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Commit 85429f1

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feat: add solutions to lc problem: No.0960 (#4294)
No.0960.Delete Columns to Make Sorted III
1 parent c4712ae commit 85429f1

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‎solution/0900-0999/0960.Delete Columns to Make Sorted III/README.md‎

Lines changed: 87 additions & 49 deletions
Original file line numberDiff line numberDiff line change
@@ -75,7 +75,15 @@ tags:
7575

7676
<!-- solution:start -->
7777

78-
### 方法一
78+
### 方法一:动态规划
79+
80+
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,ドル答案即为 $n - \max(f)$。
81+
82+
考虑计算 $f[i],ドル我们可以枚举 $j < i,ドル如果对于所有的字符串 $s,ドル有 $s[j] \le s[i],ドル那么 $f[i] = \max(f[i], f[j] + 1)$。
83+
84+
最后,我们返回 $n - \max(f)$。
85+
86+
时间复杂度 $O(n^2 \times m),ドル空间复杂度 $O(n)$。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。
7987

8088
<!-- tabs:start -->
8189

@@ -85,12 +93,12 @@ tags:
8593
class Solution:
8694
def minDeletionSize(self, strs: List[str]) -> int:
8795
n = len(strs[0])
88-
dp = [1] * n
89-
for i in range(1, n):
96+
f = [1] * n
97+
for i in range(n):
9098
for j in range(i):
9199
if all(s[j] <= s[i] for s in strs):
92-
dp[i] = max(dp[i], dp[j] + 1)
93-
return n - max(dp)
100+
f[i] = max(f[i], f[j] + 1)
101+
return n - max(f)
94102
```
95103

96104
#### Java
@@ -99,27 +107,23 @@ class Solution:
99107
class Solution {
100108
public int minDeletionSize(String[] strs) {
101109
int n = strs[0].length();
102-
int[] dp = new int[n];
103-
Arrays.fill(dp, 1);
104-
int mx = 1;
110+
int[] f = new int[n];
111+
Arrays.fill(f, 1);
105112
for (int i = 1; i < n; ++i) {
106113
for (int j = 0; j < i; ++j) {
107-
if (check(i, j, strs)) {
108-
dp[i] = Math.max(dp[i], dp[j] + 1);
114+
boolean ok = true;
115+
for (String s : strs) {
116+
if (s.charAt(j) > s.charAt(i)) {
117+
ok = false;
118+
break;
119+
}
120+
}
121+
if (ok) {
122+
f[i] = Math.max(f[i], f[j] + 1);
109123
}
110-
}
111-
mx = Math.max(mx, dp[i]);
112-
}
113-
return n - mx;
114-
}
115-
116-
private boolean check(int i, int j, String[] strs) {
117-
for (String s : strs) {
118-
if (s.charAt(i) < s.charAt(j)) {
119-
return false;
120124
}
121125
}
122-
return true;
126+
return n -Arrays.stream(f).max().getAsInt();
123127
}
124128
}
125129
```
@@ -131,24 +135,15 @@ class Solution {
131135
public:
132136
int minDeletionSize(vector<string>& strs) {
133137
int n = strs[0].size();
134-
vector<int> dp(n, 1);
135-
int mx = 1;
138+
vector<int> f(n, 1);
136139
for (int i = 1; i < n; ++i) {
137140
for (int j = 0; j < i; ++j) {
138-
if (check(i, j, strs)) {
139-
dp[i] = max(dp[i], dp[j] + 1);
141+
if (ranges::all_of(strs, [&](const string& s) { return s[j] <= s[i]; })) {
142+
f[i] = max(f[i], f[j] + 1);
140143
}
141144
}
142-
mx = max(mx, dp[i]);
143145
}
144-
return n - mx;
145-
}
146-
147-
bool check(int i, int j, vector<string>& strs) {
148-
for (string& s : strs)
149-
if (s[i] < s[j])
150-
return false;
151-
return true;
146+
return n - ranges::max(f);
152147
}
153148
};
154149
```
@@ -158,27 +153,70 @@ public:
158153
```go
159154
