Number of triangles possible with given lengths of sticks which are powers of 2

Given an array of N integers where arr[i] denotes the number of sticks of length 2i. The task is to find the number of triangles possible with given lengths having area ≥ 0.
Note: Every stick can only be used once.

Examples:

Input: a[] = {1, 2, 2, 2, 2}
Output: 3
All possible triangles are:
(20, 24, 24), (21, 23, 23), (21, 22, 22).



Input: a[] = {3, 3, 3}
Output: 3

Approach: The main observation is that the triangles with area ≥ 0 can only be formed if there are three same lengths of sticks or one different and two similar lengths of sticks. Hence greedily iterate from the back and count the number of pairs of same length sticks available which is arr[i] / 2. But if there is a stick remaining, then a pair and a stick is used to form a triangle. In the end, the total number of sticks left is calculated which is 2 * pairs and the number of triangles that can be formed with these remaining sticks will be (2 * pairs) / 3.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the
// number of positive area triangles
int countTriangles(int a[], int n)
{
  
    // To store the count of
    // total triangles
    int cnt = 0;
  
    // To store the count of pairs of sticks
    // with equal lengths
    int pairs = 0;
  
    // Back-traverse and count
    // the number of triangles
    for (int i = n - 1; i >= 0; i--) {
  
        // Count the number of pairs
        pairs += a[i] / 2;
  
        // If we have one remaining stick
        // and we have a pair
        if (a[i] % 2 == 1 && pairs > 0) {
  
            // Count 1 triangle
            cnt += 1;
  
            // Reduce one pair
            pairs -= 1;
        }
    }
  
    // Count the remaining triangles
    // that can be formed
    cnt += (2 * pairs) / 3;
    return cnt;
}
  
// Driver code
int main()
{
    int a[] = { 1, 2, 2, 2, 2 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << countTriangles(a, n);
  
    return 0;
}

chevron_right






                        filter_none









Java














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















// Java implementation of the approach 


class GFG 


{ 


      


// Function to return the 


// number of positive area triangles 


static int countTriangles(int a[], int n) 


{ 


  


    // To store the count of 


    // total triangles 


    int cnt = 0; 


  


    // To store the count of pairs of sticks 


    // with equal lengths 


    int pairs = 0; 


  


    // Back-traverse and count 


    // the number of triangles 


    for (int i = n - 1; i >= 0; i--)  


    { 


  


        // Count the number of pairs 


        pairs += a[i] / 2; 


  


        // If we have one remaining stick 


        // and we have a pair 


        if (a[i] % 2 == 1 && pairs > 0)  


        { 


  


            // Count 1 triangle 


            cnt += 1; 


  


            // Reduce one pair 


            pairs -= 1; 


        } 


    } 


  


    // Count the remaining triangles 


    // that can be formed 


    cnt += (2 * pairs) / 3; 


    return cnt; 


} 


  


// Driver code 


public static void main(String[] args) 


{ 


    int a[] = { 1, 2, 2, 2, 2 }; 


    int n = a.length; 


    System.out.println(countTriangles(a, n)); 


} 


} 


  


// This code is contributed by Code_Mech. 



















                        chevron_right







                        filter_none









Python3














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















# Python3 implementation of the approach 


  


# Function to return the 


# number of positive area triangles 


def countTriangles(a, n): 


  


    # To store the count of 


    # total triangles 


    cnt = 0


  


    # To store the count of pairs of sticks 


    # with equal lengths 


    pairs = 0


  


    # Back-traverse and count 


    # the number of triangles 


    for i in range(n - 1, -1, -1): 


  


        # Count the number of pairs 


        pairs += a[i] // 2


  


        # If we have one remaining stick 


        # and we have a pair 


        if (a[i] % 2 == 1 and pairs > 0): 


  


            # Count 1 triangle 


            cnt += 1


  


            # Reduce one pair 


            pairs -= 1


          


    # Count the remaining triangles 


    # that can be formed 


    cnt += (2 * pairs) // 3


    return cnt 


  


# Driver code 


a = [1, 2, 2, 2, 2] 


n = len(a) 


print(countTriangles(a, n)) 


  


# This code is contributed by mohit kumar 



















                        chevron_right







                        filter_none









C#














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















// C# implementation of the approach  


using System; 


  


class GFG  


{  


          


    // Function to return the  


    // number of positive area triangles  


    static int countTriangles(int []a, int n)  


    {  


      


        // To store the count of  


        // total triangles  


        int cnt = 0;  


      


        // To store the count of pairs of sticks  


        // with equal lengths  


        int pairs = 0;  


      


        // Back-traverse and count  


        // the number of triangles  


        for (int i = n - 1; i >= 0; i--)  


        {  


      


            // Count the number of pairs  


            pairs += a[i] / 2;  


      


            // If we have one remaining stick  


            // and we have a pair  


            if (a[i] % 2 == 1 && pairs > 0)  


            {  


      


                // Count 1 triangle  


                cnt += 1;  


      


                // Reduce one pair  


                pairs -= 1;  


            }  


        }  


      


        // Count the remaining triangles  


        // that can be formed  


        cnt += (2 * pairs) / 3;  


        return cnt;  


    }  


      


    // Driver code  


    public static void Main()  


    {  


        int []a = { 1, 2, 2, 2, 2 };  


        int n = a.Length;  


        Console.WriteLine(countTriangles(a, n));  


    }  


}  


  


// This code is contributed by Ryuga 



















                        chevron_right







                        filter_none









PHP














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















<?php 


// PHP implementation of the approach 


  


// Function to return the 


// number of positive area triangles 


Function countTriangles($a, $n) 


{ 


  


    // To store the count of 


    // total triangles 


    $cnt = 0; 


  


    // To store the count of pairs of sticks 


    // with equal lengths 


    $pairs = 0; 


  


    // Back-traverse and count 


    // the number of triangles 


    for ($i = $n - 1; $i >= 0; $i--)  


    { 


  


        // Count the number of pairs 


        $pairs += $a[$i] / 2; 


  


        // If we have one remaining stick 


        // and we have a pair 


        if ($a[$i] % 2 == 1 && $pairs > 0)  


        { 


  


            // Count 1 triangle 


            $cnt += 1; 


  


            // Reduce one pair 


            $pairs -= 1; 


        } 


    } 


  


    // Count the remaining triangles 


    // that can be formed 


    $cnt += (int)((2 * $pairs) / 3); 


    return $cnt; 


} 


  


// Driver code 


$a = array(1, 2, 2, 2, 2 ); 


$n = sizeof($a); 


echo(countTriangles($a, $n)); 


  


// This code is contributed by Code_Mech. 


?> 



















                        chevron_right







                        filter_none











Output:


3






          
          
          
          

            

    

        My Personal Notes
        arrow_drop_up
    

    

        
        

            
            
        

    


Recommended Posts:
Striver
Striver(underscore)79 at Codechef and codeforces D
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on  the "Improve Article" button below.