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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

// 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; 
} 

Java

// 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. 

Python3

# 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 

C#

// 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 

PHP

<?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. 
?> 

Output:

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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