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feat: add solutions to lc problem: No.1625
No.1625.Lexicographically Smallest String After Applying Operations
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‎solution/1600-1699/1625.Lexicographically Smallest String After Applying Operations/README.md

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Original file line numberDiff line numberDiff line change
@@ -85,6 +85,18 @@
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**方法一：BFS**
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本题数据规模较小，我们可以使用 BFS 暴力搜索所有可能的状态，然后取字典序最小的状态即可。
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**方法二：枚举**
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我们观察发现，对于累加操作，数字最多累加 $10$ 次，就会回到原来的状态；对于轮转操作，字符串最多轮转 $n$ 次，也会回到原来的状态。
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因此，轮转操作最多产生 $n$ 种状态，如果轮转位数 $b$ 为偶数，累加操作只会对奇数位数字产生影响，因此总共产生 $n \times 10$ 种状态；如果轮转位数 $b$ 为奇数，累加操作既会对奇数位数字产生影响，也会对偶数位数字产生影响，因此总共产生 $n \times 10 \times 10$ 种状态。
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所以，我们直接枚举所有的字符串状态，取字典序最小的状态即可。
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时间复杂度 $O(n^2 \times 10^2)$，空间复杂度 $O(n)$。其中 $n$ 为字符串 $s$ 的长度。
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<!-- tabs:start -->
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### **Python3**
@@ -99,16 +111,39 @@ class Solution:
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ans = s
100112
while q:
101113
s = q.popleft()
102-
if s < ans:
114+
if ans > s:
103115
ans = s
104-
nxt1 = ''.join(
105-
[str((int(c) + a) % 10) if i & 1 else c for i, c in enumerate(s)]
106-
)
107-
nxt2 = s[-b:] + s[:-b]
108-
for nxt in (nxt1, nxt2):
109-
if nxt not in vis:
110-
vis.add(nxt)
111-
q.append(nxt)
116+
t1 = ''.join([str((int(c) + a) % 10) if i & 1 else c for i, c in enumerate(s)])
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t2 = s[-b:] + s[:-b]
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for t in (t1, t2):
119+
if t not in vis:
120+
vis.add(t)
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q.append(t)
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return ans
123+
```
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```python
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class Solution:
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def findLexSmallestString(self, s: str, a: int, b: int) -> str:
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ans = s
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n = len(s)
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s = list(s)
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for _ in range(n):
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s = s[-b:] + s[:-b]
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for j in range(10):
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for k in range(1, n, 2):
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s[k] = str((int(s[k]) + a) % 10)
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if b & 1:
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for p in range(10):
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for k in range(0, n, 2):
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s[k] = str((int(s[k]) + a) % 10)
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t = ''.join(s)
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if ans > t:
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ans = t
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else:
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t = ''.join(s)
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if ans > t:
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ans = t
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return ans
113148
```
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@@ -119,26 +154,61 @@ class Solution:
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```java
120155
class Solution {
121156
public String findLexSmallestString(String s, int a, int b) {
122-
Queue<String> q = new ArrayDeque<>();
157+
Deque<String> q = new ArrayDeque<>();
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q.offer(s);
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Set<String> vis = new HashSet<>();
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vis.add(s);
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String ans = s;
162+
int n = s.length();
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while (!q.isEmpty()) {
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s = q.poll();
129-
if (s.compareTo(ans) < 0) {
165+
if (ans.compareTo(s) > 0) {
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ans = s;
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}
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char[] cs = s.toCharArray();
133-
for (int i = 1; i < cs.length; i += 2) {
169+
for (int i = 1; i < n; i += 2) {
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cs[i] = (char) (((cs[i] - '0' + a) % 10) + '0');
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}
136-
String nxt1 = String.valueOf(cs);
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String nxt2 = s.substring(b) + s.substring(0, b);
