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

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  • Last Updated : 25 Jan, 2023
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Given an array arr[] consisting of N distinct integers, the task is to count the number of triplets (i, j, k) possible from the array arr[] such that i < j < k and arr[i] < arr[j] > arr[k].

Examples:

Input: arr[] = {2, 3, 1, -1}
Output: 2
Explanation: From the given array, all possible triplets satisfying the property (i, j, k) and arr[i] < arr[j] > arr[k] are:

  1. (0, 1, 2): arr[0](= 2) < arr[1](= 3) > arr[2](= 1).
  2. (0, 1, 3): arr[0](= 2) < arr[1](= 3) > arr[3](= -1).

Therefore, the count of triplets is 2.

Input: arr[] = {2, 3, 4, 6, 7, 9, 1, 12, 10, 8}
Output: 41

Naive Approach: The simplest approach to solve the problem is to traverse the given array and for each element arr[i], the product of the count of smaller elements on the left side of arr[i] and the count of smaller elements on the right side of arr[i] gives the count of triplets for the element arr[i] as the middle element. The sum of all the counts obtained for each index is the required number of valid triplets. Time Complexity: O(N2)
Auxiliary Space: O(1)

Efficient Approach: The above approach can also be optimized by finding the count of smaller elements using a Policy-based data structure (PBDS). Follow the steps below to solve the problem:

  • Initialize the variable, say ans to 0 that stores the total number of possible pairs.
  • Initialize two containers of the Policy-based data structure, say P and Q.
  • Initialize a vector of pairs V, where V[i]. first and V[i].second stores the count of smaller elements on the left and the right side of every array element arr[i].
  • Traverse the given array and for each element arr[i], update the value of V[i].first as P.order_of_key(arr[i]) and insert arr[i] to set P.
  • Traverse the array from right to left and for each element arr[i], update the value of V[i].first as P.order_of_key(arr[i]) and insert arr[i] to set Q.
  • Traverse the vector of pairs V and add the value of V[i].first * V[i].second to the variable ans.
  • After completing the above steps, print the value of ans as the total number of pairs.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <functional>
#include <iostream>
using namespace __gnu_pbds;
using namespace std;
 
// Function to find the count of triplets
// satisfying the given conditions
void findTriplets(int arr[], int n)
{
    // Stores the total count of pairs
    int ans = 0;
 
    // Declare the set
    tree<int, null_type, less<int>, rb_tree_tag,
         tree_order_statistics_node_update>
        p, q;
 
    // Declare the vector of pairs
    vector<pair<int, int> > v(n);
 
    // Iterate over the array from
    // left to right
    for (int i = 0; i < n; i++) {
 
        // Find the index of element
        // in sorted array
        int index = p.order_of_key(arr[i]);
 
        // Assign to the left
        v[i].first = index;
 
        // Insert into the set
        p.insert(arr[i]);
    }
 
    // Iterate from right to left
    for (int i = n - 1; i >= 0; i--) {
 
        // Find the index of element
        // in the sorted array
        int index = q.order_of_key(arr[i]);
 
        // Assign to the right
        v[i].second = index;
 
        // Insert into the set
        q.insert(arr[i]);
    }
 
    // Traverse the vector of pairs
    for (int i = 0; i < n; i++) {
        ans += (v[i].first * v[i].second);
    }
 
    // Print the total count
    cout << ans;
}
 
// Driver Code
int main()
{
    int arr[] = { 2, 3, 1, -1 };
    int N = sizeof(arr) / sizeof(arr[0]);
    findTriplets(arr, N);
 
    return 0;
}


Java




import java.util.ArrayList;
import java.util.List;
 
public class Main
{
 
  // Function to find the count of triplets
  // satisfying the given conditions
  public static void findTriplets(int[] arr, int n)
  {
 
    // Stores the total count of pairs
    int ans = 0;
 
    // Declare the list of pairs
    List<Pair<Integer, Integer>> v = new ArrayList<>();
 
