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Commit ff3350d

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feat: add ts solution to lc problem: No.0514 (doocs#2273)
No.0514.Freedom Trail
1 parent e3f7ac0 commit ff3350d

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3 files changed

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‎solution/0500-0599/0514.Freedom Trail/README.md‎

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@@ -194,6 +194,40 @@ func abs(x int) int {
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}
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```
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```ts
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function findRotateSteps(ring: string, key: string): number {
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const m: number = key.length;
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const n: number = ring.length;
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const pos: number[][] = Array.from({ length: 26 }, () => []);
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for (let i = 0; i < n; ++i) {
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const j: number = ring.charCodeAt(i) - 'a'.charCodeAt(0);
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pos[j].push(i);
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}
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const f: number[][] = Array.from({ length: m }, () => Array(n).fill(1 << 30));
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for (const j of pos[key.charCodeAt(0) - 'a'.charCodeAt(0)]) {
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f[0][j] = Math.min(j, n - j) + 1;
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}
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for (let i = 1; i < m; ++i) {
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for (const j of pos[key.charCodeAt(i) - 'a'.charCodeAt(0)]) {
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for (const k of pos[key.charCodeAt(i - 1) - 'a'.charCodeAt(0)]) {
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f[i][j] = Math.min(
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f[i][j],
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f[i - 1][k] + Math.min(Math.abs(j - k), n - Math.abs(j - k)) + 1,
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);
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}
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}
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}
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let ans: number = 1 << 30;
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for (const j of pos[key.charCodeAt(m - 1) - 'a'.charCodeAt(0)]) {
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ans = Math.min(ans, f[m - 1][j]);
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}
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return ans;
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}
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```
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<!-- tabs:end -->
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<!-- end -->

‎solution/0500-0599/0514.Freedom Trail/README_EN.md‎

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@@ -48,7 +48,19 @@ So the final output is 4.
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## Solutions
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### Solution 1
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### Solution 1: Dynamic Programming
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First, we preprocess the positions of each character $c$ in the string $ring,ドル and record them in the array $pos[c]$. Suppose the lengths of the strings $key$ and $ring$ are $m$ and $n,ドル respectively.
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Then we define $f[i][j]$ as the minimum number of steps to spell the first $i+1$ characters of the string $key,ドル and the $j$-th character of $ring$ is aligned with the 12ドル:00$ direction. Initially, $f[i][j]=+\infty$. The answer is $\min_{0 \leq j < n} f[m - 1][j]$.
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We can first initialize $f[0][j],ドル where $j$ is the position where the character $key[0]$ appears in $ring$. Since the $j$-th character of $ring$ is aligned with the 12ドル:00$ direction, we only need 1ドル$ step to spell $key[0]$. In addition, we need $min(j, n - j)$ steps to rotate $ring$ to the 12ドル:00$ direction. Therefore, $f[0][j]=min(j, n - j) + 1$.
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Next, we consider how the state transitions when $i \geq 1$. We can enumerate the position list $pos[key[i]]$ where $key[i]$ appears in $ring,ドル and enumerate the position list $pos[key[i-1]]$ where $key[i-1]$ appears in $ring,ドル and then update $f[i][j],ドル i.e., $f[i][j]=\min_{k \in pos[key[i-1]]} f[i-1][k] + \min(\text{abs}(j - k), n - \text{abs}(j - k)) + 1$.
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Finally, we return $\min_{0 \leq j \lt n} f[m - 1][j]$.
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The time complexity is $O(m \times n^2),ドル and the space complexity is $O(m \times n)$. Here, $m$ and $n$ are the lengths of the strings $key$ and $ring,ドル respectively.
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<!-- tabs:start -->
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@@ -174,6 +186,40 @@ func abs(x int) int {
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}
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```
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```ts
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function findRotateSteps(ring: string, key: string): number {
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const m: number = key.length;
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const n: number = ring.length;
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const pos: number[][] = Array.from({ length: 26 }, () => []);
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for (let i = 0; i < n; ++i) {
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const j: number = ring.charCodeAt(i) - 'a'.charCodeAt(0);
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pos[j].push(i);
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}
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const f: number[][] = Array.from({ length: m }, () => Array(n).fill(1 << 30));
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for (const j of pos[key.charCodeAt(0) - 'a'.charCodeAt(0)]) {
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f[0][j] = Math.min(j, n - j) + 1;
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}
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for (let i = 1; i < m; ++i) {
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for (const j of pos[key.charCodeAt(i) - 'a'.charCodeAt(0)]) {
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for (const k of pos[key.charCodeAt(i - 1) - 'a'.charCodeAt(0)]) {
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f[i][j] = Math.min(
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f[i][j],
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f[i - 1][k] + Math.min(Math.abs(j - k), n - Math.abs(j - k)) + 1,
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);
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}
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}
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}
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let ans: number = 1 << 30;
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for (const j of pos[key.charCodeAt(m - 1) - 'a'.charCodeAt(0)]) {
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ans = Math.min(ans, f[m - 1][j]);
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}
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return ans;
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}
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```
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<!-- tabs:end -->
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<!-- end -->
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function findRotateSteps(ring: string, key: string): number {
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const m: number = key.length;
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const n: number = ring.length;
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const pos: number[][] = Array.from({ length: 26 }, () => []);
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for (let i = 0; i < n; ++i) {
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const j: number = ring.charCodeAt(i) - 'a'.charCodeAt(0);
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pos[j].push(i);
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}
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const f: number[][] = Array.from({ length: m }, () => Array(n).fill(1 << 30));
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for (const j of pos[key.charCodeAt(0) - 'a'.charCodeAt(0)]) {
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f[0][j] = Math.min(j, n - j) + 1;
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}
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for (let i = 1; i < m; ++i) {
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for (const j of pos[key.charCodeAt(i) - 'a'.charCodeAt(0)]) {
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for (const k of pos[key.charCodeAt(i - 1) - 'a'.charCodeAt(0)]) {
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f[i][j] = Math.min(
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f[i][j],
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f[i - 1][k] + Math.min(Math.abs(j - k), n - Math.abs(j - k)) + 1,
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);
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}
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}
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}
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let ans: number = 1 << 30;
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for (const j of pos[key.charCodeAt(m - 1) - 'a'.charCodeAt(0)]) {
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ans = Math.min(ans, f[m - 1][j]);
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}
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return ans;
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}

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