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++
// 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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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. |
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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 |
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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 |
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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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