Count quadruples of given type from given array

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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++

// 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

// 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

# 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#

// 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

Javascript

<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>

Output:

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++

// 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

// 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

# 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#

// 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

Javascript

<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> 

Output:

Time Complexity: O(N²)
Auxiliary Space: O(N²)

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