Count of triplets in an array that satisfy the given conditions

Given an array arr[] of N elements, the task is to find the count of triplets (arr[i], arr[j], arr[k]) such that (arr[i] + arr[j] + arr[k] = L) and (L % arr[i] = L % arr[j] = L % arr[k] = 0.

Examples:

Input: arr[] = {2, 4, 5, 6, 7}
Output: 1
Only possible triplet is {2, 4, 6}



Input: arr[] = {4, 4, 4, 4, 4}
Output: 10

Approach:

  • Consider the equations below:

    L = a * arr[i], L = b * arr[j] and L = c * arr[k].
    Thus, arr[i] = L / a, arr[j] = L / b and arr[k] = L / c.
    So, using this in arr[i] + arr[j] + arr[k] = L gives L / a + L / b + L / c = L
    or 1 / a + 1 / b + 1 / c = 1.
    Now, there can only be 10 possible solutions of the above equation, which are
    {2, 3, 6}
    {2, 4, 4}
    {2, 6, 3}
    {3, 2, 6}
    {3, 3, 3}
    {3, 6, 2}
    {4, 2, 4}
    {4, 4, 2}
    {6, 2, 3}
    {6, 3, 2}

    All possible triplets (arr[i], arr[j], arr[k]) will satisfy these solutions. So, all these solutions can be stored in a 2D array say test[][3]. So, that all the triplets which satisfy these solutions can be found.

  • For all i from 1 to N. Consider arr[i] as the middle element of the triplet. And find corresponding first and third elements of the triplet for all possible solutions of the equation 1 / a + 1 / b + 1 / c = 1. Find the answer for all the cases and add them to the final answer.
  • Maintain the indices where arr[i] occurs in the array. For a given possible solution of equation X and for every number arr[i] keep the number as middle element of triplet and find the first and the third element. Apply binary search on the available set of indices for the first and the third element to find the number of occurrence of the first element from 1 to i – 1 and the number of occurrences of the third element from i + 1 to N. Multiply both the values and add it to the final answer.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
#define MAX 100001
#define ROW 10
#define COl 3
  
vector<int> indices[MAX];
  
// All possible solutions of the
// equation 1/a + 1/b + 1/c = 1
int test[ROW][COl] = { { 2, 3, 6 },
                       { 2, 4, 4 },
                       { 2, 6, 3 },
                       { 3, 2, 6 },
                       { 3, 3, 3 },
                       { 3, 6, 2 },
                       { 4, 2, 4 },
                       { 4, 4, 2 },
                       { 6, 2, 3 },
                       { 6, 3, 2 } };
  
// Function to find the triplets
int find_triplet(int array[], int n)
{
    int answer = 0;
  
    // Storing indices of the elements
    for (int i = 0; i < n; i++) {
        indices[array[i]].push_back(i);
    }
  
    for (int i = 0; i < n; i++) {
        int y = array[i];
  
        for (int j = 0; j < ROW; j++) {
            int s = test[j][1] * y;
  
            // Check if y can act as the middle
            // element of triplet with the given
            // solution of 1/a + 1/b + 1/c = 1
            if (s % test[j][0] != 0)
                continue;
            if (s % test[j][2] != 0)
                continue;
  
            int x = s / test[j][0];
            ll z = s / test[j][2];
            if (x > MAX || z > MAX)
                continue;
  
            int l = 0;
            int r = indices[x].size() - 1;
  
            int first = -1;
  
            // Binary search to find the number of
            // possible values of the first element
            while (l <= r) {
                int m = (l + r) / 2;
  
                if (indices[x][m] < i) {
                    first = m;
                    l = m + 1;
                }
                else {
                    r = m - 1;
                }
            }
  
            l = 0;
            r = indices[z].size() - 1;
  
            int third = -1;
  
            // Binary search to find the number of
            // possible values of the third element
            while (l <= r) {
                int m = (l + r) / 2;
  
                if (indices[z][m] > i) {
                    third = m;
                    r = m - 1;
                }
                else {
                    l = m + 1;
                }
            }
  
            if (first != -1 && third != -1) {
  
                // Contribution to the answer would
                // be the multiplication of the possible
                // values for the first and the third element
                answer += (first + 1) * (indices[z].size() - third);
            }
        }
    }
  
    return answer;
}
  
// Driver code
int main()
{
    int array[] = { 2, 4, 5, 6, 7 };
    int n = sizeof(array) / sizeof(array[0]);
  
    cout << find_triplet(array, n);
  
    return 0;
}

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// Java implementation of the approach 


import java.util.*; 


