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feat: add solutions to lc problem: No.0753
No.0753.Cracking the Safe
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‎solution/0700-0799/0753.Cracking the Safe/README.md

Lines changed: 98 additions & 1 deletion
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@@ -48,22 +48,119 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：欧拉回路**
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我们可以对题目中所描述的内容构建有向图：将每个点看作一个长度为 $n-1$ 的 $k$ 字符串，每条边都带有一个从 $0$ 到 $k-1$ 的字符。如果点 $u$ 到点 $v$ 之间有一条有向边 $e$，假设 $e$ 携带的字符为 $c$，那么 $u+c$ 的末尾 $k-1$ 个字符形成的字符串等于 $v$，此时边 $u+c$ 就表示了一个长度为 $n$ 的密码。
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在这个有向图中，一共有 $k^{n-1}$ 个点，每个点都有 $k$ 条出边，也有 $k$ 条入边，因此，该有向图存在欧拉回路，欧拉回路所经过的路径拼接起来就是题目中的答案。
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时间复杂度 $O(k^n)$，空间复杂度 $O(k^n)$。
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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 crackSafe(self, n: int, k: int) -> str:
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def dfs(u):
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for x in range(k):
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e = u * 10 + x
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if e not in vis:
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vis.add(e)
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v = e % mod
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dfs(v)
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ans.append(str(x))
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mod = 10 ** (n - 1)
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vis = set()
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ans = []
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dfs(0)
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ans.append("0" * (n - 1))
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return "".join(ans)
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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
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class Solution {
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private Set<Integer> vis = new HashSet<>();
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private StringBuilder ans = new StringBuilder();
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private int mod;
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public String crackSafe(int n, int k) {
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mod = (int) Math.pow(10, n - 1);
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dfs(0, k);
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ans.append("0".repeat(n - 1));
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return ans.toString();
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}
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private void dfs(int u, int k) {
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for (int x = 0; x < k; ++x) {
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int e = u * 10 + x;
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if (vis.add(e)) {
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int v = e % mod;
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dfs(v, k);
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ans.append(x);
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}
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}
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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string crackSafe(int n, int k) {
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unordered_set<int> vis;
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int mod = pow(10, n - 1);
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string ans;
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function<void(int)> dfs = [&](int u) {
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for (int x = 0; x < k; ++x) {
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int e = u * 10 + x;
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if (!vis.count(e)) {
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vis.insert(e);
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dfs(e % mod);
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ans += (x + '0');
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}
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}
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};
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dfs(0);
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ans += string(n - 1, '0');
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return ans;
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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 crackSafe(n int, k int) string {
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mod := int(math.Pow(10, float64(n-1)))
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vis := map[int]bool{}
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ans := &strings.Builder{}
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var dfs func(int)
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dfs = func(u int) {
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for x := 0; x < k; x++ {
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e := u*10 + x
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if !vis[e] {
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vis[e] = true
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v := e % mod
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dfs(v)
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ans.WriteByte(byte('0' + x))
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}
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}
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}
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dfs(0)
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ans.WriteString(strings.Repeat("0", n-1))
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return ans.String()
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}
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```
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### **...**

