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Mar 11, 2024
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feat: add solutions to lc problems: No.0921~0923 #2433

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bee6ea4
feat: add ts solution to lc problem: No.0921
yanglbme Mar 11, 2024
219cc09
feat: add solutions to lc problems: No.0922~0923
yanglbme Mar 11, 2024
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33 changes: 32 additions & 1 deletion solution/0900-0999/0921.Minimum Add to Make Parentheses Valid/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -124,6 +124,20 @@ func minAddToMakeValid(s string) int {
}
```

```ts
function minAddToMakeValid(s: string): number {
const stk: string[] = [];
for (const c of s) {
if (c === ')' && stk.length > 0 && stk.at(-1)! === '(') {
stk.pop();
} else {
stk.push(c);
}
}
return stk.length;
}
```

<!-- tabs:end -->

### 方法二：贪心 + 计数
Expand All @@ -139,7 +153,7 @@ func minAddToMakeValid(s string) int {

遍历结束后，将 `cnt` 的值加到 `ans` 中，即为答案。

时间复杂度为 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为字符串 $s$ 的长度。
时间复杂度为 $O(n)$，其中 $n$ 为字符串 $s$ 的长度。空间复杂度 $O(1)$。

<!-- tabs:start -->

Expand Down Expand Up @@ -213,6 +227,23 @@ func minAddToMakeValid(s string) int {
}
```

```ts
function minAddToMakeValid(s: string): number {
let [ans, cnt] = [0, 0];
for (const c of s) {
if (c === '(') {
++cnt;
} else if (cnt) {
--cnt;
} else {
++ans;
}
}
ans += cnt;
return ans;
}
```

<!-- tabs:end -->

<!-- end -->
59 changes: 57 additions & 2 deletions solution/0900-0999/0921.Minimum Add to Make Parentheses Valid/README_EN.md
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Expand Up @@ -47,7 +47,18 @@

## Solutions

### Solution 1
### Solution 1: Greedy + Stack

This problem is a classic parenthesis matching problem, which can be solved using "Greedy + Stack".

Iterate through each character $c$ in the string $s$:

- If $c$ is a left parenthesis, directly push $c$ into the stack;
- If $c$ is a right parenthesis, at this point if the stack is not empty, and the top element of the stack is a left parenthesis, then pop the top element of the stack, indicating a successful match; otherwise, push $c$ into the stack.

After the iteration ends, the number of remaining elements in the stack is the number of parentheses that need to be added.

The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the length of the string $s$.

<!-- tabs:start -->

Expand Down Expand Up @@ -109,9 +120,36 @@ func minAddToMakeValid(s string) int {
}
```

```ts
function minAddToMakeValid(s: string): number {
const stk: string[] = [];
for (const c of s) {
if (c === ')' && stk.length > 0 && stk.at(-1)! === '(') {
stk.pop();
} else {
stk.push(c);
}
}
return stk.length;
}
```

<!-- tabs:end -->

### Solution 2
### Solution 2: Greedy + Counting

Solution 1 uses a stack to implement parenthesis matching, but we can also directly implement it through counting.

Define a variable `cnt` to represent the current number of left parentheses to be matched, and a variable `ans` to record the answer. Initially, both variables are set to $0$.

Iterate through each character $c$ in the string $s$:

- If $c$ is a left parenthesis, increase the value of `cnt` by $1$;
- If $c$ is a right parenthesis, at this point if $cnt > 0$, it means that there are left parentheses that can be matched, so decrease the value of `cnt` by $1$; otherwise, it means that the current right parenthesis cannot be matched, so increase the value of `ans` by $1$.

After the iteration ends, add the value of `cnt` to `ans`, which is the answer.

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

<!-- tabs:start -->

Expand Down Expand Up @@ -185,6 +223,23 @@ func minAddToMakeValid(s string) int {
}
```

```ts
function minAddToMakeValid(s: string): number {
let [ans, cnt] = [0, 0];
for (const c of s) {
if (c === '(') {
++cnt;
} else if (cnt) {
--cnt;
} else {
++ans;
}
}
ans += cnt;
return ans;
}
```

<!-- tabs:end -->

<!-- end -->
11 changes: 11 additions & 0 deletions solution/0900-0999/0921.Minimum Add to Make Parentheses Valid/Solution.ts
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Original file line number Diff line number Diff line change
@@ -0,0 +1,11 @@
function minAddToMakeValid(s: string): number {
const stk: string[] = [];
for (const c of s) {
if (c === ')' && stk.length > 0 && stk.at(-1)! === '(') {
stk.pop();
} else {
stk.push(c);
}
}
return stk.length;
}
14 changes: 14 additions & 0 deletions solution/0900-0999/0921.Minimum Add to Make Parentheses Valid/Solution2.ts
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Original file line number Diff line number Diff line change
@@ -0,0 +1,14 @@
function minAddToMakeValid(s: string): number {
let [ans, cnt] = [0, 0];
for (const c of s) {
if (c === '(') {
++cnt;
} else if (cnt) {
--cnt;
} else {
++ans;
}
}
ans += cnt;
return ans;
}
42 changes: 31 additions & 11 deletions solution/0900-0999/0922.Sort Array By Parity II/README.md
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Expand Up @@ -48,7 +48,13 @@

