Handshakes That Don't Cross - Practice Coding Problems

Handshakes That Don't Cross - Problem

Non-Crossing Handshakes Problem

Imagine numPeople individuals standing in a perfect circle at a social gathering. Each person must shake hands with exactly one other person, creating numPeople / 2 handshake pairs total.

The challenge: How many different ways can these handshakes occur such that none of the handshake "lines" cross each other?

🤝 Key Constraints:
• numPeople is always even
• Every person shakes hands with exactly one other person
• Handshakes cannot cross when viewed from above the circle
• Return the count modulo 10⁹ + 7

Example: With 4 people (A, B, C, D) in a circle, valid arrangements include A-B & C-D, but A-C & B-D would cross!

Input & Output

example_1.py — Basic Case

$ Input: numPeople = 2

› Output: 1

💡 Note: With 2 people in a circle, there's only one way to pair them without crossing: person 0 shakes hands with person 1.

example_2.py — Four People

$ Input: numPeople = 4

› Output: 2

💡 Note: With 4 people (0,1,2,3) in a circle: Way 1: (0,1) and (2,3) - adjacent pairs. Way 2: (0,3) and (1,2) - opposite pairs. The pairing (0,2) and (1,3) would cross, so it's invalid.

example_3.py — Six People

$ Input: numPeople = 6

› Output: 5

💡 Note: With 6 people, we get the 3rd Catalan number: C(3) = 5. This follows the pattern: C(0)=1, C(1)=1, C(2)=2, C(3)=5, representing the number of non-crossing handshake arrangements.

Constraints

2 ≤ numPeople ≤ 1000
numPeople is always even
Return result modulo 10⁹ + 7

Visualization

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Understanding the Visualization

Recognize the pattern

This is actually the famous Catalan number sequence in disguise

Dynamic programming

Build up solutions from smaller subproblems using C(n) recurrence

Optimal calculation

Use the mathematical formula or DP table to compute the nth Catalan number

Key Takeaway

🎯 Key Insight: Non-crossing handshakes in a circle follow the Catalan number sequence - a beautiful connection between geometry and combinatorics!

Asked in

G Google 45 a Amazon 38 f Meta 29 ⊞ Microsoft 22

This problem is equivalent to finding the nth Catalan number where n = numPeople/2. The optimal approach uses dynamic programming with the recurrence relation C(n) = Σ C(i) × C(n-1-i), achieving O(n²) time complexity. Each valid handshake arrangement corresponds to a way of properly nesting parentheses or building a binary search tree.

Common Approaches

Approach	Time	Space	Notes
✓ Dynamic Programming (Bottom-Up)	O(n³)	O(n²)	Build solution using DP to recognize Catalan number pattern

Dynamic Programming (Bottom-Up) — Algorithm Steps

Recognize this as a divide-and-conquer problem
For each person, try pairing with every other person
Split remaining people into two groups (left and right)
Recursively solve for both groups
Use memoization to avoid recalculating subproblems
Build up from base case: 0 people = 1 way, 2 people = 1 way

Visualization

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Step-by-Step Walkthrough

Base cases

C(0) = 1 (empty arrangement), C(1) = 1 (one pair)

Recurrence relation

C(n) = sum of C(i) * C(n-1-i) for all valid splits

Build table

Fill DP table from bottom up using smaller subproblems

Return result

The nth Catalan number gives us the answer

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define MOD 1000000007

int numberOfWays(int numPeople) {
    int n = numPeople / 2;
    if (n == 0) return 1;
    
    long long* catalan = (long long*)calloc(n + 1, sizeof(long long));
    catalan[0] = 1;
    
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j < i; j++) {
            catalan[i] = (catalan[i] + (catalan[j] * catalan[i - 1 - j]) % MOD) % MOD;
        }
    }
    
    int result = catalan[n];
    free(catalan);
    return result;
}

int main() {
    int numPeople;
    scanf("%d", &numPeople);
    printf("%d\n", numberOfWays(numPeople));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

For each of n people, we try n pairings and solve two subproblems

⚠ Quadratic Growth

Space Complexity

O(n²)

Memoization table stores results for all subproblem states

⚠ Quadratic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Dynamic Programming (Bottom-Up) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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