Skip to content
Related Articles

Related Articles

Count quadruples of given type from given array
  • Difficulty Level : Medium
  • Last Updated : 22 Apr, 2021

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(1, 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, 0, -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
if __name__ == '__main__':
    a = [1, 2, 3, 2, 1, 3, 2];
    print(countSubsequence(a, len(a)));
 
# This code contributed by gauravrajput1
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 chitranayal
Output:
5
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: 
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++



// 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
Output: 
5
 
Time Complexity: O(N2)
Auxiliary Space: O(N2)
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
My Personal Notes
arrow_drop_up
Previous
Sum of absolute differences of indices of occurrences of each array element
Next

Check if given point lies in range of any of the given towers
Recommended Articles
Page :
Count quadruples (i, j, k, l) in an array such that i < j < k < l and arr[i] = arr[k] and arr[j] = arr[l]
17, Sep 20
Count quadruples from four sorted arrays whose sum is equal to a given value x
04, Jul 17
Count of quadruples with product of a pair equal to the product of the remaining pair
08, Feb 21
Count all Quadruples from four arrays such that their XOR equals to 'x'
04, Mar 17
Maximize sum of second minimums in all quadruples of a given array
25, Jan 21
Number of quadruples where the first three terms are in AP and last three terms are in GP
01, Mar 19
Count of replacements required to make the sum of all Pairs of given type from the Array equal
21, Aug 20
Minimum operations of given type required to empty given array
05, Jan 21
Sort an array by swapping elements of different type specified by another array
18, Jun 20
Make all the array elements odd with minimum operations of given type
13, Jan 20
Find the modified array after performing k operations of given type
08, Mar 19
Minimize consecutive removals of elements of the same type to empty given array
07, Dec 20
Queries to check if sweets of given type can be eaten on given day or not
19, Dec 20
Type of array and its maximum element
08, Mar 17
Maximum length of subarray consisting of same type of element on both halves of sub-array
25, Jul 20
Minimum, maximum and average price values for all the items of given type
23, Oct 19
Check if a string can be converted to another by swapping of adjacent characters of given type
01, Sep 20
Count of pairs {X, Y} from an array such that sum of count of set bits in X ⊕ Y and twice the count of set bits in X & Y is M
12, Mar 21
Minimum moves required to type a word in QWERTY based keyboard
19, Aug 20
Rearrange string such that no pair of adjacent characters are of the same type
21, Sep 20
Count clockwise array rotations required to maximize count of array elements present at indices same as their value
24, Nov 20
Modify array to another given array by replacing array elements with the sum of the array
12, Feb 21
Count pairs from an array with even product of count of distinct prime factors
22, Feb 21
Count array elements whose count of divisors is a prime number
19, Apr 21
Article Contributed By :
https://media.geeksforgeeks.org/auth/avatar.png
sudheer_121
@sudheer_121
Vote for difficulty
Current difficulty :
Medium





Improved By :
Article Tags :
Practice Tags :
Report Issue

We use cookies to ensure you have the best browsing experience on our website. By using our site, you
acknowledge that you have read and understood our
Cookie Policy &
        Privacy Policy

Lightbox