## 解法

### 方法一
### 方法一：双指针

我们用两个指针 $i$ 和 $j$ 分别指向偶数下标和奇数下标。

当 $i$ 指向偶数下标时，如果 $nums[i]$ 是奇数，那么我们需要找到一个奇数下标 $j$，使得 $nums[j]$ 是偶数，然后交换 $nums[i]$ 和 $nums[j]$。继续遍历，直到 $i$ 指向数组末尾。

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

<!-- tabs:start -->

Expand All @@ -57,8 +63,8 @@ class Solution:
def sortArrayByParityII(self, nums: List[int]) -> List[int]:
n, j = len(nums), 1
for i in range(0, n, 2):
if (nums[i] & 1) == 1:
while (nums[j] & 1) == 1:
if nums[i] % 2:
while nums[j] % 2:
j += 2
nums[i], nums[j] = nums[j], nums[i]
return nums
Expand All @@ -68,8 +74,8 @@ class Solution:
class Solution {
public int[] sortArrayByParityII(int[] nums) {
for (int i = 0, j = 1; i < nums.length; i += 2) {
if ((nums[i] & 1) == 1) {
while ((nums[j] & 1) == 1) {
if (nums[i] % 2 == 1) {
while (nums[j] % 2 == 1) {
j += 2;
}
int t = nums[i];
Expand All @@ -87,8 +93,8 @@ class Solution {
public:
vector<int> sortArrayByParityII(vector<int>& nums) {
for (int i = 0, j = 1; i < nums.size(); i += 2) {
if ((nums[i] & 1) == 1) {
while ((nums[j] & 1) == 1) {
if (nums[i] % 2) {
while (nums[j] % 2) {
j += 2;
}
swap(nums[i], nums[j]);
Expand All @@ -102,8 +108,8 @@ public:
```go
func sortArrayByParityII(nums []int) []int {
for i, j := 0, 1; i < len(nums); i += 2 {
if (nums[i] & 1) == 1 {
for (nums[j] & 1) == 1 {
if nums[i]%2 == 1 {
for nums[j]%2 == 1 {
j += 2
}
nums[i], nums[j] = nums[j], nums[i]
Expand All @@ -113,15 +119,29 @@ func sortArrayByParityII(nums []int) []int {
}
```

```ts
function sortArrayByParityII(nums: number[]): number[] {
for (let i = 0, j = 1; i < nums.length; i += 2) {
if (nums[i] % 2) {
while (nums[j] % 2) {
j += 2;
}
[nums[i], nums[j]] = [nums[j], nums[i]];
}
}
return nums;
}
```

```js
/**
* @param {number[]} nums
* @return {number[]}
*/
var sortArrayByParityII = function (nums) {
for (let i = 0, j = 1; i < nums.length; i += 2) {
if ((nums[i] & 1) == 1) {
while ((nums[j] & 1) == 1) {
if (nums[i] % 2) {
while (nums[j] % 2) {
j += 2;
}
[nums[i], nums[j]] = [nums[j], nums[i]];
Expand Down
42 changes: 31 additions & 11 deletions solution/0900-0999/0922.Sort Array By Parity II/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -43,7 +43,13 @@

## Solutions

### Solution 1
### Solution 1: Two Pointers

We use two pointers $i$ and $j$ to point to even and odd indices respectively.

When $i$ points to an even index, if $nums[i]$ is odd, then we need to find an odd index $j$ such that $nums[j]$ is even, and then swap $nums[i]$ and $nums[j]$. Continue to iterate until $i$ points to the end of the array.

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

<!-- tabs:start -->

Expand All @@ -52,8 +58,8 @@ class Solution:
def sortArrayByParityII(self, nums: List[int]) -> List[int]:
n, j = len(nums), 1
for i in range(0, n, 2):
if (nums[i] & 1) == 1:
while (nums[j] & 1) == 1:
if nums[i] % 2:
while nums[j] % 2:
j += 2
nums[i], nums[j] = nums[j], nums[i]
return nums
Expand All @@ -63,8 +69,8 @@ class Solution:
class Solution {
public int[] sortArrayByParityII(int[] nums) {
for (int i = 0, j = 1; i < nums.length; i += 2) {
if ((nums[i] & 1) == 1) {
while ((nums[j] & 1) == 1) {
if (nums[i] % 2 == 1) {
while (nums[j] % 2 == 1) {
j += 2;
}
int t = nums[i];
Expand All @@ -82,8 +88,8 @@ class Solution {
public:
vector<int> sortArrayByParityII(vector<int>& nums) {
for (int i = 0, j = 1; i < nums.size(); i += 2) {
if ((nums[i] & 1) == 1) {
while ((nums[j] & 1) == 1) {
if (nums[i] % 2) {
while (nums[j] % 2) {
j += 2;
}
swap(nums[i], nums[j]);
Expand All @@ -97,8 +103,8 @@ public:
```go
func sortArrayByParityII(nums []int) []int {
for i, j := 0, 1; i < len(nums); i += 2 {
if (nums[i] & 1) == 1 {
for (nums[j] & 1) == 1 {
if nums[i]%2 == 1 {
for nums[j]%2 == 1 {
j += 2
}
nums[i], nums[j] = nums[j], nums[i]
Expand All @@ -108,15 +114,29 @@ func sortArrayByParityII(nums []int) []int {
}
```

