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Aug 29, 2023
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feat: add solutions to lc problem: No.0823 #1536

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52 changes: 38 additions & 14 deletions solution/0800-0899/0823.Binary Trees With Factors/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -44,11 +44,11 @@

**方法一：动态规划**

我们可以枚举 `arr` 中的每一个数 $a$ 作为二叉树的根节点（根节点一定最大），然后枚举枚举左子树的值 $b$，若 $a$ 能被 $b$ 整除，则右子树的值为 $a / b$，若 $a / b$ 也在 `arr` 中，则可以构成一棵二叉树。此时，以 $a$ 为根节点的二叉树的个数为 $f(a) = f(b) \times f(a / b)$，其中 $f(b)$ 和 $f(a / b)$ 分别为左子树和右子树的二叉树个数。
我们可以枚举 $arr$ 中的每一个数 $a$ 作为二叉树的根节点（根节点一定最大），然后枚举枚举左子树的值 $b$，若 $a$ 能被 $b$ 整除，则右子树的值为 $a / b$，若 $a / b$ 也在 $arr$ 中，则可以构成一棵二叉树。此时，以 $a$ 为根节点的二叉树的个数为 $f(a) = f(b) \times f(a / b)$，其中 $f(b)$ 和 $f(a / b)$ 分别为左子树和右子树的二叉树个数。

因此，我们先将 `arr` 排序，然后用 $f[i]$ 表示以 $arr[i]$ 为根节点的二叉树的个数，最终答案即为 $f[0] + f[1] + \cdots + f[n - 1]$。
因此，我们先将 $arr$ 排序，然后用 $f[i]$ 表示以 $arr[i]$ 为根节点的二叉树的个数，最终答案即为 $f[0] + f[1] + \cdots + f[n - 1]$。注意答案可能很大，需要对 $10^9 + 7$ 取模。

时间复杂度为 $O(n^2)$，空间复杂度为 $O(n)$。其中 $n$ 为 `arr` 的长度。
时间复杂度为 $O(n^2)$，空间复杂度为 $O(n)$。其中 $n$ 为 $arr$ 的长度。

<!-- tabs:start -->

Expand Down Expand Up @@ -78,9 +78,8 @@ class Solution:

```java
class Solution {
private static final int MOD = (int) 1e9 + 7;

public int numFactoredBinaryTrees(int[] arr) {
final int mod = (int) 1e9 + 7;
Arrays.sort(arr);
int n = arr.length;
long[] f = new long[n];
Expand All @@ -97,14 +96,14 @@ class Solution {
int c = a / b;
if (idx.containsKey(c)) {
int k = idx.get(c);
f[i] = (f[i] + f[j] * f[k]) % MOD;
f[i] = (f[i] + f[j] * f[k]) % mod;
}
}
}
}
long ans = 0;
for (long v : f) {
ans = (ans + v) % MOD;
ans = (ans + v) % mod;
}
return (int) ans;
}
Expand All @@ -116,9 +115,8 @@ class Solution {
```cpp
class Solution {
public:
const int mod = 1e9 + 7;

int numFactoredBinaryTrees(vector<int>& arr) {
const int mod = 1e9 + 7;
sort(arr.begin(), arr.end());
unordered_map<int, int> idx;
int n = arr.size();
Expand Down Expand Up @@ -155,18 +153,15 @@ 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 {
f[i] = 1
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 c := a / b; a%b == 0 {
if k, ok := idx[c]; ok {
f[i] = (f[i] + f[j]*f[k]) % mod
}
Expand All @@ -181,6 +176,35 @@ func numFactoredBinaryTrees(arr []int) int {
}
```

### **TypeScript**

```ts
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;
}
```

### **...**

```
Expand Down
46 changes: 35 additions & 11 deletions solution/0800-0899/0823.Binary Trees With Factors/README_EN.md
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -60,9 +60,8 @@ class Solution:

```java
class Solution {
private static final int MOD = (int) 1e9 + 7;

public int numFactoredBinaryTrees(int[] arr) {
final int mod = (int) 1e9 + 7;
Arrays.sort(arr);
int n = arr.length;
long[] f = new long[n];
Expand All @@ -79,14 +78,14 @@ class Solution {
int c = a / b;
if (idx.containsKey(c)) {
int k = idx.get(c);
f[i] = (f[i] + f[j] * f[k]) % MOD;
f[i] = (f[i] + f[j] * f[k]) % mod;
}
}
}
}
long ans = 0;
for (long v : f) {
ans = (ans + v) % MOD;
ans = (ans + v) % mod;
}
return (int) ans;
}
Expand All @@ -98,9 +97,8 @@ class Solution {
```cpp
class Solution {
public:
const int mod = 1e9 + 7;

int numFactoredBinaryTrees(vector<int>& arr) {
const int mod = 1e9 + 7;
sort(arr.begin(), arr.end());
unordered_map<int, int> idx;
int n = arr.size();
Expand Down Expand Up @@ -137,18 +135,15 @@ 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 {
f[i] = 1
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 c := a / b; a%b == 0 {
if k, ok := idx[c]; ok {
f[i] = (f[i] + f[j]*f[k]) % mod
}
Expand All @@ -163,6 +158,35 @@ func numFactoredBinaryTrees(arr []int) int {
}
```

### **TypeScript**

```ts
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;
}
```

### **...**

```
Expand Down
61 changes: 30 additions & 31 deletions solution/0800-0899/0823.Binary Trees With Factors/Solutioin.java
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Original file line number Diff line number Diff line change
@@ -1,32 +1,31 @@
class Solution {
private static final int MOD = (int) 1e9 + 7;

public int numFactoredBinaryTrees(int[] arr) {
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;
}
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;
}
}
61 changes: 30 additions & 31 deletions solution/0800-0899/0823.Binary Trees With Factors/Solution.cpp
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,32 +1,31 @@
class Solution {
public:
const int mod = 1e9 + 7;

int numFactoredBinaryTrees(vector<int>& arr) {
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;
}
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;
}
};
7 changes: 2 additions & 5 deletions solution/0800-0899/0823.Binary Trees With Factors/Solution.go
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -2,18 +2,15 @@ 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 {
f[i] = 1
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 c := a / b; a%b == 0 {
if k, ok := idx[c]; ok {
f[i] = (f[i] + f[j]*f[k]) % mod
}
Expand Down
24 changes: 24 additions & 0 deletions solution/0800-0899/0823.Binary Trees With Factors/Solution.ts
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Original file line number Diff line number Diff line change
@@ -0,0 +1,24 @@
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;
}