  


class GFG 


{ 


  


static int MAX = 100001; 


static int ROW = 10; 


static int COl = 3; 


  


static Vector<Integer> []indices = new Vector[MAX]; 


  


// All possible solutions of the 


// equation 1/a + 1/b + 1/c = 1 


static int test[][] = { { 2, 3, 6 }, { 2, 4, 4 }, 


                        { 2, 6, 3 }, { 3, 2, 6 }, 


                        { 3, 3, 3 }, { 3, 6, 2 }, 


                        { 4, 2, 4 }, { 4, 4, 2 }, 


                        { 6, 2, 3 }, { 6, 3, 2 } }; 


  


// Function to find the triplets 


static int find_triplet(int array[], int n) 


{ 


    int answer = 0; 


    for (int i = 0; i < MAX; i++) 


    { 


        indices[i] = new Vector<>(); 


    } 


      


    // Storing indices of the elements 


    for (int i = 0; i < n; i++) 


    { 


        indices[array[i]].add(i); 


    } 


  


    for (int i = 0; i < n; i++) 


    { 


        int y = array[i]; 


  


        for (int j = 0; j < ROW; j++)  


        { 


            int s = test[j][1] * y; 


  


            // Check if y can act as the middle 


            // element of triplet with the given 


            // solution of 1/a + 1/b + 1/c = 1 


            if (s % test[j][0] != 0) 


                continue; 


            if (s % test[j][2] != 0) 


                continue; 


  


            int x = s / test[j][0]; 


            int z = s / test[j][2]; 


            if (x > MAX || z > MAX) 


                continue; 


  


            int l = 0; 


            int r = indices[x].size() - 1; 


  


            int first = -1; 


  


            // Binary search to find the number of 


            // possible values of the first element 


            while (l <= r) 


            { 


                int m = (l + r) / 2; 


  


                if (indices[x].get(m) < i) 


                { 


                    first = m; 


                    l = m + 1; 


                } 


                else 


                { 


                    r = m - 1; 


                } 


            } 


  


            l = 0; 


            r = indices[z].size() - 1; 


  


            int third = -1; 


  


            // Binary search to find the number of 


            // possible values of the third element 


            while (l <= r)  


            { 


                int m = (l + r) / 2; 


  


                if (indices[z].get(m) > i)  


                { 


                    third = m; 


                    r = m - 1; 


                } 


                else 


                { 


                    l = m + 1; 


                } 


            } 


  


            if (first != -1 && third != -1)  


            { 


  


                // Contribution to the answer would 


                // be the multiplication of the possible 


                // values for the first and the third element 


                answer += (first + 1) * (indices[z].size() - third); 


            } 


        } 


    } 


    return answer; 


} 


  


// Driver code 


public static void main(String []args)  


{ 


    int array[] = { 2, 4, 5, 6, 7 }; 


    int n = array.length; 


  


    System.out.println(find_triplet(array, n)); 


} 


} 


  


// This code is contributed by Rajput-Ji 



















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# Python3 implementation of the approach 


MAX = 100001


ROW = 10


COL = 3


  


indices = [0] * MAX


  


# All possible solutions of the 


# equation 1/a + 1/b + 1/c = 1 


test = [[2, 3, 6], [2, 4, 4],  


        [2, 6, 3], [3, 2, 6],  


        [3, 3, 3], [3, 6, 2], 


        [4, 2, 4], [4, 4, 2],  


        [6, 2, 3], [6, 3, 2]] 


  


# Function to find the triplets 


def find_triplet(array, n): 


    answer = 0


  


    for i in range(MAX): 


        indices[i] = [] 


  


    # Storing indices of the elements 


    for i in range(n): 


        indices[array[i]].append(i) 


  


    for i in range(n): 


        y = array[i] 


  


        for j in range(ROW): 


            s = test[j][1] * y 


  


