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Mar 24, 2025
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feat: add solutions to lc problem: No.0960 #4294

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136 changes: 87 additions & 49 deletions solution/0900-0999/0960.Delete Columns to Make Sorted III/README.md
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -75,7 +75,15 @@ tags:

<!-- solution:start -->

### 方法一
### 方法一：动态规划

我们定义 $f[i]$ 表示以第 $i$ 列结尾的最长不下降子序列的长度，初始时 $f[i] = 1$，答案即为 $n - \max(f)$。

考虑计算 $f[i]$，我们可以枚举 $j < i$，如果对于所有的字符串 $s$，有 $s[j] \le s[i]$，那么 $f[i] = \max(f[i], f[j] + 1)$。

最后，我们返回 $n - \max(f)$。

时间复杂度 $O(n^2 \times m)$，空间复杂度 $O(n)$。其中 $n$ 和 $m$ 分别是数组 $\textit{strs}$ 每个字符串的长度和数组的长度。

<!-- tabs:start -->

Expand All @@ -85,12 +93,12 @@ tags:
class Solution:
def minDeletionSize(self, strs: List[str]) -> int:
n = len(strs[0])
dp = [1] * n
for i in range(1, n):
f = [1] * n
for i in range(n):
for j in range(i):
if all(s[j] <= s[i] for s in strs):
dp[i] = max(dp[i], dp[j] + 1)
return n - max(dp)
f[i] = max(f[i], f[j] + 1)
return n - max(f)
```

#### Java
Expand All @@ -99,27 +107,23 @@ class Solution:
class Solution {
public int minDeletionSize(String[] strs) {
int n = strs[0].length();
int[] dp = new int[n];
Arrays.fill(dp, 1);
int mx = 1;
int[] f = new int[n];
Arrays.fill(f, 1);
for (int i = 1; i < n; ++i) {
for (int j = 0; j < i; ++j) {
if (check(i, j, strs)) {
dp[i] = Math.max(dp[i], dp[j] + 1);
boolean ok = true;
for (String s : strs) {
if (s.charAt(j) > s.charAt(i)) {
ok = false;
break;
}
}
if (ok) {
f[i] = Math.max(f[i], f[j] + 1);
}
}
mx = Math.max(mx, dp[i]);
}
return n - mx;
}

private boolean check(int i, int j, String[] strs) {
for (String s : strs) {
if (s.charAt(i) < s.charAt(j)) {
return false;
}
}
return true;
return n - Arrays.stream(f).max().getAsInt();
}
}
```
Expand All @@ -131,24 +135,15 @@ class Solution {
public:
int minDeletionSize(vector<string>& strs) {
int n = strs[0].size();
vector<int> dp(n, 1);
int mx = 1;
vector<int> f(n, 1);
for (int i = 1; i < n; ++i) {
for (int j = 0; j < i; ++j) {
if (check(i, j, strs)) {
dp[i] = max(dp[i], dp[j] + 1);
if (ranges::all_of(strs, [&](const string& s) { return s[j] <= s[i]; })) {
f[i] = max(f[i], f[j] + 1);
}
}
mx = max(mx, dp[i]);
}
return n - mx;
}

bool check(int i, int j, vector<string>& strs) {
for (string& s : strs)
if (s[i] < s[j])
return false;
return true;
return n - ranges::max(f);
}
};
```
Expand All @@ -158,27 +153,70 @@ public:
```go
func minDeletionSize(strs []string) int {
n := len(strs[0])
dp := make([]int, n)
mx := 1
dp[0] = 1
check := func(i, j int) bool {
for _, s := range strs {
if s[i] < s[j] {
return false
}
}
return true
f := make([]int, n)
for i := range f {
f[i] = 1
}
for i := 1; i < n; i++ {
dp[i] = 1
for j := 0; j < i; j++ {
if check(i, j) {
dp[i] = max(dp[i], dp[j]+1)
ok := true
for _, s := range strs {
if s[j] > s[i] {
ok = false
break
}
}
if ok {
f[i] = max(f[i], f[j]+1)
}
}
mx = max(mx, dp[i])
}
return n - mx
return n - slices.Max(f)
}
```

#### TypeScript

```ts
function minDeletionSize(strs: string[]): number {
const n = strs[0].length;
const f: number[] = Array(n).fill(1);
for (let i = 1; i < n; i++) {
for (let j = 0; j < i; j++) {
let ok = true;
for (const s of strs) {
if (s[j] > s[i]) {
ok = false;
break;
}
}
if (ok) {
f[i] = Math.max(f[i], f[j] + 1);
}
}
}
return n - Math.max(...f);
}
```

#### Rust

```rust
impl Solution {
pub fn min_deletion_size(strs: Vec<String>) -> i32 {
let n = strs[0].len();
let mut f = vec![1; n];

for i in 1..n {
for j in 0..i {
if strs.iter().all(|s| s.as_bytes()[j] <= s.as_bytes()[i]) {
f[i] = f[i].max(f[j] + 1);
}
}
}

