Number of triangles after N moves
Last Updated : 17 Feb, 2023
Find the number of triangles in Nth step,
Rules: Draw an equilateral triangle at the start. In the i-th move, take the uncolored triangles, divides each of them in 4 parts of equal areas and color the central part. Keep a count of triangles till the Nth step.
Examples:
Input : 1 Output : 5 Explanation: In 1st move we get
Input : 2 Output : 17 Explanation: In 2nd move we get
Naive approach:
The number of triangles in the nth figure are 3 times the number of triangles in the (n-1)th figure+2. We can see by observation the nth figure is made by placing 3 triangles similar to that in (n-1) figure and an inverted triangle. We also take into account the bigger triangle that has been formed. Hence the number of triangles in the nth figure becomes (number of triangles in the (n-1)th figure)*3 + 2.
C++ // C++ program to calculate the number of equilateral // triangles #include <bits/stdc++.h> using namespace std; // function to calculate number of triangles in Nth step int numberOfTriangles(int n) { int answer[n + 1] = { 0 }; answer[0] = 1; for (int i = 1; i <= n; i++) answer[i] = answer[i - 1] * 3 + 2; return answer[n]; } // driver program int main() { int n = 2; cout << numberOfTriangles(n); return 0; } // Java program to find middle of three // distinct numbers to calculate the // number of equilateral triangles import java.util.*; class Triangle { // function to calculate number of // triangles in Nth step public static int numberOfTriangles(int n) { int[] answer = new int[n+1]; answer[0] = 1; for (int i = 1; i <= n; i++) answer[i] = answer[i - 1] * 3 + 2; return answer[n]; } // driver code public static void main(String[] args) { int n = 2; System.out.println(numberOfTriangles(n)); } } // This code is contributed by rishabh_jain # Python3 code to calculate the # number of equilateral triangles # function to calculate number # of triangles in Nth step def numberOfTriangles (n) : answer = [None] * (n + 1); answer[0] = 1; i = 1 while i <= n: answer[i] = answer[i - 1] * 3 + 2; i = i + 1 return answer[n]; # Driver code n = 2 print(numberOfTriangles(n)) # This code is contributed by "rishabh_jain".
// C# program to find middle of three // distinct numbers to calculate the // number of equilateral triangles using System; class Triangle { // function to calculate number of // triangles in Nth step public static int numberOfTriangles(int n) { int[] answer = new int[n+1]; answer[0] = 1; for (int i = 1; i <= n; i++) answer[i] = answer[i - 1] * 3 + 2; return answer[n]; } // Driver code public static void Main() { int n = 2; Console.WriteLine(numberOfTriangles(n)); } } // This code is contributed by vt_m <?php // PHP program to calculate // the number of equilateral // triangles // function to calculate number // of triangles in Nth step function numberOfTriangles($n) { $answer = array(); $answer[0] = 1; for ($i = 1; $i <= $n; $i++) $answer[$i] = $answer[$i - 1] * 3 + 2; return $answer[$n]; } // Driver Code $n = 2; echo numberOfTriangles($n); // This code is contributed by anuj_67. ?> <script> // Javascript program to calculate // the number of equilateral // triangles // Function to calculate number // of triangles in Nth step function numberOfTriangles(n) { let answer = new Uint8Array(n + 1); answer[0] = 1; for(let i = 1; i <= n; i++) answer[i] = answer[i - 1] * 3 + 2; return answer[n]; } // Driver code let n = 2; document.write(numberOfTriangles(n)); // This code is contributed by Mayank Tyagi </script> Time Complexity: O(n)
Auxiliary Space: O(n)
An efficient solution will be to derive a formula for Nth step:
If we follow the naive approach for every step, then we get for Nth step the number of triangles to be
(2*(3^n))-1.
C++ // C++ program to calculate the number of // equilateral triangles #include <bits/stdc++.h> using namespace std; // function to calculate number of triangles // in Nth step int numberOfTriangles(int n) { int ans = 2 * (pow(3, n)) - 1; return ans; } // driver program int main() { int n = 2; cout << numberOfTriangles(n); return 0; } // Java program to find middle of three // distinct numbers to calculate the // number of equilateral triangles import java.util.*; import static java.lang.Math.pow; class Triangle { // function to calculate number // of triangles in Nth step public static double numberOfTriangles(int n) { double ans = 2 * (pow(3, n)) - 1; return ans; } // driver code public static void main(String[] args) { int n = 2; System.out.println(numberOfTriangles(n)); } } // This code is contributed by rishabh_jain # Python3 code to calculate the # number of equilateral triangles # function to calculate number # of triangles in Nth step def numberOfTriangles (n) : ans = 2 * (pow(3, n)) - 1; return ans; # Driver code n = 2 print (numberOfTriangles(n)) # This code is contributed by "rishabh_jain".
//C# program to find middle of three // distinct numbers to calculate the // number of equilateral triangles using System; class Triangle { // function to calculate number // of triangles in Nth step public static double numberOfTriangles(int n) { double ans = 2 * (Math.Pow(3, n)) - 1; return ans; } // Driver code public static void Main() { int n = 2; Console.WriteLine(numberOfTriangles(n)); } } // This code is contributed by vt_m <?php // PHP program to calculate the // number of equilateral triangles // function to calculate // number of triangles // in Nth step function numberOfTriangles($n) { $ans = 2 * (pow(3, $n)) - 1; return $ans; } // Driver Code $n = 2; echo numberOfTriangles($n); // This code is contributed by anuj_67. ?> <script> // javascript program to find middle of three // distinct numbers to calculate the // number of equilateral triangles // function to calculate number // of triangles in Nth step function numberOfTriangles(n) { var ans = 2 * (Math.pow(3, n)) - 1; return ans; } // Driver code var n = 2; document.write(numberOfTriangles(n)); // This code is contributed by aashish1995 </script> Time Complexity: O(log n), due to the inbuilt pow() function.
Auxiliary Space: O(1)