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Mar 23, 2024
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feat: update solutions to lc problems: No.633~635 #2481

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52 changes: 30 additions & 22 deletions solution/0600-0699/0633.Sum of Square Numbers/README.md
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Expand Up @@ -37,7 +37,11 @@

## 解法

### 方法一
### 方法一：数学 + 双指针

我们可以使用双指针的方法来解决这个问题，定义两个指针 $a$ 和 $b$，分别指向 $0$ 和 $\sqrt{c}$。在每一步中，我们会计算 $s = a^2 + b^2$ 的值，然后比较 $s$ 与 $c$ 的大小。如果 $s = c$，我们就找到了两个整数 $a$ 和 $b$，使得 $a^2 + b^2 = c$。如果 $s < c$，我们将 $a$ 的值增加 $1$，如果 $s > c$，我们将 $b$ 的值减小 $1$。我们不断进行这个过程直到找到答案，或者 $a$ 的值大于 $b$ 的值，返回 `false`。

时间复杂度 $O(\sqrt{c})$，其中 $c$ 是给定的非负整数。空间复杂度 $O(1)$。

<!-- tabs:start -->

Expand Down Expand Up @@ -80,14 +84,17 @@ class Solution {
class Solution {
public:
bool judgeSquareSum(int c) {
long a = 0, b = (long) sqrt(c);
long long a = 0, b = sqrt(c);
while (a <= b) {
long s = a * a + b * b;
if (s == c) return true;
if (s < c)
long long s = a * a + b * b;
if (s == c) {
return true;
}
if (s < c) {
++a;
else
} else {
--b;
}
}
return false;
}
Expand All @@ -114,12 +121,13 @@ func judgeSquareSum(c int) bool {

```ts
function judgeSquareSum(c: number): boolean {
let a = 0,
b = Math.floor(Math.sqrt(c));
let [a, b] = [0, Math.floor(Math.sqrt(c))];
while (a <= b) {
let sum = a ** 2 + b ** 2;
if (sum == c) return true;
if (sum < c) {
const s = a * a + b * b;
if (s === c) {
return true;
}
if (s < c) {
++a;
} else {
--b;
Expand All @@ -131,22 +139,22 @@ function judgeSquareSum(c: number): boolean {

```rust
use std::cmp::Ordering;

impl Solution {
pub fn judge_square_sum(c: i32) -> bool {
let c = c as i64;
let mut left = 0;
let mut right = (c as f64).sqrt() as i64;
while left <= right {
let num = left * left + right * right;
match num.cmp(&c) {
let mut a: i64 = 0;
let mut b: i64 = (c as f64).sqrt() as i64;
while a <= b {
let s = a * a + b * b;
match s.cmp(&(c as i64)) {
Ordering::Equal => {
return true;
}
Ordering::Less => {
left += 1;
a += 1;
}
Ordering::Greater => {
right -= 1;
}
Ordering::Equal => {
return true;
b -= 1;
}
}
}
Expand Down
52 changes: 30 additions & 22 deletions solution/0600-0699/0633.Sum of Square Numbers/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -33,7 +33,11 @@

## Solutions

### Solution 1
### Solution 1: Mathematics + Two Pointers

We can use the two-pointer method to solve this problem. Define two pointers $a$ and $b$, pointing to $0$ and $\sqrt{c}$ respectively. In each step, we calculate the value of $s = a^2 + b^2$, and then compare the size of $s$ and $c$. If $s = c$, we have found two integers $a$ and $b$ such that $a^2 + b^2 = c$. If $s < c$, we increase the value of $a$ by $1$. If $s > c$, we decrease the value of $b$ by $1$. We continue this process until we find the answer, or the value of $a$ is greater than the value of $b$, and return `false`.

The time complexity is $O(\sqrt{c})$, where $c$ is the given non-negative integer. The space complexity is $O(1)$.

<!-- tabs:start -->

Expand Down Expand Up @@ -76,14 +80,17 @@ class Solution {
class Solution {
public:
bool judgeSquareSum(int c) {
long a = 0, b = (long) sqrt(c);
long long a = 0, b = sqrt(c);
while (a <= b) {
long s = a * a + b * b;
if (s == c) return true;
if (s < c)
long long s = a * a + b * b;
if (s == c) {
return true;
}
if (s < c) {
++a;
else
} else {
--b;
}
}
return false;
}
Expand All @@ -110,12 +117,13 @@ func judgeSquareSum(c int) bool {

```ts
function judgeSquareSum(c: number): boolean {
let a = 0,
b = Math.floor(Math.sqrt(c));
let [a, b] = [0, Math.floor(Math.sqrt(c))];
while (a <= b) {
let sum = a ** 2 + b ** 2;
if (sum == c) return true;
if (sum < c) {
const s = a * a + b * b;
if (s === c) {
return true;
}
if (s < c) {
++a;
} else {
--b;
Expand All @@ -127,22 +135,22 @@ function judgeSquareSum(c: number): boolean {

