Problem Name: Nodes and Depth

Problem setter: Mb. Abdul Alim

Limits 1s, 512 MB

To solve this problem, we can observe that the number of nodes at each depth follows a geometric progression with a common ratio of 3. Specifically, the number of nodes at depth D can be calculated as 3^D.

We can find the total number of nodes up to depth N by summing the number of nodes at each depth from 1 to N. This sum can be calculated using the formula for the sum of a geometric series:

Total nodes = a1 (r^N - 1) / (r-1) + 1; (Depth 0 = 1 Node)

where:

  • a1 is the first term in the series (number of nodes at depth 1),

  • r is the common ratio (3 in this case)

  • N is the depth.

Here's the approach in a paragraph:

We start by reading the input value of N. Then, we apply the formula for the sum of a geometric series to calculate the total number of nodes up to depth N, using a1 = 3 ( number of nodes at depth 1) and r = 3 (common ratio). Finally, we output the result, which represents the total number of nodes within depth N.

Complexity:

Time Complexity: O(N)

Space complexity: O(1)

Statistics

82% Solution Ratio
TanbirTopEarliest, 3M ago
Al_NahianFastest, 0.0s
faheeedLightest, 5.1 MB
Nusab19Shortest, 35B
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