(n - *f.iter().max().unwrap()) as i32
}
}
```

Expand Down
137 changes: 88 additions & 49 deletions solution/0900-0999/0960.Delete Columns to Make Sorted III/README_EN.md
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -72,7 +72,16 @@ Note that strs[0] &gt; strs[1] - the array strs is not necessarily in lexicograp

<!-- solution:start -->

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

We define $f[i]$ as the length of the longest non-decreasing subsequence ending at column $i$. Initially, $f[i] = 1$, and the final answer is $n - \max(f)$.

To compute $f[i]$, we iterate over all $j < i$. If for all strings $s$, we have $s[j] \leq s[i]$, then we update $f[i]$ as follows:
$$ f[i] = \max(f[i], f[j] + 1) $$

Finally, we return $n - \max(f)$.

The time complexity is $O(n^2 \times m)$, and the space complexity is $O(n)$, where $n$ is the length of each string in the array $\textit{strs}$, and $m$ is the number of strings in the array.

<!-- tabs:start -->

Expand All @@ -82,12 +91,12 @@ Note that strs[0] &gt; strs[1] - the array strs is not necessarily in lexicograp
class Solution:
def minDeletionSize(self, strs: List[str]) -> int:
n = len(strs[0])
dp = [1] * n
for i in range(1, n):
f = [1] * n
for i in range(n):
for j in range(i):
if all(s[j] <= s[i] for s in strs):
dp[i] = max(dp[i], dp[j] + 1)
return n - max(dp)
f[i] = max(f[i], f[j] + 1)
return n - max(f)
```

#### Java
Expand All @@ -96,27 +105,23 @@ class Solution:
class Solution {
public int minDeletionSize(String[] strs) {
int n = strs[0].length();
int[] dp = new int[n];
Arrays.fill(dp, 1);
int mx = 1;
int[] f = new int[n];
Arrays.fill(f, 1);
for (int i = 1; i < n; ++i) {
for (int j = 0; j < i; ++j) {
if (check(i, j, strs)) {
dp[i] = Math.max(dp[i], dp[j] + 1);
boolean ok = true;
for (String s : strs) {
if (s.charAt(j) > s.charAt(i)) {
ok = false;
break;
}
}
if (ok) {
f[i] = Math.max(f[i], f[j] + 1);
}
}
mx = Math.max(mx, dp[i]);
}
return n - mx;
}

private boolean check(int i, int j, String[] strs) {
for (String s : strs) {
if (s.charAt(i) < s.charAt(j)) {
return false;
}
}
return true;
return n - Arrays.stream(f).max().getAsInt();
}
}
```
Expand All @@ -128,24 +133,15 @@ class Solution {
public:
int minDeletionSize(vector<string>& strs) {
int n = strs[0].size();
vector<int> dp(n, 1);
int mx = 1;
vector<int> f(n, 1);
for (int i = 1; i < n; ++i) {
for (int j = 0; j < i; ++j) {
if (check(i, j, strs)) {
dp[i] = max(dp[i], dp[j] + 1);
if (ranges::all_of(strs, [&](const string& s) { return s[j] <= s[i]; })) {
f[i] = max(f[i], f[j] + 1);
}
}
mx = max(mx, dp[i]);
}
return n - mx;
}

bool check(int i, int j, vector<string>& strs) {
for (string& s : strs)
if (s[i] < s[j])
return false;
return true;
return n - ranges::max(f);
}
};
```
Expand All @@ -155,27 +151,70 @@ public:
```go
func minDeletionSize(strs []string) int {
n := len(strs[0])
dp := make([]int, n)
mx := 1
dp[0] = 1
check := func(i, j int) bool {
for _, s := range strs {
if s[i] < s[j] {
return false
}
}
return true
f := make([]int, n)
for i := range f {
f[i] = 1
}
for i := 1; i < n; i++ {
dp[i] = 1
for j := 0; j < i; j++ {
if check(i, j) {
dp[i] = max(dp[i], dp[j]+1)
ok := true
for _, s := range strs {
if s[j] > s[i] {
ok = false
break
}
}
if ok {
f[i] = max(f[i], f[j]+1)
}
}
mx = max(mx, dp[i])
}
return n - mx
return n - slices.Max(f)
}
```

#### TypeScript

```ts
function minDeletionSize(strs: string[]): number {
const n = strs[0].length;
const f: number[] = Array(n).fill(1);
for (let i = 1; i < n; i++) {
for (let j = 0; j < i; j++) {
let ok = true;
for (const s of strs) {
if (s[j] > s[i]) {
ok = false;
break;
}
}
if (ok) {
f[i] = Math.max(f[i], f[j] + 1);
}
}
}
return n - Math.max(...f);
}
```

#### Rust

```rust
impl Solution {
pub fn min_deletion_size(strs: Vec<String>) -> i32 {
let n = strs[0].len();
let mut f = vec![1; n];

for i in 1..n {
for j in 0..i {
if strs.iter().all(|s| s.as_bytes()[j] <= s.as_bytes()[i]) {
f[i] = f[i].max(f[j] + 1);
}
}
}

(n - *f.iter().max().unwrap()) as i32
}
}
```

Expand Down
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