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feat: add solutions to lc problems: No.1771,1772 (#2124)
* No.1771.Maximize Palindrome Length From Subsequences * No.1772.Sort Features by Popularity
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‎solution/1700-1799/1771.Maximize Palindrome Length From Subsequences/README.md

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@@ -65,7 +65,7 @@
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最后我们返回答案即可。
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时间复杂度为 $O(n^2)$，其中 $n$ 是字符串 $s$ 的长度。
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时间复杂度 $O(n^2)$，空间复杂度 $O(n^2)$，其中 $n$ 为字符串 $s$ 的长度。
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<!-- tabs:start -->
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@@ -82,11 +82,11 @@ class Solution:
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for i in range(n):
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f[i][i] = 1
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ans = 0
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for i in range(n - 1, -1, -1):
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for i in range(n - 2, -1, -1):
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for j in range(i + 1, n):
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if s[i] == s[j]:
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f[i][j] = f[i + 1][j - 1] + 2
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if i < len(word1) and j >= len(word1):
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if i < len(word1) <= j:
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ans = max(ans, f[i][j])
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else:
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f[i][j] = max(f[i + 1][j], f[i][j - 1])
@@ -180,6 +180,62 @@ func longestPalindrome(word1 string, word2 string) (ans int) {
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}
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```
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### **TypeScript**
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```ts
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function longestPalindrome(word1: string, word2: string): number {
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const s = word1 + word2;
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const n = s.length;
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const f: number[][] = Array.from({ length: n }, () => Array.from({ length: n }, () => 0));
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for (let i = 0; i < n; ++i) {
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f[i][i] = 1;
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}
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let ans = 0;
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for (let i = n - 2; ~i; --i) {
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for (let j = i + 1; j < n; ++j) {
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if (s[i] === s[j]) {
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f[i][j] = f[i + 1][j - 1] + 2;
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if (i < word1.length && j >= word1.length) {
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ans = Math.max(ans, f[i][j]);
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}
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} else {
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f[i][j] = Math.max(f[i + 1][j], f[i][j - 1]);
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}
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}
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}
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return ans;
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}
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```
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### **Rust**
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```rust
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impl Solution {
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pub fn longest_palindrome(word1: String, word2: String) -> i32 {
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let s: Vec<char> = format!("{}{}", word1, word2).chars().collect();
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let n = s.len();
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let mut f = vec![vec![0; n]; n];
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for i in 0..n {
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f[i][i] = 1;
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}
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let mut ans = 0;
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for i in (0..n - 1).rev() {
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for j in i + 1..n {
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if s[i] == s[j] {
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f[i][j] = f[i + 1][j - 1] + 2;
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if i < word1.len() && j >= word1.len() {
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ans = ans.max(f[i][j]);
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}
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} else {
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f[i][j] = f[i + 1][j].max(f[i][j - 1]);
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}
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}
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}
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ans
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}
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}
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```
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### **...**
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```

‎solution/1700-1799/1771.Maximize Palindrome Length From Subsequences/README_EN.md

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## Solutions
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**Solution 1: Dynamic Programming**
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First, we concatenate strings `word1` and `word2` to get string $s$. Then we can transform the problem into finding the length of the longest palindromic subsequence in string $s$. However, when calculating the final answer, we need to ensure that at least one character in the palindrome string comes from `word1` and another character comes from `word2`.
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We define $f[i][j]$ as the length of the longest palindromic subsequence in the substring of string $s$ with index range $[i, j]$.
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If $s[i] = s[j]$, then $s[i]$ and $s[j]$ must be in the longest palindromic subsequence, at this time $f[i][j] = f[i + 1][j - 1] + 2$. At this point, we also need to judge whether $s[i]$ and $s[j]$ come from `word1` and `word2`. If so, we update the maximum value of the answer to $ans=\max(ans, f[i][j])$.
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If $s[i] \neq s[j]$, then $s[i]$ and $s[j]$ will definitely not appear in the longest palindromic subsequence at the same time, at this time $f[i][j] = max(f[i + 1][j], f[i][j - 1])$.
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Finally, we return the answer.
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The time complexity is $O(n^2)$, and the space complexity is $O(n^2)$. Here, $n$ is the length of string $s$.
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<!-- tabs:start -->
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### **Python3**
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for i in range(n):
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f[i][i] = 1
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ans = 0
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for i in range(n - 1, -1, -1):
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for i in range(n - 2, -1, -1):
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for j in range(i + 1, n):
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if s[i] == s[j]:
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f[i][j] = f[i + 1][j - 1] + 2
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if i < len(word1) and j >= len(word1):
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if i < len(word1) <= j:
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ans = max(ans, f[i][j])
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else:
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f[i][j] = max(f[i + 1][j], f[i][j - 1])
@@ -159,6 +173,62 @@ func longestPalindrome(word1 string, word2 string) (ans int) {
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}
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```
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### **TypeScript**
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```ts
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function longestPalindrome(word1: string, word2: string): number {
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const s = word1 + word2;
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const n = s.length;
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const f: number[][] = Array.from({ length: n }, () => Array.from({ length: n }, () => 0));
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for (let i = 0; i < n; ++i) {
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f[i][i] = 1;
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}
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let ans = 0;
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for (let i = n - 2; ~i; --i) {
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for (let j = i + 1; j < n; ++j) {
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if (s[i] === s[j]) {
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f[i][j] = f[i + 1][j - 1] + 2;
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if (i < word1.length && j >= word1.length) {
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ans = Math.max(ans, f[i][j]);
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}
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} else {
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f[i][j] = Math.max(f[i + 1][j], f[i][j - 1]);
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}
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}
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}
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return ans;
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}
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```
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### **Rust**
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```rust
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impl Solution {
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pub fn longest_palindrome(word1: String, word2: String) -> i32 {
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let s: Vec<char> = format!("{}{}", word1, word2).chars().collect();
209+
let n = s.len();
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let mut f = vec![vec![0; n]; n];
211+
for i in 0..n {
212+
f[i][i] = 1;
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}
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let mut ans = 0;
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for i in (0..n - 1).rev() {
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for j in i + 1..n {
217+
if s[i] == s[j] {
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f[i][j] = f[i + 1][j - 1] + 2;
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if i < word1.len() && j >= word1.len() {
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ans = ans.max(f[i][j]);
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}
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} else {
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f[i][j] = f[i + 1][j].max(f[i][j - 1]);
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}
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}
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}
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ans
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}
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}
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```
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### **...**
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```

