Leetcode #1259: Handshakes That Don't Cross
In this guide, we solve Leetcode #1259 Handshakes That Don't Cross in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given an even number of people numPeople that stand around a circle and each person shakes hands with someone else so that there are numPeople / 2 handshakes total. Return the number of ways these handshakes could occur such that none of the handshakes cross.
Quick Facts
- Difficulty: Hard
- Premium: Yes
- Tags: Math, Dynamic Programming
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: numPeople = 4
Output: 2
Explanation: There are two ways to do it, the first way is [(1,2),(3,4)] and the second one is [(2,3),(4,1)].
Python Solution
class Solution:
def numberOfWays(self, numPeople: int) -> int:
def dfs(i: int) -> int:
if i < 2:
return 1
ans = 0
for l in range(0, i, 2):
r = i - l - 2
ans += dfs(l) * dfs(r)
ans %= mod
return ans
mod = 10**9 + 7
return dfs(numPeople)
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.