    // Iterate over the array from
    // left to right
    for (int i = 0; i < n; i++) {
      // Find the index of element
      // in sorted array
      int index = 0;
      for (int j = 0; j < i; j++) {
        if (arr[j] < arr[i]) {
          index++;
        }
      }
 
      // Assign to the left
      v.add(new Pair<>(index, 0));
    }
 
    // Iterate from right to left
    for (int i = n - 1; i >= 0; i--) {
      // Find the index of element
      // in the sorted array
      int index = 0;
      for (int j = n - 1; j > i; j--) {
        if (arr[j] < arr[i]) {
          index++;
        }
      }
 
      // Assign to the right
      v.get(i).setValue(index);
    }
 
    // Traverse the list of pairs
    for (int i = 0; i < n; i++) {
      ans += (v.get(i).getKey() * v.get(i).getValue());
    }
 
    // Print the total count
    System.out.println(ans);
  }
 
  public static void main(String[] args) {
    int[] arr = { 2, 3, 1, -1 };
    int N = arr.length;
    findTriplets(arr, N);
  }
}
 
class Pair<K, V> {
  private K key;
  private V value;
 
  public Pair(K key, V value) {
    this.key = key;
    this.value = value;
  }
 
  public void setKey(K key) {
    this.key = key;
  }
 
  public void setValue(V value) {
    this.value = value;
  }
 
  public K getKey() {
    return key;
  }
 
  public V getValue() {
    return value;
  }
}
 
// This code is contributed by aadityaburujwale.


Python3




import bisect
 
def findTriplets(arr, n):
    # Stores the total count of pairs
    ans = 0
    # Declare the lists
    p = []
    q = []
 
    # Iterate over the array from left to right
    for i in range(n):
        # Find the index of element in sorted array
        index = bisect.bisect_left(p, arr[i])
        # Insert into the list
        p.insert(index, arr[i])
    # Iterate from right to left
    for i in range(n-1, -1, -1):
        # Find the index of element in the sorted array
        index = bisect.bisect_left(q, arr[i])
        # Insert into the list
        q.insert(index, arr[i])
 
    ans = 0
    for i in range(n):
        for j in range(i+1, n):
            for k in range(j+1, n):
                if arr[i] < arr[j] > arr[k]:
                    ans += 1
    print(ans)
 
# Driver Code
arr = [2, 3, 1, -1]
n = len(arr)
findTriplets(arr, n)
 
# This code is contributed by Vikram_Shirsat


C#




using System;
using System.Collections.Generic;
 
class GFG {
 
  public static void findTriplets(int[] arr, int n)
  {
 
    // Stores the total count of pairs
    int ans = 0;
 
    // Declare the list of pairs
    List<KeyValuePair<int, int>> v = new List<KeyValuePair<int, int>>();
 
    // Iterate over the array from
    // left to right
    for (int i = 0; i < n; i++) {
      // Find the index of element
      // in sorted array
      int index = 0;
      for (int j = 0; j < i; j++) {
        if (arr[j] < arr[i]) {
          index++;
        }
      }
 
      // Assign to the left
      v.Add(new KeyValuePair<int, int>(index, 0));
    }
 
    // Iterate from right to left
    for (int i = n - 1; i >= 0; i--) {
      // Find the index of element
      // in the sorted array
      int index = 0;
      for (int j = n - 1; j > i; j--) {
        if (arr[j] < arr[i]) {
          index++;
        }
      }
 
      // Assign to the right
      v[i] = new KeyValuePair<int, int>(v[i].Key, index);
    }
 
    // Traverse the list of pairs
    for (int i = 0; i < n; i++) {
      ans += (v[i].Key * v[i].Value);
    }
 
    // Print the total count
    Console.WriteLine(ans);
  }
 
  public static void Main(string[] args) {
    int[] arr = { 2, 3, 1, -1 };
    int N = arr.Length;
    findTriplets(arr, N);
  }
}
 
// This code is contributed by phasing17.


Output:

2

Time Complexity: O(N * log N)
Auxiliary Space: O(N)


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