Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/823.html

823. Binary Trees With Factors (Medium)

Given an array of unique integers, each integer 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 it's children.

How many binary trees can we make?  Return the answer modulo 10 ** 9 + 7.

Example 1:

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

Example 2:

Input: A = [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].

 

Note:

  1. 1 <= A.length <= 1000.
  2. 2 <= A[i] <= 10 ^ 9.

Solution 1. DP

// OJ: https://leetcode.com/problems/binary-trees-with-factors/
// Time: O(N^2)
// Space: O(N)
class Solution {
public:
    int numFactoredBinaryTrees(vector<int>& A) {
        int mod = 1e9 + 7, N = A.size();
        sort(A.begin(), A.end());
        vector<long long> dp(N, 1);
        unordered_map<int, long long> m;
        for (int i = 0; i < N; ++i) m[A[i]] = i;
        for (int i = 0; i < N; ++i) {
            for (int j = 0; j < i; ++j) {
                if (A[i] % A[j] || m.find(A[i] / A[j]) == m.end()) continue;
                dp[i] = (dp[i] + dp[j] * dp[m[A[i] / A[j]]]) % mod;
            }
        }
        int ans = 0;
        for (auto n : dp) ans = (ans + n) % mod;
        return ans;
    }
};

  • class Solution {
    public int numFactoredBinaryTrees(int[] A) {
        final int MODULO = 1000000007;
        Arrays.sort(A);
        int length = A.length;
        long[] counts = new long[length];
        Arrays.fill(counts, 1);
        Map<Integer, Integer> indexMap = new HashMap<Integer, Integer>();
        for (int i = 0; i < length; i++)
            indexMap.put(A[i], i);
        for (int i = 1; i < length; i++) {
            for (int j = 0; j < i; j++) {
                if (A[i] % A[j] == 0) {
                    int leftChild = A[j];
                    int rightChild = A[i] / A[j];
                    if (indexMap.containsKey(rightChild)) {
                        int rightIndex = indexMap.get(rightChild);
                        counts[i] = (counts[i] + counts[j] * counts[rightIndex]) % MODULO;
                    }
                }
            }
        }
        long trees = 0;
        for (int i = 0; i < length; i++)
            trees = (trees + counts[i]) % MODULO;
        return (int) trees;
    }
}

  • // OJ: https://leetcode.com/problems/binary-trees-with-factors/
// Time: O(N^2)
// Space: O(N)
class Solution {
public:
    int numFactoredBinaryTrees(vector<int>& A) {
        int mod = 1e9 + 7, N = A.size();
        sort(A.begin(), A.end());
        vector<long long> dp(N, 1);
        unordered_map<int, long long> m;
        for (int i = 0; i < N; ++i) m[A[i]] = i;
        for (int i = 0; i < N; ++i) {
            for (int j = 0; j < i; ++j) {
                if (A[i] % A[j] || m.find(A[i] / A[j]) == m.end()) continue;
                dp[i] = (dp[i] + dp[j] * dp[m[A[i] / A[j]]]) % mod;
            }
        }
        int ans = 0;
        for (auto n : dp) ans = (ans + n) % mod;
        return ans;
    }
};

  • 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

############

class Solution(object):
    def numFactoredBinaryTrees(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        A.sort()
        dp = {}
        for i, a in enumerate(A):
            dp[a] = 1
            for j in range(i):
                if a % A[j] == 0 and a / A[j] in dp:
                    dp[a] += dp[A[j]] * dp[a / A[j]]
        return sum(dp.values()) % (10**9 + 7)

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

  • 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;
}

All Problems

All Solutions