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Merged
yanglbme merged 2 commits into from
Mar 28, 2026
Merged

feat: add solutions for lc No.2574 #5120

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0a02a4b
feat: add solutions for lc No.2574
yanglbme Mar 27, 2026
dcc1938
fix: code style
yanglbme Mar 27, 2026
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52 changes: 51 additions & 1 deletion solution/2500-2599/2573.Find the String with LCP/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -48,7 +48,7 @@ tags:
<pre>
<strong>输入：</strong>lcp = [[4,3,2,1],[3,3,2,1],[2,2,2,1],[1,1,1,1]]
<strong>输出：</strong>"aaaa"
<strong>解释：</strong>lcp 对应只有一个不同字母的任意 4 字母字符串，字典序最小的是 "aaaa" 。
<strong>解释：</strong>lcp 对应只有一个不同字母的任意 4 字母字符串，字典序最小的是 "aaaa" 。
</pre>

<p><strong>示例 3：</strong></p>
Expand Down Expand Up @@ -298,6 +298,56 @@ function findTheString(lcp: number[][]): string {
}
```

#### Rust

```rust
impl Solution {
pub fn find_the_string(lcp: Vec<Vec<i32>>) -> String {
let n = lcp.len();
let mut s = vec!['\0'; n];
let mut i = 0;

for c in b'a'..=b'z' {
while i < n && s[i] != '\0' {
i += 1;
}
if i == n {
break;
}
for j in i..n {
if lcp[i][j] > 0 {
s[j] = c as char;
}
}
}

for i in 0..n {
if s[i] == '\0' {
return "".to_string();
}
}

for i in (0..n).rev() {
for j in (0..n).rev() {
if s[i] == s[j] {
if i == n - 1 || j == n - 1 {
if lcp[i][j] != 1 {
return "".to_string();
}
} else if lcp[i][j] != lcp[i + 1][j + 1] + 1 {
return "".to_string();
}
} else if lcp[i][j] > 0 {
return "".to_string();
}
}
}

s.into_iter().collect()
}
}
```

<!-- tabs:end -->

<!-- solution:end -->
Expand Down
52 changes: 51 additions & 1 deletion solution/2500-2599/2573.Find the String with LCP/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -47,7 +47,7 @@ tags:
<pre>
<strong>Input:</strong> lcp = [[4,3,2,1],[3,3,2,1],[2,2,2,1],[1,1,1,1]]
<strong>Output:</strong> &quot;aaaa&quot;
<strong>Explanation:</strong> lcp corresponds to any 4 letter string with a single distinct letter. The lexicographically smallest of them is &quot;aaaa&quot;.
<strong>Explanation:</strong> lcp corresponds to any 4 letter string with a single distinct letter. The lexicographically smallest of them is &quot;aaaa&quot;.
</pre>

<p><strong class="example">Example 3:</strong></p>
Expand Down Expand Up @@ -296,6 +296,56 @@ function findTheString(lcp: number[][]): string {
}
```

#### Rust

```rust
impl Solution {
pub fn find_the_string(lcp: Vec<Vec<i32>>) -> String {
let n = lcp.len();
let mut s = vec!['\0'; n];
let mut i = 0;

for c in b'a'..=b'z' {
while i < n && s[i] != '\0' {
i += 1;
}
if i == n {
break;
}
for j in i..n {
if lcp[i][j] > 0 {
s[j] = c as char;
}
}
}

for i in 0..n {
if s[i] == '\0' {
return "".to_string();
}
}

for i in (0..n).rev() {
for j in (0..n).rev() {
if s[i] == s[j] {
if i == n - 1 || j == n - 1 {
if lcp[i][j] != 1 {
return "".to_string();
}
} else if lcp[i][j] != lcp[i + 1][j + 1] + 1 {
return "".to_string();
}
} else if lcp[i][j] > 0 {
return "".to_string();
}
}
}

s.into_iter().collect()
}
}
```

<!-- tabs:end -->

<!-- solution:end -->
Expand Down
45 changes: 45 additions & 0 deletions solution/2500-2599/2573.Find the String with LCP/Solution.rs
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Original file line number Diff line number Diff line change
@@ -0,0 +1,45 @@
impl Solution {
pub fn find_the_string(lcp: Vec<Vec<i32>>) -> String {
let n = lcp.len();
Comment on lines +1 to +3
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Copilot AI Mar 28, 2026

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The PR title references adding solutions for LC 2574, but this PR also adds a new Rust solution for LC 2573 and updates its documentation. Please update the PR title (or description) to reflect LC 2573 additions and the LC 2574 consolidation/renaming so the change scope is accurately represented.

Copilot uses AI. Check for mistakes.
let mut s = vec!['\0'; n];
let mut i = 0;

for c in b'a'..=b'z' {
while i < n && s[i] != '\0' {
i += 1;
}
if i == n {
break;
}
for j in i..n {
if lcp[i][j] > 0 {
s[j] = c as char;
}
}
}

for i in 0..n {
if s[i] == '\0' {
return "".to_string();
}
}

for i in (0..n).rev() {
for j in (0..n).rev() {
if s[i] == s[j] {
if i == n - 1 || j == n - 1 {
if lcp[i][j] != 1 {
return "".to_string();
}
} else if lcp[i][j] != lcp[i + 1][j + 1] + 1 {
return "".to_string();
}
} else if lcp[i][j] > 0 {
return "".to_string();
}
}
Comment on lines +27 to +40
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Copilot AI Mar 28, 2026

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This validation iterates over the full n x n matrix even though the conditions are symmetric in i/j (i.e., lcp[i][j] mirrors lcp[j][i]). You can cut the work roughly in half by iterating only one triangle (e.g., j from 0..=i) while keeping the same checks, which helps if n is large.

Copilot uses AI. Check for mistakes.
}

s.into_iter().collect()
}
}
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