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Commit 110e86a

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feat: add solutions to lc problem: No.0753 (doocs#3939)
No.0753.Cracking the Safe
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‎solution/0700-0799/0753.Cracking the Safe/README.md‎

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@@ -189,6 +189,32 @@ func crackSafe(n int, k int) string {
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}
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```
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#### TypeScript
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```ts
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function crackSafe(n: number, k: number): string {
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function dfs(u: number): void {
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for (let x = 0; x < k; x++) {
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const e = u * 10 + x;
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if (!vis.has(e)) {
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vis.add(e);
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const v = e % mod;
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dfs(v);
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ans.push(x.toString());
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}
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}
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}
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const mod = Math.pow(10, n - 1);
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const vis = new Set<number>();
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const ans: string[] = [];
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dfs(0);
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ans.push('0'.repeat(n - 1));
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return ans.join('');
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}
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```
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<!-- solution:end -->

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

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<!-- solution:start -->
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### Solution 1
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### Solution 1: Eulerian Circuit
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We can construct a directed graph based on the description in the problem: each point is considered as a length $n-1$ $k$-string, and each edge carries a character from 0ドル$ to $k-1$. If there is a directed edge $e$ from point $u$ to point $v,ドル and the character carried by $e$ is $c,ドル then the last $k-1$ characters of $u+c$ form the string $v$. At this point, the edge $u+c$ represents a password of length $n$.
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In this directed graph, there are $k^{n-1}$ points, each point has $k$ outgoing edges and $k$ incoming edges. Therefore, this directed graph has an Eulerian circuit, and the path traversed by the Eulerian circuit is the answer to the problem.
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The time complexity is $O(k^n),ドル and the space complexity is $O(k^n)$.
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}
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```
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#### TypeScript
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```ts
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function crackSafe(n: number, k: number): string {
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function dfs(u: number): void {
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for (let x = 0; x < k; x++) {
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const e = u * 10 + x;
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if (!vis.has(e)) {
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vis.add(e);
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const v = e % mod;
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dfs(v);
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ans.push(x.toString());
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}
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}
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}
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const mod = Math.pow(10, n - 1);
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const vis = new Set<number>();
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const ans: string[] = [];
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dfs(0);
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ans.push('0'.repeat(n - 1));
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return ans.join('');
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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function crackSafe(n: number, k: number): string {
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function dfs(u: number): void {
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for (let x = 0; x < k; x++) {
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const e = u * 10 + x;
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if (!vis.has(e)) {
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vis.add(e);
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const v = e % mod;
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dfs(v);
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ans.push(x.toString());
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}
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}
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}
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const mod = Math.pow(10, n - 1);
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const vis = new Set<number>();
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const ans: string[] = [];
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dfs(0);
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ans.push('0'.repeat(n - 1));
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return ans.join('');
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}

‎solution/0700-0799/0754.Reach a Number/README.md‎

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### 方法一:数学分析
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由于对称性,每次可以选择向左或向右移动,因此,我们可以将 $target$ 统一取绝对值。
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由于对称性,每次可以选择向左或向右移动,因此,我们可以将 $\textit{target}$ 统一取绝对值。
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定义 $s$ 表示当前所处的位置,用变量 $k$ 记录移动的次数。初始时 $s$ 和 $k$ 均为 0ドル$。
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我们将 $s$ 一直循环累加,直到满足 $s\ge target$ 并且 $(s-target)\mod 2 = 0,ドル此时的移动次数 $k$ 就是答案,直接返回。
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我们将 $s$ 一直循环累加,直到满足 $s\ge \textit{target}$ 并且 $(s-\textit{target}) \bmod 2 = 0,ドル此时的移动次数 $k$ 就是答案,直接返回。
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为什么?因为如果 $s\ge target$ 且 $(s-target)\mod 2 = 0,ドル我们只需要把前面 $\frac{s-target}{2}$ 这个正整数变为负数,就能使得 $s$ 与 $target$ 相等。正整数变负数的过程,实际上是将移动的方向改变,但实际移动次数仍然不变。
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为什么?因为如果 $s\ge \textit{target}$ 且 $(s-\textit{target})\mod 2 = 0,ドル我们只需要把前面 $\frac{s-\textit{target}}{2}$ 这个正整数变为负数,就能使得 $s$ 与 $\textit{target}$ 相等。正整数变负数的过程,实际上是将移动的方向改变,但实际移动次数仍然不变。
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时间复杂度 $O(\sqrt{\left | target \right | }),ドル空间复杂度 $O(1)$。
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时间复杂度 $O(\sqrt{\left | \textit{target} \right | }),ドル空间复杂度 $O(1)$。
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‎solution/0700-0799/0754.Reach a Number/README_EN.md‎

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<!-- solution:start -->
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### Solution 1
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### Solution 1: Mathematical Analysis
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Due to symmetry, each time we can choose to move left or right, so we can take the absolute value of $\textit{target}$.
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Define $s$ as the current position, and use the variable $k$ to record the number of moves. Initially, both $s$ and $k$ are 0ドル$.
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We keep adding to $s$ in a loop until $s \ge \textit{target}$ and $(s - \textit{target}) \bmod 2 = 0$. At this point, the number of moves $k$ is the answer, and we return it directly.
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Why? Because if $s \ge \textit{target}$ and $(s - \textit{target}) \bmod 2 = 0,ドル we only need to change the sign of the positive integer $\frac{s - \textit{target}}{2}$ to negative, so that $s$ equals $\textit{target}$. Changing the sign of a positive integer essentially means changing the direction of the move, but the actual number of moves remains the same.
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The time complexity is $O(\sqrt{\left | \textit{target} \right | }),ドル and the space complexity is $O(1)$.
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‎solution/0700-0799/0755.Pour Water/README_EN.md‎

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### Solution 1
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### Solution 1: Simulation
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We can simulate the process of each unit of water dropping. Each time a drop falls, we first try to move left. If it can move to a lower height, it moves to the lowest height; if it cannot move to a lower height, we try to move right. If it can move to a lower height, it moves to the lowest height; if it cannot move to a lower height, it rises at the current position.
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The time complexity is $O(v \times n),ドル and the space complexity is $O(1),ドル where $v$ and $n$ are the number of water drops and the length of the height array, respectively.
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