```rust
use std::cmp::Ordering;

impl Solution {
pub fn judge_square_sum(c: i32) -> bool {
let c = c as i64;
let mut left = 0;
let mut right = (c as f64).sqrt() as i64;
while left <= right {
let num = left * left + right * right;
match num.cmp(&c) {
let mut a: i64 = 0;
let mut b: i64 = (c as f64).sqrt() as i64;
while a <= b {
let s = a * a + b * b;
match s.cmp(&(c as i64)) {
Ordering::Equal => {
return true;
}
Ordering::Less => {
left += 1;
a += 1;
}
Ordering::Greater => {
right -= 1;
}
Ordering::Equal => {
return true;
b -= 1;
}
}
}
Expand Down
13 changes: 8 additions & 5 deletions solution/0600-0699/0633.Sum of Square Numbers/Solution.cpp
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Original file line number Diff line number Diff line change
@@ -1,14 +1,17 @@
class Solution {
public:
bool judgeSquareSum(int c) {
long a = 0, b = (long) sqrt(c);
long long a = 0, b = sqrt(c);
while (a <= b) {
long s = a * a + b * b;
if (s == c) return true;
if (s < c)
long long s = a * a + b * b;
if (s == c) {
return true;
}
if (s < c) {
++a;
else
} else {
--b;
}
}
return false;
}
Expand Down
22 changes: 11 additions & 11 deletions solution/0600-0699/0633.Sum of Square Numbers/Solution.rs
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Original file line number Diff line number Diff line change
@@ -1,20 +1,20 @@
use std::cmp::Ordering;

impl Solution {
pub fn judge_square_sum(c: i32) -> bool {
let c = c as i64;
let mut left = 0;
let mut right = (c as f64).sqrt() as i64;
while left <= right {
let num = left * left + right * right;
match num.cmp(&c) {
let mut a: i64 = 0;
let mut b: i64 = (c as f64).sqrt() as i64;
while a <= b {
let s = a * a + b * b;
match s.cmp(&(c as i64)) {
Ordering::Equal => {
return true;
}
Ordering::Less => {
left += 1;
a += 1;
}
Ordering::Greater => {
right -= 1;
}
Ordering::Equal => {
return true;
b -= 1;
}
}
}
Expand Down
11 changes: 6 additions & 5 deletions solution/0600-0699/0633.Sum of Square Numbers/Solution.ts
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Original file line number Diff line number Diff line change
@@ -1,10 +1,11 @@
function judgeSquareSum(c: number): boolean {
let a = 0,
b = Math.floor(Math.sqrt(c));
let [a, b] = [0, Math.floor(Math.sqrt(c))];
while (a <= b) {
let sum = a ** 2 + b ** 2;
if (sum == c) return true;
if (sum < c) {
const s = a * a + b * b;
if (s === c) {
return true;
}
if (s < c) {
++a;
} else {
--b;
Expand Down
8 changes: 4 additions & 4 deletions solution/0600-0699/0634.Find the Derangement of An Array/README.md
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Expand Up @@ -56,9 +56,7 @@ $$

最终答案即为 $f[n]$。注意答案的取模操作。

我们发现，状态转移方程中只与 $f[i - 1]$ 和 $f[i - 2]$ 有关，因此我们可以使用两个变量 $a$ 和 $b$ 来分别表示 $f[i - 1]$ 和 $f[i - 2]$，从而将空间复杂度降低到 $O(1)$。

时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为数组的长度。
时间复杂度 $O(n)$，其中 $n$ 为数组的长度。空间复杂度 $O(1)$。

<!-- tabs:start -->

Expand Down Expand Up @@ -116,7 +114,9 @@ func findDerangement(n int) int {

<!-- tabs:end -->

### 方法二
### 方法二：动态规划（空间优化）

我们发现，状态转移方程中只与 $f[i - 1]$ 和 $f[i - 2]$ 有关，因此我们可以使用两个变量 $a$ 和 $b$ 来分别表示 $f[i - 1]$ 和 $f[i - 2]$，从而将空间复杂度降低到 $O(1)$。

<!-- tabs:start -->

Expand Down
23 changes: 21 additions & 2 deletions solution/0600-0699/0634.Find the Derangement of An Array/README_EN.md
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Expand Up @@ -35,7 +35,24 @@

## Solutions

### Solution 1
### Solution 1: Dynamic Programming

We define $f[i]$ as the number of derangement of an array of length $i$. Initially, $f[0] = 1$, $f[1] = 0$. The answer is $f[n]$.

For an array of length $i$, we consider where to place the number $1$. Suppose it is placed in the $j$-th position, where there are $i-1$ choices. Then, the number $j$ has two choices:

- Placed in the first position, then the remaining $i - 2$ positions have $f[i - 2]$ derangements, so there are a total of $(i - 1) \times f[i - 2]$ derangements;
- Not placed in the first position, which is equivalent to the derangement of an array of length $i - 1$, so there are a total of $(i - 1) \times f[i - 1]$ derangements.

In summary, we have the following state transition equation:

$$
f[i] = (i - 1) \times (f[i - 1] + f[i - 2])
$$

The final answer is $f[n]$. Note the modulo operation in the answer.

The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

<!-- tabs:start -->

Expand Down Expand Up @@ -93,7 +110,9 @@ func findDerangement(n int) int {

<!-- tabs:end -->

### Solution 2
### Solution 2: Dynamic Programming (Space Optimization)

We notice that the state transition equation only relates to $f[i - 1]$ and $f[i - 2]$. Therefore, we can use two variables $a$ and $b$ to represent $f[i - 1]$ and $f[i - 2]$ respectively, thereby reducing the space complexity to $O(1)$.

<!-- tabs:start -->

Expand Down
6 changes: 5 additions & 1 deletion solution/0600-0699/0635.Design Log Storage System/README_EN.md
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Expand Up @@ -56,7 +56,11 @@ logSystem.retrieve(&quot;2016:01:01:01:01:01&quot;, &quot;2017:01:01:23:00:00&qu

## Solutions

### Solution 1
### Solution 1: String Comparison

Store the `id` and `timestamp` of the logs as tuples in an array. Then in the `retrieve()` method, truncate the corresponding parts of `start` and `end` based on `granularity`, and traverse the array, adding the `id` that meets the conditions to the result array.

In terms of time complexity, the time complexity of the `put()` method is $O(1)$, and the time complexity of the `retrieve()` method is $O(n)$, where $n$ is the length of the array.

<!-- tabs:start -->

Expand Down