‎solution/1700-1799/1771.Maximize Palindrome Length From Subsequences/Solution.py

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Original file line numberDiff line numberDiff line change
@@ -1,17 +1,17 @@
1-
class Solution:
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def longestPalindrome(self, word1: str, word2: str) -> int:
3-
s = word1 + word2
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n = len(s)
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f = [[0] * n for _ in range(n)]
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for i in range(n):
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f[i][i] = 1
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ans = 0
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for i in range(n - 1, -1, -1):
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for j in range(i + 1, n):
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if s[i] == s[j]:
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f[i][j] = f[i + 1][j - 1] + 2
13-
if i < len(word1) and j >= len(word1):
14-
ans = max(ans, f[i][j])
15-
else:
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f[i][j] = max(f[i + 1][j], f[i][j - 1])
17-
return ans
1+
class Solution:
2+
def longestPalindrome(self, word1: str, word2: str) -> int:
3+
s = word1 + word2
4+
n = len(s)
5+
f = [[0] * n for _ in range(n)]
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for i in range(n):
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f[i][i] = 1
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ans = 0
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for i in range(n - 2, -1, -1):
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for j in range(i + 1, n):
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if s[i] == s[j]:
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f[i][j] = f[i + 1][j - 1] + 2
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if i < len(word1) <= j:
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ans = max(ans, f[i][j])
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else:
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f[i][j] = max(f[i + 1][j], f[i][j - 1])
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return ans

‎solution/1700-1799/1771.Maximize Palindrome Length From Subsequences/Solution.rs

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,24 @@
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impl Solution {
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pub fn longest_palindrome(word1: String, word2: String) -> i32 {
3+
let s: Vec<char> = format!("{}{}", word1, word2).chars().collect();
4+
let n = s.len();
5+
let mut f = vec![vec![0; n]; n];
6+
for i in 0..n {
7+
f[i][i] = 1;
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}
9+
let mut ans = 0;
10+
for i in (0..n - 1).rev() {
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for j in i + 1..n {
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if s[i] == s[j] {
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f[i][j] = f[i + 1][j - 1] + 2;
14+
if i < word1.len() && j >= word1.len() {
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ans = ans.max(f[i][j]);
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}
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} else {
18+
f[i][j] = f[i + 1][j].max(f[i][j - 1]);
19+
}
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}
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}
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ans
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}
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}

‎solution/1700-1799/1771.Maximize Palindrome Length From Subsequences/Solution.ts

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,22 @@
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function longestPalindrome(word1: string, word2: string): number {
2+
const s = word1 + word2;
3+
const n = s.length;
4+
const f: number[][] = Array.from({ length: n }, () => Array.from({ length: n }, () => 0));
5+
for (let i = 0; i < n; ++i) {
6+
f[i][i] = 1;
7+
}
8+
let ans = 0;
9+
for (let i = n - 2; ~i; --i) {
10+
for (let j = i + 1; j < n; ++j) {
11+
if (s[i] === s[j]) {
12+
f[i][j] = f[i + 1][j - 1] + 2;
13+
if (i < word1.length && j >= word1.length) {
14+
ans = Math.max(ans, f[i][j]);
15+
}
16+
} else {
17+
f[i][j] = Math.max(f[i + 1][j], f[i][j - 1]);
18+
}
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}
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}
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return ans;
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}

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