func minDeletionSize(strs []string) int {
160155
n := len(strs[0])
161-
dp := make([]int, n)
162-
mx := 1
163-
dp[0] = 1
164-
check := func(i, j int) bool {
165-
for _, s := range strs {
166-
if s[i] < s[j] {
167-
return false
168-
}
169-
}
170-
return true
156+
f := make([]int, n)
157+
for i := range f {
158+
f[i] = 1
171159
}
172160
for i := 1; i < n; i++ {
173-
dp[i] = 1
174161
for j := 0; j < i; j++ {
175-
if check(i, j) {
176-
dp[i] = max(dp[i], dp[j]+1)
162+
ok := true
163+
for _, s := range strs {
164+
if s[j] > s[i] {
165+
ok = false
166+
break
167+
}
168+
}
169+
if ok {
170+
f[i] = max(f[i], f[j]+1)
177171
}
178172
}
179-
mx = max(mx, dp[i])
180173
}
181-
return n - mx
174+
return n - slices.Max(f)
175+
}
176+
```
177+
178+
#### TypeScript
179+
180+
```ts
181+
function minDeletionSize(strs: string[]): number {
182+
const n = strs[0].length;
183+
const f: number[] = Array(n).fill(1);
184+
for (let i = 1; i < n; i++) {
185+
for (let j = 0; j < i; j++) {
186+
let ok = true;
187+
for (const s of strs) {
188+
if (s[j] > s[i]) {
189+
ok = false;
190+
break;
191+
}
192+
}
193+
if (ok) {
194+
f[i] = Math.max(f[i], f[j] + 1);
195+
}
196+
}
197+
}
198+
return n - Math.max(...f);
199+
}
200+
```
201+
202+
#### Rust
203+
204+
```rust
205+
impl Solution {
206+
pub fn min_deletion_size(strs: Vec<String>) -> i32 {
207+
let n = strs[0].len();
208+
let mut f = vec![1; n];
209+
210+
for i in 1..n {
211+
for j in 0..i {
212+
if strs.iter().all(|s| s.as_bytes()[j] <= s.as_bytes()[i]) {
213+
f[i] = f[i].max(f[j] + 1);
214+
}
215+
}
216+
}
217+
218+
(n - *f.iter().max().unwrap()) as i32
219+
}
182220
}
183221
```
184222

‎solution/0900-0999/0960.Delete Columns to Make Sorted III/README_EN.md‎

Lines changed: 88 additions & 49 deletions
Original file line numberDiff line numberDiff line change
@@ -72,7 +72,16 @@ Note that strs[0] &gt; strs[1] - the array strs is not necessarily in lexicograp
7272

7373
<!-- solution:start -->
7474

75-
### Solution 1
75+
### Solution 1: Dynamic Programming
76+
77+
We define $f[i]$ as the length of the longest non-decreasing subsequence ending at column $i$. Initially, $f[i] = 1,ドル and the final answer is $n - \max(f)$.
78+
79+
To compute $f[i],ドル we iterate over all $j < i$. If for all strings $s,ドル we have $s[j] \leq s[i],ドル then we update $f[i]$ as follows:
80+
$$ f[i] = \max(f[i], f[j] + 1) $$
81+
82+
Finally, we return $n - \max(f)$.
83+
84+
The time complexity is $O(n^2 \times m),ドル and the space complexity is $O(n),ドル where $n$ is the length of each string in the array $\textit{strs},ドル and $m$ is the number of strings in the array.
7685

7786
<!-- tabs:start -->
7887

@@ -82,12 +91,12 @@ Note that strs[0] &gt; strs[1] - the array strs is not necessarily in lexicograp
8291
class Solution:
8392
def minDeletionSize(self, strs: List[str]) -> int:
8493
n = len(strs[0])
85-
dp = [1] * n
86-
for i in range(1, n):
94+
f = [1] * n
95+
for i in range(n):
8796
for j in range(i):
8897
if all(s[j] <= s[i] for s in strs):
89-
dp[i] = max(dp[i], dp[j] + 1)
90-
return n - max(dp)
98+
f[i] = max(f[i], f[j] + 1)
99+
return n - max(f)
91100
```
92101

