Welcome to Subscribe On Youtube

1259. Handshakes That Don’t Cross

Description

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.

Since the answer could be very large, return it modulo 109 + 7.

 

Example 1:

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)].

Example 2:

Input: numPeople = 6
Output: 5

 

Constraints:

  • 2 <= numPeople <= 1000
  • numPeople is even.

Solutions

Solution 1: Memoization Search

We design a function $dfs(i)$, which represents the number of handshake schemes for $i$ people. The answer is $dfs(n)$.

The execution logic of the function $dfs(i)$ is as follows:

  • If $i \lt 2$, then there is only one handshake scheme, which is not to shake hands, so return $1$.
  • Otherwise, we can enumerate who the first person shakes hands with. Let the number of remaining people on the left be $l$, and the number of people on the right be $r=i-l-2$. Then we have $dfs(i)= \sum_{l=0}^{i-1} dfs(l) \times dfs(r)$.

To avoid repeated calculations, we use the method of memoization search.

The time complexity is $O(n^2)$, and the space complexity is $O(n)$. Where $n$ is the size of $numPeople$.

  • class Solution {
    private int[] f;
    private final int mod = (int) 1e9 + 7;

    public int numberOfWays(int numPeople) {
        f = new int[numPeople + 1];
        return dfs(numPeople);
    }

    private int dfs(int i) {
        if (i < 2) {
            return 1;
        }
        if (f[i] != 0) {
            return f[i];
        }
        for (int l = 0; l < i; l += 2) {
            int r = i - l - 2;
            f[i] = (int) ((f[i] + (1L * dfs(l) * dfs(r) % mod)) % mod);
        }
        return f[i];
    }
}

  • class Solution {
public:
    int numberOfWays(int numPeople) {
        const int mod = 1e9 + 7;
        int f[numPeople + 1];
        memset(f, 0, sizeof(f));
        function<int(int)> dfs = [&](int i) {
            if (i < 2) {
                return 1;
            }
            if (f[i]) {
                return f[i];
            }
            for (int l = 0; l < i; l += 2) {
                int r = i - l - 2;
                f[i] = (f[i] + 1LL * dfs(l) * dfs(r) % mod) % mod;
            }
            return f[i];
        };
        return dfs(numPeople);
    }
};

  • class Solution:
    def numberOfWays(self, numPeople: int) -> int:
        @cache
        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)


  • func numberOfWays(numPeople int) int {
	const mod int = 1e9 + 7
	f := make([]int, numPeople+1)
	var dfs func(int) int
	dfs = func(i int) int {
		if i < 2 {
			return 1
		}
		if f[i] != 0 {
			return f[i]
		}
		for l := 0; l < i; l += 2 {
			r := i - l - 2
			f[i] = (f[i] + dfs(l)*dfs(r)) % mod
		}
		return f[i]
	}
	return dfs(numPeople)
}

  • function numberOfWays(numPeople: number): number {
    const mod = 10 ** 9 + 7;
    const f: number[] = Array(numPeople + 1).fill(0);
    const dfs = (i: number): number => {
        if (i < 2) {
            return 1;
        }
        if (f[i] !== 0) {
            return f[i];
        }
        for (let l = 0; l < i; l += 2) {
            const r = i - l - 2;
            f[i] += Number((BigInt(dfs(l)) * BigInt(dfs(r))) % BigInt(mod));
            f[i] %= mod;
        }
        return f[i];
    };
    return dfs(numPeople);
}

All Problems

All Solutions