Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Move from brute-force thinking to an efficient approach using array strategy.
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 <= 10002 <= arr[i] <= 109arr are unique.Problem summary: 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Hash Map · Dynamic Programming
[2,4]
[2,4,5,10]
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #823: Binary Trees With Factors
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;
}
}
// Accepted solution for LeetCode #823: Binary Trees With Factors
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
}
# Accepted solution for LeetCode #823: Binary Trees With Factors
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
// Accepted solution for LeetCode #823: Binary Trees With Factors
struct Solution;
use std::collections::HashMap;
impl Solution {
fn num_factored_binary_trees(mut a: Vec<i32>) -> i32 {
let n = a.len();
let mut dp: Vec<i64> = vec![1; n];
let modulo = 1_000_000_007;
let mut hm: HashMap<i32, usize> = HashMap::new();
a.sort_unstable();
let mut res = 0;
for i in 0..n {
hm.insert(a[i], i);
for j in 0..i {
if a[i] % a[j] == 0 {
if let Some(&k) = hm.get(&(a[i] / a[j])) {
dp[i] += dp[j] * dp[k];
}
}
}
res = (res + dp[i]) % modulo;
}
res as i32
}
}
#[test]
fn test() {
let a = vec![2, 4];
let res = 3;
assert_eq!(Solution::num_factored_binary_trees(a), res);
let a = vec![2, 4, 5, 10];
let res = 7;
assert_eq!(Solution::num_factored_binary_trees(a), res);
}
// Accepted solution for LeetCode #823: Binary Trees With Factors
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;
}
Use this to step through a reusable interview workflow for this problem.
Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.
Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Wrong move: Zero-count keys stay in map and break distinct/count constraints.
Usually fails on: Window/map size checks are consistently off by one.
Fix: Delete keys when count reaches zero.
Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.
Usually fails on: Correctness breaks on cases that differ only in hidden state.
Fix: Define state so each unique subproblem maps to one DP cell.