93102
#### Java
@@ -96,27 +105,23 @@ class Solution:
96105
class Solution {
97106
public int minDeletionSize(String[] strs) {
98107
int n = strs[0].length();
99-
int[] dp = new int[n];
100-
Arrays.fill(dp, 1);
101-
int mx = 1;
108+
int[] f = new int[n];
109+
Arrays.fill(f, 1);
102110
for (int i = 1; i < n; ++i) {
103111
for (int j = 0; j < i; ++j) {
104-
if (check(i, j, strs)) {
105-
dp[i] = Math.max(dp[i], dp[j] + 1);
112+
boolean ok = true;
113+
for (String s : strs) {
114+
if (s.charAt(j) > s.charAt(i)) {
115+
ok = false;
116+
break;
117+
}
118+
}
119+
if (ok) {
120+
f[i] = Math.max(f[i], f[j] + 1);
106121
}
107-
}
108-
mx = Math.max(mx, dp[i]);
109-
}
110-
return n - mx;
111-
}
112-
113-
private boolean check(int i, int j, String[] strs) {
114-
for (String s : strs) {
115-
if (s.charAt(i) < s.charAt(j)) {
116-
return false;
117122
}
118123
}
119-
return true;
124+
return n -Arrays.stream(f).max().getAsInt();
120125
}
121126
}
122127
```
@@ -128,24 +133,15 @@ class Solution {
128133
public:
129134
int minDeletionSize(vector<string>& strs) {
130135
int n = strs[0].size();
131-
vector<int> dp(n, 1);
132-
int mx = 1;
136+
vector<int> f(n, 1);
133137
for (int i = 1; i < n; ++i) {
134138
for (int j = 0; j < i; ++j) {
135-
if (check(i, j, strs)) {
136-
dp[i] = max(dp[i], dp[j] + 1);
139+
if (ranges::all_of(strs, [&](const string& s) { return s[j] <= s[i]; })) {
140+
f[i] = max(f[i], f[j] + 1);
137141
}
138142
}
139-
mx = max(mx, dp[i]);
140143
}
141-
return n - mx;
142-
}
143-
144-
bool check(int i, int j, vector<string>& strs) {
145-
for (string& s : strs)
146-
if (s[i] < s[j])
147-
return false;
148-
return true;
144+
return n - ranges::max(f);
149145
}
150146
};
151147
```
@@ -155,27 +151,70 @@ public:
155151
```go
156152
func minDeletionSize(strs []string) int {
157153
n := len(strs[0])
158-
dp := make([]int, n)
159-
mx := 1
160-
dp[0] = 1
161-
check := func(i, j int) bool {
162-
for _, s := range strs {
163-
if s[i] < s[j] {
164-
return false
165-
}
166-
}
167-
return true
154+
f := make([]int, n)
155+
for i := range f {
156+
f[i] = 1
168157
}
169158
for i := 1; i < n; i++ {
170-
dp[i] = 1
171159
for j := 0; j < i; j++ {
172-
if check(i, j) {
173-
dp[i] = max(dp[i], dp[j]+1)
160+
ok := true
161+
for _, s := range strs {
162+
if s[j] > s[i] {
163+
ok = false
164+
break
165+
}
166+
}
167+
if ok {
168+
f[i] = max(f[i], f[j]+1)
174169
}
175170
}
176-
mx = max(mx, dp[i])
177171
}
178-
return n - mx
172+
return n - slices.Max(f)
173+
}
174+
```
175+
176+
#### TypeScript
177+
178+
```ts
179+
function minDeletionSize(strs: string[]): number {
180+
const n = strs[0].length;
181+
const f: number[] = Array(n).fill(1);
182+
for (let i = 1; i < n; i++) {
183+
for (let j = 0; j < i; j++) {
184+
let ok = true;
185+
for (const s of strs) {
186+
if (s[j] > s[i]) {
187+
ok = false;
188+
break;
189+
}
190+
}
191+
if (ok) {
192+
f[i] = Math.max(f[i], f[j] + 1);
193+
}
194+
}
195+
}
196+
return n - Math.max(...f);
197+
}
198+
```
199+
200+
#### Rust
201+
202+
```rust
203+
impl Solution {
204+
pub fn min_deletion_size(strs: Vec<String>) -> i32 {
205+
let n = strs[0].len();
206+
let mut f = vec![1; n];
207+
208+
for i in 1..n {
209+
for j in 0..i {
210+
if strs.iter().all(|s| s.as_bytes()[j] <= s.as_bytes()[i]) {
211+
f[i] = f[i].max(f[j] + 1);
212+
}
213+
}
214+
}
215+
216+
(n - *f.iter().max().unwrap()) as i32
217+
}
179218
}
180219
```
181220

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