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feat: add solutions to lc problem: No.2573
No.2573.Find the String with LCP
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‎solution/2500-2599/2573.Find the String with LCP/README.md

Lines changed: 159 additions & 4 deletions
Original file line numberDiff line numberDiff line change
@@ -55,34 +55,189 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：贪心 + 构造**
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由于构造的字符串要求字典序最小，因此我们可以从字符 `'a'` 开始，填充到字符串 $s$ 中。
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如果当前位置 $i$ 还未填充字符，那么我们可以将字符 `'a'` 填充到 $i$ 位置，然后枚举所有 $j \gt i$ 的位置，如果 $lcp[i][j] \gt 0$，那么位置 $j$ 也应该填充字符 `'a'`。然后我们将字符 `'a'` 的 ASCII 码加一，继续填充剩余未填充的位置。
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填充结束后，如果字符串中存在未填充的位置，说明无法构造出对应的字符串，返回空字符串。
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接下来，我们可以从大到小枚举字符串中的每个位置 $i$ 和 $j$，然后判断 $s[i]$ 和 $s[j]$ 是否相等：
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- 如果 $s[i] = s[j]$，此时我们需要判断 $i$ 和 $j$ 是否为字符串的最后一个位置，如果是，那么 $lcp[i][j]$ 应该等于 $1$，否则 $lcp[i][j]$ 应该等于 $0$。如果不满足上述条件，说明无法构造出对应的字符串，返回空字符串。如果 $i$ 和 $j$ 不是字符串的最后一个位置，那么 $lcp[i][j]$ 应该等于 $lcp[i + 1][j + 1] + 1$，否则说明无法构造出对应的字符串，返回空字符串。
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- 否则，如果 $lcp[i][j] \gt 0$，说明无法构造出对应的字符串，返回空字符串。
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如果字符串中的每个位置都满足上述条件，那么我们就可以构造出对应的字符串，返回即可。
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时间复杂度为 $O(n^2)$，空间复杂度为 $O(n)$。其中 $n$ 为字符串的长度。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def findTheString(self, lcp: List[List[int]]) -> str:
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n = len(lcp)
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s = [""] * n
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i = 0
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for c in ascii_lowercase:
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while i < n and s[i]:
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i += 1
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if i == n:
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break
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for j in range(i, n):
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if lcp[i][j]:
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s[j] = c
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if "" in s:
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return ""
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for i in range(n - 1, -1, -1):
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for j in range(n - 1, -1, -1):
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if s[i] == s[j]:
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if i == n - 1 or j == n - 1:
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if lcp[i][j] != 1:
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return ""
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elif lcp[i][j] != lcp[i + 1][j + 1] + 1:
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return ""
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elif lcp[i][j]:
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return ""
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return "".join(s)
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
73-
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class Solution {
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public String findTheString(int[][] lcp) {
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int n = lcp.length;
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char[] s = new char[n];
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int i = 0;
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for (char c = 'a'; c <= 'z'; ++c) {
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while (i < n && s[i] != '\0') {
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++i;
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}
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if (i == n) {
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break;
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}
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for (int j = i; j < n; ++j) {
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if (lcp[i][j] > 0) {
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s[j] = c;
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}
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}
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}
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for (i = 0; i < n; ++i) {
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if (s[i] == '\0') {
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return "";
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}
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}
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for (i = n - 1; i >= 0; --i) {
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for (int j = n - 1; j >= 0; --j) {
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if (s[i] == s[j]) {
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if (i == n - 1 || j == n - 1) {
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if (lcp[i][j] != 1) {
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return "";
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}
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} else if (lcp[i][j] != lcp[i + 1][j + 1] + 1) {
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return "";
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}
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} else if (lcp[i][j] > 0) {
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return "";
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}
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}
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}
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return String.valueOf(s);
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}
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}
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```
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### **C++**
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```cpp
79-
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class Solution {
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public:
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string findTheString(vector<vector<int>>& lcp) {
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int i = 0, n = lcp.size();
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string s(n, '\0');
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for (char c = 'a'; c <= 'z'; ++c) {
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while (i < n && s[i]) {
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++i;
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}
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if (i == n) {
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break;
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}
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for (int j = i; j < n; ++j) {
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if (lcp[i][j]) {
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s[j] = c;
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}
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}
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}
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if (s.find('\0') != -1) {
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return "";
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}
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for (i = n - 1; ~i; --i) {
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for (int j = n - 1; ~j; --j) {
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if (s[i] == s[j]) {
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if (i == n - 1 || j == n - 1) {
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if (lcp[i][j] != 1) {
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return "";
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}
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} else if (lcp[i][j] != lcp[i + 1][j + 1] + 1) {
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return "";
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}
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} else if (lcp[i][j]) {
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return "";
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}
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}
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}
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return s;
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}
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};
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```
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### **Go**
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```go
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func findTheString(lcp [][]int) string {
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i, n := 0, len(lcp)