            # Check if y can act as the middle 


            # element of triplet with the given 


            # solution of 1/a + 1/b + 1/c = 1 


            if s % test[j][0] != 0: 


                continue


            if s % test[j][2] != 0: 


                continue


  


            x = s // test[j][0] 


            z = s // test[j][2] 


            if x > MAX or z > MAX: 


                continue


  


            l = 0


            r = len(indices[x]) - 1


            first = -1


  


            # Binary search to find the number of 


            # possible values of the first element 


            while l <= r: 


                m = (l + r) // 2


  


                if indices[x][m] < i: 


                    first = m 


                    l = m + 1


                else: 


                    r = m - 1


  


            l = 0


            r = len(indices[z]) - 1


            third = -1


  


            # Binary search to find the number of 


            # possible values of the third element 


            while l <= r: 


                m = (l + r) // 2


  


                if indices[z][m] > i: 


                    third = m 


                    r = m - 1


                else: 


                    l = m + 1


  


            if first != -1 and third != -1: 


  


                # Contribution to the answer would 


                # be the multiplication of the possible 


                # values for the first and the third element 


                answer += (first + 1) * (len(indices[z]) - third) 


    return answer 


  


# Driver Code 


if __name__ == "__main__": 


    array = [2, 4, 5, 6, 7] 


    n = len(array) 


  


    print(find_triplet(array, n)) 


  


# This code is contributed by 


# sanjeev2552 



















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// C# implementation of the approach 


using System; 


using System.Collections.Generic; 


  


class GFG 


{ 


  


static int MAX = 100001; 


static int ROW = 10; 


static int COl = 3; 


  


static List<int> []indices = new List<int>[MAX]; 


  


// All possible solutions of the 


// equation 1/a + 1/b + 1/c = 1 


static int [,]test = { { 2, 3, 6 }, { 2, 4, 4 }, 


                       { 2, 6, 3 }, { 3, 2, 6 }, 


                       { 3, 3, 3 }, { 3, 6, 2 }, 


                       { 4, 2, 4 }, { 4, 4, 2 }, 


                       { 6, 2, 3 }, { 6, 3, 2 } }; 


  


// Function to find the triplets 


static int find_triplet(int []array, int n) 


{ 


    int answer = 0; 


    for (int i = 0; i < MAX; i++) 


    { 


        indices[i] = new List<int>(); 


    } 


      


    // Storing indices of the elements 


    for (int i = 0; i < n; i++) 


    { 


        indices[array[i]].Add(i); 


    } 


  


    for (int i = 0; i < n; i++) 


    { 


        int y = array[i]; 


  


        for (int j = 0; j < ROW; j++)  


        { 


            int s = test[j, 1] * y; 


  


            // Check if y can act as the middle 


            // element of triplet with the given 


            // solution of 1/a + 1/b + 1/c = 1 


            if (s % test[j, 0] != 0) 


                continue; 


            if (s % test[j, 2] != 0) 


                continue; 


  


            int x = s / test[j, 0]; 


            int z = s / test[j, 2]; 


            if (x > MAX || z > MAX) 


                continue; 


  


            int l = 0; 


            int r = indices[x].Count - 1; 


  


            int first = -1; 


  


            // Binary search to find the number of 


            // possible values of the first element 


            while (l <= r) 


            { 


                int m = (l + r) / 2; 


  


                if (indices[x][m] < i) 


                { 


                    first = m; 


                    l = m + 1; 


                } 


                else


                { 


                    r = m - 1; 


                } 


            } 


  


            l = 0; 


            r = indices[z].Count - 1; 


  


            int third = -1; 


  


            // Binary search to find the number of 


            // possible values of the third element 


            while (l <= r)  


            { 


                int m = (l + r) / 2; 


  


                if (indices[z][m] > i)  


                { 


                    third = m; 


                    r = m - 1; 


                } 


                else


                { 


                    l = m + 1; 


                } 


            } 


  


            if (first != -1 && third != -1)  


            { 


  


                // Contribution to the answer would 


                // be the multiplication of the possible 


                // values for the first and the third element 


                answer += (first + 1) * (indices[z].Count - third); 


            } 


        } 


    } 


    return answer; 


} 


  


// Driver code 


public static void Main(String []args)  


{ 


    int []array = { 2, 4, 5, 6, 7 }; 


    int n = array.Length; 


  


    Console.WriteLine(find_triplet(array, n)); 


} 


} 


  


// This code is contributed by Rajput-Ji 



















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