README_EN.md

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823. Binary Trees With Factors

中文文档

Description

Given an array of unique integers, arr, where each integer arr[i] is strictly greater than 1.

We make a binary tree using these integers, and each number may be used for any number of times. Each non-leaf node's value should be equal to the product of the values of its children.

Return the number of binary trees we can make. The answer may be too large so return the answer modulo 109 + 7.

 

Example 1:

Input: arr = [2,4]
Output: 3
Explanation: We can make these trees: [2], [4], [4, 2, 2]

Example 2:

Input: arr = [2,4,5,10]
Output: 7
Explanation: We can make these trees: [2], [4], [5], [10], [4, 2, 2], [10, 2, 5], [10, 5, 2].

 

Constraints:

• 1 <= arr.length <= 1000
• 2 <= arr[i] <= 109
• All the values of arr are unique.

Solutions

Solution 1

Python3

class Solution:
    def numFactoredBinaryTrees(self, arr: List[int]) -> int:
        mod = 10**9 + 7
        n = len(arr)
        arr.sort()
        idx = {v: i for i, v in enumerate(arr)}
        f = [1] * n
        for i, a in enumerate(arr):
            for j in range(i):
                b = arr[j]
                if a % b == 0 and (c := (a // b)) in idx:
                    f[i] = (f[i] + f[j] * f[idx[c]]) % mod
        return sum(f) % mod

Java

class Solution {
    public int numFactoredBinaryTrees(int[] arr) {
        final int mod = (int) 1e9 + 7;
        Arrays.sort(arr);
        int n = arr.length;
        long[] f = new long[n];
        Arrays.fill(f, 1);
        Map<Integer, Integer> idx = new HashMap<>(n);
        for (int i = 0; i < n; ++i) {
            idx.put(arr[i], i);
        }
        for (int i = 0; i < n; ++i) {
            int a = arr[i];
            for (int j = 0; j < i; ++j) {
                int b = arr[j];
                if (a % b == 0) {
                    int c = a / b;
                    if (idx.containsKey(c)) {
                        int k = idx.get(c);
                        f[i] = (f[i] + f[j] * f[k]) % mod;
                    }
                }
            }
        }
        long ans = 0;
        for (long v : f) {
            ans = (ans + v) % mod;
        }
        return (int) ans;
    }
}

C++

class Solution {
public:
    int numFactoredBinaryTrees(vector<int>& arr) {
        const int mod = 1e9 + 7;
        sort(arr.begin(), arr.end());
        unordered_map<int, int> idx;
        int n = arr.size();
        for (int i = 0; i < n; ++i) {
            idx[arr[i]] = i;
        }
        vector<long> f(n, 1);
        for (int i = 0; i < n; ++i) {
            int a = arr[i];
            for (int j = 0; j < i; ++j) {
                int b = arr[j];
                if (a % b == 0) {
                    int c = a / b;
                    if (idx.count(c)) {
                        int k = idx[c];
                        f[i] = (f[i] + 1l * f[j] * f[k]) % mod;
                    }
                }
            }
        }
        long ans = 0;
        for (long v : f) {
            ans = (ans + v) % mod;
        }
        return ans;
    }
};

Go

func numFactoredBinaryTrees(arr []int) int {
	const mod int = 1e9 + 7
	sort.Ints(arr)
	f := make([]int, len(arr))
	idx := map[int]int{}
	for i, v := range arr {
		f[i] = 1
		idx[v] = i
	}
	for i, a := range arr {
		for j := 0; j < i; j++ {
			b := arr[j]
			if c := a / b; a%b == 0 {
				if k, ok := idx[c]; ok {
					f[i] = (f[i] + f[j]*f[k]) % mod
				}
			}
		}
	}
	ans := 0
	for _, v := range f {
		ans = (ans + v) % mod
	}
	return ans
}

TypeScript

function numFactoredBinaryTrees(arr: number[]): number {
    const mod = 10 ** 9 + 7;
    arr.sort((a, b) => a - b);
    const idx: Map<number, number> = new Map();
    const n = arr.length;
    for (let i = 0; i < n; ++i) {
        idx.set(arr[i], i);
    }
    const f: number[] = new Array(n).fill(1);
    for (let i = 0; i < n; ++i) {
        const a = arr[i];
        for (let j = 0; j < i; ++j) {
            const b = arr[j];
            if (a % b === 0) {
                const c = a / b;
                if (idx.has(c)) {
                    const k = idx.get(c)!;
                    f[i] = (f[i] + f[j] * f[k]) % mod;
                }
            }
        }
    }
    return f.reduce((a, b) => a + b) % mod;
}