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for (String nxt : new String[] {nxt1, nxt2}) {
139-
if (!vis.contains(nxt)) {
140-
vis.add(nxt);
141-
q.offer(nxt);
172+
String t1 = String.valueOf(cs);
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String t2 = s.substring(b) + s.substring(0, b);
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for (String t : List.of(t1, t2)) {
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if (vis.add(t)) {
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q.offer(t);
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}
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}
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}
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return ans;
181+
}
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}
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```
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```java
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class Solution {
187+
public String findLexSmallestString(String s, int a, int b) {
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int n = s.length();
189+
String ans = s;
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for (int i = 0; i < n; ++i) {
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s = s.substring(b) + s.substring(0, b);
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char[] cs = s.toCharArray();
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for (int j = 0; j < 10; ++j) {
194+
for (int k = 1; k < n; k += 2) {
195+
cs[k] = (char) (((cs[k] - '0' + a) % 10) + '0');
196+
}
197+
if ((b & 1) == 1) {
198+
for (int p = 0; p < 10; ++p) {
199+
for (int k = 0; k < n; k += 2) {
200+
cs[k] = (char) (((cs[k] - '0' + a) % 10) + '0');
201+
}
202+
s = String.valueOf(cs);
203+
if (ans.compareTo(s) > 0) {
204+
ans = s;
205+
}
206+
}
207+
} else {
208+
s = String.valueOf(cs);
209+
if (ans.compareTo(s) > 0) {
210+
ans = s;
211+
}
142212
}
143213
}
144214
}
@@ -153,21 +223,52 @@ class Solution {
153223
class Solution {
154224
public:
155225
string findLexSmallestString(string s, int a, int b) {
156-
unordered_set<string> vis {{s}};
157-
queue<string> q {{s}};
226+
queue<string> q{{s}};
227+
unordered_set<string> vis{{s}};
158228
string ans = s;
159229
int n = s.size();
160230
while (!q.empty()) {
161231
s = q.front();
162232
q.pop();
163-
if (s < ans) ans = s;
164-
string nxt1 = s;
165-
for (int i = 1; i < n; i += 2) nxt1[i] = ((nxt1[i] - '0' + a) % 10) + '0';
166-
string nxt2 = s.substr(n - b) + s.substr(0, n - b);
167-
for (string nxt : {nxt1, nxt2}) {
168-
if (!vis.count(nxt)) {
169-
vis.insert(nxt);
170-
q.push(nxt);
233+
ans = min(ans, s);
234+
string t1 = s;
235+
for (int i = 1; i < n; i += 2) {
236+
t1[i] = (t1[i] - '0' + a) % 10 + '0';
237+
}
238+
string t2 = s.substr(n - b) + s.substr(0, n - b);
239+
for (auto& t : {t1, t2}) {
240+
if (!vis.count(t)) {
241+
vis.insert(t);
242+
q.emplace(t);
243+
}
244+
}
245+
}
246+
return ans;
247+
}
248+
};
249+
```
250+
251+
```cpp
252+
class Solution {
253+
public:
254+
string findLexSmallestString(string s, int a, int b) {
255+
int n = s.size();
256+
string ans = s;
257+
for (int i = 0; i < n; ++i) {
258+
s = s.substr(n - b) + s.substr(0, n - b);
259+
for (int j = 0; j < 10; ++j) {
260+
for (int k = 1; k < n; k += 2) {
261+
s[k] = (s[k] - '0' + a) % 10 + '0';
262+
}
263+
if (b & 1) {
264+
for (int p = 0; p < 10; ++p) {
265+
for (int k = 0; k < n; k += 2) {
266+
s[k] = (s[k] - '0' + a) % 10 + '0';
267+
}
268+
ans = min(ans, s);
269+
}
270+
} else {
271+
ans = min(ans, s);
171272
}
172273
}
173274
}
@@ -183,32 +284,59 @@ func findLexSmallestString(s string, a int, b int) string {
183284
q := []string{s}
184285
vis := map[string]bool{s: true}
185286
ans := s
287+
n := len(s)
186288
for len(q) > 0 {
187289
s = q[0]
188290
q = q[1:]
189-
if s < ans {
291+
if ans > s {
190292
ans = s
191293
}
192-
for _, nxt := range []string{op1(s, a), op2(s, b)} {
193-
if !vis[nxt] {
194-
vis[nxt] = true
195-
q = append(q, nxt)
294+
t1 := []byte(s)
295+
for i := 1; i < n; i += 2 {
296+
t1[i] = byte((int(t1[i]-'0')+a)%10 + '0')
297+
}
298+
t2 := s[n-b:] + s[:n-b]
299+
for _, t := range []string{string(t1), t2} {
300+
if !vis[t] {
301+
vis[t] = true
302+
q = append(q, t)
196303
}
197304
}
198305
}
199306
return ans
200307
}
308+
```
201309

202-
func op1(s string, a int) string {
203-
res := []byte(s)
204-
for i := 1; i < len(s); i += 2 {
205-
res[i] = byte((int(res[i]-'0')+a)%10 + '0')
310+
```go
311+
func findLexSmallestString(s string, a int, b int) string {
312+
n := len(s)
313+
ans := s
314+
for _ = range s {
315+
s = s[n-b:] + s[:n-b]
316+
cs := []byte(s)
317+
for j := 0; j < 10; j++ {
318+
for k := 1; k < n; k += 2 {
319+
cs[k] = byte((int(cs[k]-'0')+a)%10 + '0')
320+
}
321+
if b&1 == 1 {
322+
for p := 0; p < 10; p++ {
323+
for k := 0; k < n; k += 2 {
324+
cs[k] = byte((int(cs[k]-'0')+a)%10 + '0')
325+
}
326+
s = string(cs)
327+
if ans > s {
328+
ans = s
329+
}
330+
}
331+
} else {
332+
s = string(cs)
333+
if ans > s {
334+
ans = s
335+
}
336+
}
337+
}
206338
}
207-
return string(res)
208-
}
209-
210-
func op2(s string, b int) string {
211-
return s[len(s)-b:] + s[:len(s)-b]
339+
return ans
212340
}
213341
```
214342

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