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++

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// 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;
}

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Python3














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# 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 



















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// 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. 



















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Output:


3






          
          
          
          

            

    

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Improved By :  mohit kumar 29, Code_Mech