```ts
function sortArrayByParityII(nums: number[]): number[] {
for (let i = 0, j = 1; i < nums.length; i += 2) {
if (nums[i] % 2) {
while (nums[j] % 2) {
j += 2;
}
[nums[i], nums[j]] = [nums[j], nums[i]];
}
}
return nums;
}
```

```js
/**
* @param {number[]} nums
* @return {number[]}
*/
var sortArrayByParityII = function (nums) {
for (let i = 0, j = 1; i < nums.length; i += 2) {
if ((nums[i] & 1) == 1) {
while ((nums[j] & 1) == 1) {
if (nums[i] % 2) {
while (nums[j] % 2) {
j += 2;
}
[nums[i], nums[j]] = [nums[j], nums[i]];
Expand Down
4 changes: 2 additions & 2 deletions solution/0900-0999/0922.Sort Array By Parity II/Solution.cpp
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Original file line number Diff line number Diff line change
Expand Up @@ -2,8 +2,8 @@ class Solution {
public:
vector<int> sortArrayByParityII(vector<int>& nums) {
for (int i = 0, j = 1; i < nums.size(); i += 2) {
if ((nums[i] & 1) == 1) {
while ((nums[j] & 1) == 1) {
if (nums[i] % 2) {
while (nums[j] % 2) {
j += 2;
}
swap(nums[i], nums[j]);
Expand Down
4 changes: 2 additions & 2 deletions solution/0900-0999/0922.Sort Array By Parity II/Solution.go
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Original file line number Diff line number Diff line change
@@ -1,7 +1,7 @@
func sortArrayByParityII(nums []int) []int {
for i, j := 0, 1; i < len(nums); i += 2 {
if (nums[i] & 1) == 1 {
for (nums[j] & 1) == 1 {
if nums[i]%2 == 1 {
for nums[j]%2 == 1 {
j += 2
}
nums[i], nums[j] = nums[j], nums[i]
Expand Down
4 changes: 2 additions & 2 deletions solution/0900-0999/0922.Sort Array By Parity II/Solution.java
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Original file line number Diff line number Diff line change
@@ -1,8 +1,8 @@
class Solution {
public int[] sortArrayByParityII(int[] nums) {
for (int i = 0, j = 1; i < nums.length; i += 2) {
if ((nums[i] & 1) == 1) {
while ((nums[j] & 1) == 1) {
if (nums[i] % 2 == 1) {
while (nums[j] % 2 == 1) {
j += 2;
}
int t = nums[i];
Expand Down
4 changes: 2 additions & 2 deletions solution/0900-0999/0922.Sort Array By Parity II/Solution.js
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Original file line number Diff line number Diff line change
Expand Up @@ -4,8 +4,8 @@
*/
var sortArrayByParityII = function (nums) {
for (let i = 0, j = 1; i < nums.length; i += 2) {
if ((nums[i] & 1) == 1) {
while ((nums[j] & 1) == 1) {
if (nums[i] % 2) {
while (nums[j] % 2) {
j += 2;
}
[nums[i], nums[j]] = [nums[j], nums[i]];
Expand Down
4 changes: 2 additions & 2 deletions solution/0900-0999/0922.Sort Array By Parity II/Solution.py
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Original file line number Diff line number Diff line change
Expand Up @@ -2,8 +2,8 @@ class Solution:
def sortArrayByParityII(self, nums: List[int]) -> List[int]:
n, j = len(nums), 1
for i in range(0, n, 2):
if (nums[i] & 1) == 1:
while (nums[j] & 1) == 1:
if nums[i] % 2:
while nums[j] % 2:
j += 2
nums[i], nums[j] = nums[j], nums[i]
return nums
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