]lcount = new int [maxN, maxN];
// rcount[i,j]: Stores the // count of i on right of index j static int [,]rcount = new int [maxN, maxN];
// Function to count unique // elements on left and right // of any index static void fill_counts( int []a,
int n)
{ int i, j;
// Find the maximum
// array element
int maxA = a[0];
for (i = 0; i < n; i++)
{
if (a[i] > maxA)
{
maxA = a[i];
}
}
for (i = 0; i < n; i++)
{
lcount[a[i], i] = 1;
rcount[a[i], i] = 1;
}
for (i = 0; i <= maxA; i++)
{
// Calculate prefix sum of
// counts of each value
for (j = 1; j < n; j++)
{
lcount[i, j] = lcount[i, j - 1] +
lcount[i, j];
}
// Calculate suffix sum of
// counts of each value
for (j = n - 2; j >= 0; j--)
{
rcount[i, j] = rcount[i, j + 1] +
rcount[i, j];
}
}
} // Function to count quadruples // of the required type static int countSubsequence( int []a,
int n)
{ int i, j;
fill_counts(a, n);
int answer = 0;
for (i = 1; i < n; i++)
{
for (j = i + 1; j < n - 1; j++)
{
answer += lcount[a[j], i - 1] *
rcount[a[i], j + 1];
}
}
return answer;
} // Driver Code public static void Main(String[] args)
{ int []a = { 1, 2, 3, 2, 1, 3, 2 };
Console.Write(
countSubsequence(a, a.Length));
} } // This code is contributed by Princi Singh |
<script> // Javascript program of the // above approach let maxN = 2002; // lcount[i][j]: Stores the // count of i on left of index j let lcount = new Array(maxN);
// rcount[i][j]: Stores the // count of i on right of index j let rcount = new Array(maxN);
for (let i = 0; i < maxN; i++)
{ lcount[i] = new Array(maxN);
rcount[i] = new Array(maxN);
for (let j = 0; j < maxN; j++)
{
lcount[i][j] = 0;
rcount[i][j] = 0;
}
} // Function to count unique // elements on left and right // of any index function fill_counts(a,n)
{ let i, j;
// Find the maximum
// array element
let maxA = a[0];
for (i = 0; i < n; i++)
{
if (a[i] > maxA)
{
maxA = a[i];
}
}
for (i = 0; i < n; i++)
{
lcount[a[i]][i] = 1;
rcount[a[i]][i] = 1;
}
for (i = 0; i <= maxA; i++)
{
// Calculate prefix sum of
// counts of each value
for (j = 1; j < n; j++)
{
lcount[i][j] = lcount[i][j - 1] +
lcount[i][j];
}
// Calculate suffix sum of
// counts of each value
for (j = n - 2; j >= 0; j--)
{
rcount[i][j] = rcount[i][j + 1] +
rcount[i][j];
}
}
} // Function to count quadruples // of the required type function countSubsequence(a,n)
{ let i, j;
fill_counts(a, n);
let answer = 0;
for (i = 1; i < n; i++)
{
for (j = i + 1; j < n - 1; j++)
{
answer += lcount[a[j]][i - 1] *
rcount[a[i]][j + 1];
}
}
return answer;
} // Driver Code let a=[1, 2, 3, 2, 1, 3, 2]; document.write(countSubsequence(a, a.length)); // This code is contributed by rag2127 </script> |
5
Time Complexity: O(N2)
Auxiliary Space: O(N2)