Count of triplets in an Array (i, j, k) such that i < j < k and a[k] < a[i] < a[j]

Given an array arr[] of N integers, the task is to count number of triplets (i, j, k) in the array such that a[k] < a[i] < a[j] and i < j < k.

Examples:

Input: arr[] = {2, 5, 1, 3, 0}
Output: 4
Explanation:
Below are the triplets (i, j, k) such that i < j < k and a[k] < a[i] < a[j]:
1. (0, 1, 2) and arr[2] < arr[0] 1 < 2 < 5.
2. (0, 1, 4) and arr[4] < arr[0] 0 < 2 < 5.
3. (0, 3, 4) and arr[4] < arr[0] 0 < 2 < 3.
4. (2, 3, 4) and arr[4] < arr[2] 0 < 1 < 3.

Input: arr[] = {2, 5, 1, 2, 0, 3, 10, 1, 5, 0 }
Output: 25

Naive Approach: The idea is to iterate 3 loops and check for each triplet (i, j, k) satisfy the given conditions or not. If yes then increment for that triplet and print the final count after checking all the triplets.



Below is the implementation of the above approach:

C++

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to count triplets with the
// given conditions
int CountTriplets(int arr[], int n)
{
    int cnt = 0;
  
    for (int i = 0; i < n; i++)
        for (int j = i + 1; j < n; j++)
            for (int k = j + 1; k < n; k++)
  
                // If it satisfy the
                // given conditions
                if (arr[k] < arr[i]
                    && arr[i] < arr[j]) {
  
                    cnt += 1;
                }
  
    // Return the final count
    return cnt;
}
  
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 2, 5, 1, 3, 0 };
  
    int n = sizeof(arr) / sizeof(arr[0]);
  
    // Function Call
    cout << CountTriplets(arr, n)
         << endl;
    return 0;
}

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Java

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// Java program for the above approach
class GFG{ 
  
// Function to count triplets 
// with the given conditions
static int CountTriplets(int arr[], int n)
{
    int cnt = 0;
  
    for(int i = 0; i < n; i++)
       for(int j = i + 1; j < n; j++)
          for(int k = j + 1; k < n; k++)
               
             // If it satisfy the
             // given conditions
             if (arr[k] < arr[i] && 
                 arr[i] < arr[j])
             {
                 cnt += 1;
             }
               
    // Return the final count
    return cnt;
}
  
// Driver Code 
public static void main(String[] args) 
      
    // Given array arr[]
    int arr[] = new int[]{ 2, 5, 1, 3, 0 };
  
    int n = arr.length;
      
    System.out.print(CountTriplets(arr, n));
  
// This code is contributed by Pratima Pandey

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Python3

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# Python3 program for the above approach
  
# Function to count triplets with the
# given conditions
def CountTriplets(arr, n):
  
    cnt = 0;
  
    for i in range(0, n):
        for j in range(i + 1, n):
            for k in range(j + 1, n):
  
                # If it satisfy the
                # given conditions
                if (arr[k] < arr[i] and arr[i] < arr[j]):
                    cnt += 1;
                  
    # Return the final count
    return cnt;
  
# Driver Code
  
# Given array arr[]
arr = [ 2, 5, 1, 3, 0 ];
  
n = len(arr);
  
# Function Call
print(CountTriplets(arr, n))
  
# This code is contributed by Code_Mech

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C#

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// C# program for the above approach
using System;
class GFG{ 
  
// Function to count triplets 
// with the given conditions
static int CountTriplets(int []arr, int n)
{
    int cnt = 0;
  
    for(int i = 0; i < n; i++)
       for(int j = i + 1; j < n; j++)
          for(int k = j + 1; k < n; k++)
               
             // If it satisfy the
             // given conditions
             if (arr[k] < arr[i] &&
                 arr[i] < arr[j])
             {
                 cnt += 1;
             }
              
    // Return the final count
    return cnt;
}
  
// Driver Code 
public static void Main(string[] args) 
      
    // Given array arr[]
    int []arr = new int[]{ 2, 5, 1, 3, 0 };
  
    int n = arr.Length;
      
    Console.Write(CountTriplets(arr, n));
}
  
// This code is contributed by Ritik Bansal

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Output:

4

Time Complexity: O(N3)
Auxiliary Space: O(1)

Efficient Approach: We can reduce the complexity from N^3 to N^2, using below steps:

  1. Run two loops to find pairs (i, j) such that i < j and arr[j] > arr[i] and keep the count of these pairs as cnt.
  2. While in the above loop if there exists any element such arr[j] < arr[i] then increment the count of triplets by cnt as as the current element is the Kth element such that a[k] < a[i] < a[j] for triplet i < j < k.

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to count  triplets
int CountTriplets(int a[], int n)
{
  
    // To store count of total triplets
    int ans = 0;
  
    for (int i = 0; i < n; i++) {
  
        // Initialize count to zero
        int cnt = 0;
  
        for (int j = i + 1; j < n; j++) {
  
            // If a[j] > a[i] then,
            // increment cnt
            if (a[j] > a[i])
                cnt++;
  
            // If a[j] < a[i], then
            // it mean we have found a[k]
            // such that a[k] < a[i] < a[j]
            else
                ans += cnt;
        }
    }
  
    // Return the final count
    return ans;
}
  
// Driver code
int main()
{
    int arr[] = { 2, 5, 1, 3, 0 };
  
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << CountTriplets(arr, n) << endl;
  
    return 0;
}

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Java

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// Java program for the above approach
class GFG{
  
// Function to count triplets
static int CountTriplets(int a[], int n)
{
  
    // To store count of total triplets
    int ans = 0;
  
    for (int i = 0; i < n; i++) 
    {
  
        // Initialize count to zero
        int cnt = 0;
  
        for (int j = i + 1; j < n; j++)
        {
  
            // If a[j] > a[i] then,
            // increment cnt
            if (a[j] > a[i])
                cnt++;
  
            // If a[j] < a[i], then
            // it mean we have found a[k]
            // such that a[k] < a[i] < a[j]
            else
                ans += cnt;
        }
    }
  
    // Return the final count
    return ans;
}
  
// Driver code
public static void main(String[] args) 
{
    int arr[] = { 2, 5, 1, 3, 0 };
  
    int n = arr.length;
  
    System.out.print(CountTriplets(arr, n));
}
}
  
// This code is contributed by shivanisinghss2110

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C#

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// C# program for the above approach
using System;
class GFG{
  
// Function to count triplets
static int CountTriplets(int []a, int n)
{
  
    // To store count of total triplets
    int ans = 0;
  
    for (int i = 0; i < n; i++) 
    {
  
        // Initialize count to zero
        int cnt = 0;
  
        for (int j = i + 1; j < n; j++)
        {
  
            // If a[j] > a[i] then,
            // increment cnt
            if (a[j] > a[i])
                cnt++;
  
            // If a[j] < a[i], then
            // it mean we have found a[k]
            // such that a[k] < a[i] < a[j]
            else
                ans += cnt;
        }
    }
  
    // Return the final count
    return ans;
}
  
// Driver code
public static void Main() 
{
    int []arr = { 2, 5, 1, 3, 0 };
  
    int n = arr.Length;
  
    Console.Write(CountTriplets(arr, n));
}
}
  
// This code is contributed by Code_Mech

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Output:

4

Time Complexity: O(N2)
Auxiliary Space: O(1)

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