‎solution/0700-0799/0753.Cracking the Safe/README_EN.md

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Original file line numberDiff line numberDiff line change
@@ -61,13 +61,102 @@ Thus &quot;01100&quot; will unlock the safe. &quot;01100&quot;, &quot;10011&quot
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### **Python3**
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```python
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class Solution:
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def crackSafe(self, n: int, k: int) -> str:
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def dfs(u):
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for x in range(k):
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e = u * 10 + x
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if e not in vis:
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vis.add(e)
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v = e % mod
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dfs(v)
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ans.append(str(x))
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mod = 10 ** (n - 1)
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vis = set()
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ans = []
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dfs(0)
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ans.append("0" * (n - 1))
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return "".join(ans)
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```
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### **Java**
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```java
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class Solution {
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private Set<Integer> vis = new HashSet<>();
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private StringBuilder ans = new StringBuilder();
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private int mod;
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public String crackSafe(int n, int k) {
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mod = (int) Math.pow(10, n - 1);
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dfs(0, k);
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ans.append("0".repeat(n - 1));
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return ans.toString();
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}
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private void dfs(int u, int k) {
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for (int x = 0; x < k; ++x) {
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int e = u * 10 + x;
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if (vis.add(e)) {
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int v = e % mod;
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dfs(v, k);
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ans.append(x);
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}
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}
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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string crackSafe(int n, int k) {
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unordered_set<int> vis;
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int mod = pow(10, n - 1);
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string ans;
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function<void(int)> dfs = [&](int u) {
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for (int x = 0; x < k; ++x) {
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int e = u * 10 + x;
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if (!vis.count(e)) {
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vis.insert(e);
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dfs(e % mod);
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ans += (x + '0');
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}
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}
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};
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dfs(0);
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ans += string(n - 1, '0');
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return ans;
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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 crackSafe(n int, k int) string {
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mod := int(math.Pow(10, float64(n-1)))
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vis := map[int]bool{}
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ans := &strings.Builder{}
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var dfs func(int)
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dfs = func(u int) {
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for x := 0; x < k; x++ {
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e := u*10 + x
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if !vis[e] {
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vis[e] = true
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v := e % mod
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dfs(v)
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ans.WriteByte(byte('0' + x))
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}
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}
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}
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dfs(0)
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ans.WriteString(strings.Repeat("0", n-1))
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return ans.String()
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}
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```
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### **...**

‎solution/0700-0799/0753.Cracking the Safe/Solution.cpp

Lines changed: 21 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,21 @@
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class Solution {
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public:
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string crackSafe(int n, int k) {
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unordered_set<int> vis;
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int mod = pow(10, n - 1);
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string ans;
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function<void(int)> dfs = [&](int u) {
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for (int x = 0; x < k; ++x) {
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int e = u * 10 + x;
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if (!vis.count(e)) {
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vis.insert(e);
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dfs(e % mod);
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ans += (x + '0');
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}
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}
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};
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dfs(0);
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ans += string(n - 1, '0');
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return ans;
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}
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};

‎solution/0700-0799/0753.Cracking the Safe/Solution.go

Lines changed: 20 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,20 @@
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func crackSafe(n int, k int) string {
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mod := int(math.Pow(10, float64(n-1)))
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vis := map[int]bool{}
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ans := &strings.Builder{}
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var dfs func(int)
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dfs = func(u int) {
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for x := 0; x < k; x++ {
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e := u*10 + x
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if !vis[e] {
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vis[e] = true
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v := e % mod
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dfs(v)
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ans.WriteByte(byte('0' + x))
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}
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}
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}
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dfs(0)
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ans.WriteString(strings.Repeat("0", n-1))
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return ans.String()
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}

‎solution/0700-0799/0753.Cracking the Safe/Solution.java

Lines changed: 23 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,23 @@
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class Solution {
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private Set<Integer> vis = new HashSet<>();
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private StringBuilder ans = new StringBuilder();
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private int mod;
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public String crackSafe(int n, int k) {
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mod = (int) Math.pow(10, n - 1);
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dfs(0, k);
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ans.append("0".repeat(n - 1));
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return ans.toString();
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}
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private void dfs(int u, int k) {
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for (int x = 0; x < k; ++x) {
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int e = u * 10 + x;
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if (vis.add(e)) {
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int v = e % mod;
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dfs(v, k);
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ans.append(x);
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}
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}
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}
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}

‎solution/0700-0799/0753.Cracking the Safe/Solution.py

Lines changed: 17 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,17 @@
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class Solution:
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def crackSafe(self, n: int, k: int) -> str:
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def dfs(u):
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for x in range(k):
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e = u * 10 + x
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if e not in vis:
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vis.add(e)
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v = e % mod
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dfs(v)
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ans.append(str(x))
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mod = 10 ** (n - 1)
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vis = set()
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ans = []
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dfs(0)
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ans.append("0" * (n - 1))
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return "".join(ans)

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