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s := make([]byte, n)
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for c := 'a'; c <= 'z'; c++ {
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for i < n && s[i] != 0 {
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i++
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}
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if i == n {
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break
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}
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for j := i; j < n; j++ {
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if lcp[i][j] > 0 {
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s[j] = byte(c)
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}
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}
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}
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if bytes.IndexByte(s, 0) >= 0 {
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return ""
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}
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for i := n - 1; i >= 0; i-- {
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for j := n - 1; j >= 0; j-- {
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if s[i] == s[j] {
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if i == n-1 || j == n-1 {
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if lcp[i][j] != 1 {
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return ""
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}
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} else if lcp[i][j] != lcp[i+1][j+1]+1 {
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return ""
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}
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} else if lcp[i][j] > 0 {
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return ""
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}
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}
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}
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return string(s)
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}
86241
```
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### **...**

‎solution/2500-2599/2573.Find the String with LCP/README_EN.md

Lines changed: 142 additions & 4 deletions
Original file line numberDiff line numberDiff line change
@@ -54,25 +54,163 @@
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### **Python3**
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```python
57-
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class Solution:
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def findTheString(self, lcp: List[List[int]]) -> str:
59+
n = len(lcp)
60+
s = [""] * n
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i = 0
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for c in ascii_lowercase:
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while i < n and s[i]:
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i += 1
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if i == n:
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break
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for j in range(i, n):
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if lcp[i][j]:
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s[j] = c
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if "" in s:
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return ""
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for i in range(n - 1, -1, -1):
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for j in range(n - 1, -1, -1):
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if s[i] == s[j]:
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if i == n - 1 or j == n - 1:
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if lcp[i][j] != 1:
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return ""
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elif lcp[i][j] != lcp[i + 1][j + 1] + 1:
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return ""
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elif lcp[i][j]:
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return ""
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return "".join(s)
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```
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### **Java**
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```java
63-
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class Solution {
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public String findTheString(int[][] lcp) {
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int n = lcp.length;
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char[] s = new char[n];
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int i = 0;
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for (char c = 'a'; c <= 'z'; ++c) {
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while (i < n && s[i] != '\0') {
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++i;
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}
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if (i == n) {
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break;
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}
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for (int j = i; j < n; ++j) {
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if (lcp[i][j] > 0) {
102+
s[j] = c;
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}
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}
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}
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for (i = 0; i < n; ++i) {
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if (s[i] == '\0') {
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return "";
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}
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}
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for (i = n - 1; i >= 0; --i) {
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for (int j = n - 1; j >= 0; --j) {
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if (s[i] == s[j]) {
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if (i == n - 1 || j == n - 1) {
115+
if (lcp[i][j] != 1) {
116+
return "";
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}
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} else if (lcp[i][j] != lcp[i + 1][j + 1] + 1) {
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return "";
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}
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} else if (lcp[i][j] > 0) {
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return "";
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}
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}
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}
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return String.valueOf(s);
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}
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}
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```
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### **C++**
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```cpp
69-
134+
class Solution {
135+
public:
136+
string findTheString(vector<vector<int>>& lcp) {
137+
int i = 0, n = lcp.size();
138+
string s(n, '\0');
139+
for (char c = 'a'; c <= 'z'; ++c) {
140+
while (i < n && s[i]) {
141+
++i;
142+
}
143+
if (i == n) {
144+
break;
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}
146+
for (int j = i; j < n; ++j) {
147+
if (lcp[i][j]) {
148+
s[j] = c;
149+
}
150+
}
151+
}
152+
if (s.find('\0') != -1) {
153+
return "";
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}
155+
for (i = n - 1; ~i; --i) {
156+
for (int j = n - 1; ~j; --j) {
157+
if (s[i] == s[j]) {
158+
if (i == n - 1 || j == n - 1) {
159+
if (lcp[i][j] != 1) {
160+
return "";
161+
}
162+
} else if (lcp[i][j] != lcp[i + 1][j + 1] + 1) {
163+
return "";
164+
}
165+
} else if (lcp[i][j]) {
166+
return "";
167+
}
168+
}
169+
}
170+
return s;
171+
}
172+
};
70173
```
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### **Go**
73176
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```go
75-
178+
func findTheString(lcp [][]int) string {
179+
i, n := 0, len(lcp)
180+
s := make([]byte, n)
181+
for c := 'a'; c <= 'z'; c++ {
182+
for i < n && s[i] != 0 {
183+
i++
184+
}
185+
if i == n {
186+
break
187+
}
188+
for j := i; j < n; j++ {
189+
if lcp[i][j] > 0 {
190+
s[j] = byte(c)
191+
}
192+
}
193+
}
194+
if bytes.IndexByte(s, 0) >= 0 {
195+
return ""
196+
}
197+
for i := n - 1; i >= 0; i-- {
198+
for j := n - 1; j >= 0; j-- {
199+
if s[i] == s[j] {
200+
if i == n-1 || j == n-1 {
201+
if lcp[i][j] != 1 {
202+
return ""
203+
}
204+
} else if lcp[i][j] != lcp[i+1][j+1]+1 {
205+
return ""
206+
}
207+
} else if lcp[i][j] > 0 {
208+
return ""
209+
}
210+
}
211+
}
212+
return string(s)
213+
}
76214
```
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### **...**

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