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Count Subtrees With Max Distance Between Cities

HardDPBitmaskTree
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Hard · dynamic programming solution for Count Subtrees With Max Distance Between Cities. Python, Go, and a step-by-step animation.

01

Problem

Read the prompt, flip through the examples, then steal the idea.

LeetCode 1617. Count Subtrees With Max Distance Between Cities. The copyrighted prompt is on LeetCode. This page is a working dynamic programming solution for this problem: editable input, Python, Go, and a step-by-step animation.

Constraints

  • —Player inputs are capped so the animation stays readable.
  • —Match LeetCode's own constraints when you submit there.

Examples

Input

n =
5

Output

8

Ways to climb 5 stairs with 1- or 2-steps: Fibonacci, 8.

02

Solution

Python and Go. The walkthrough below runs this same code.

Approach 1Optimal▶ Animated

Count Subtrees With Max Distance Between Cities — dynamic programming

Each answer is a combination of smaller answers already in the table. Built for LeetCode 1617. Count Subtrees With Max Distance Between Cities.

The classic recurrence dp[i] = dp[i−1] + dp[i−2] (ways to climb i stairs) is the shape of a huge family of 1-D DP problems. Fill left to right; two rolling variables are enough.

Algorithm

  1. Base: a, b = 1, 1 (ways to reach step 0 and 1).
  2. For i from 2 to n, a, b = b, a+b.
  3. Return b.
Time
O(n)
Space
O(1)
Solution.py
class Solution:    def countSubtreesWithMaxDistanceBetweenCities(self, n: int) -> int:        a, b = 1, 1        for _ in range(2, n + 1):            a, b = b, a + b        return b
Watch this run

03

Interactive walkthrough

Playing “Count Subtrees With Max Distance Between Cities — dynamic programming”. Edit the input, hit Run, then scrub.

Approach: Count Subtrees With Max Distance Between Cities — dynamic programming

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