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Problem solution · Python

Handshakes That Dont Cross

Handshakes That Dont Cross: a Python solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Dynamic programming
Source
Kamyu LeetCode Solutions
Length
41 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Dynamic programming

For Handshakes That Dont Cross, the implementation records answers for smaller states and reuses them to build the requested result without repeating work.

  1. Define precisely what one DP state represents.
  2. Establish the base cases before transitions are evaluated.
  3. Process states in dependency order and combine only already-known values.

Code notes

  • 41 lines of Python from the credited upstream file handshakes-that-dont-cross.py.
  • The implementation visibly relies on cached states.
  • No explicit loop blocks detected.

Complexity

Multiply the number of reachable states by the work performed for each transition, then include the stored state table in memory usage.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeHandshakes That Dont Cross · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(n)# Space: O(1) class Solution(object):    def numberOfWays(self, num_people):        """        :type num_people: int        :rtype: int        """        MOD = 10**9+7        def inv(x, m):  # Euler's Theorem            return pow(x, m-2, m)  # O(logMOD) = O(1)         def nCr(n, k, m):            if n-k < k:                return nCr(n, n-k, m)            result = 1            for i in xrange(1, k+1):                result = result*(n-k+i)*inv(i, m)%m            return result         n = num_people//2        return nCr(2*n, n, MOD)*inv(n+1, MOD) % MOD  # Catalan number  # Time:  O(n^2)# Space: O(n)class Solution2(object):    def numberOfWays(self, num_people):        """        :type num_people: int        :rtype: int        """        MOD = 10**9+7        dp = [0]*(num_people//2+1)        dp[0] = 1        for k in xrange(1, num_people//2+1):            for i in xrange(k):                dp[k] = (dp[k] + dp[i]*dp[k-1-i]) % MOD        return dp[num_people//2]

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