# 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 ` `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); ` `    ``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

 `= 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:

```3
```

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