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Commit bf9633f

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feat: add solutions to lc problem: No.1478 (doocs#4283)
No.1478.Allocate Mailboxes
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‎images/leetcode-doocs-old.png‎

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‎images/leetcode-doocs.png‎

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‎solution/1400-1499/1478.Allocate Mailboxes/README.md‎

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@@ -92,7 +92,7 @@ $$
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其中 $g[i][j]$ 的计算方法如下:
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$$
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g[i][j] = g[i + 1][j - 1] + houses[j] - houses[i]
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g[i][j] = g[i + 1][j - 1] + \textit{houses}[j] - \textit{houses}[i]
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$$
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时间复杂度 $O(n^2 \times k),ドル空间复杂度 $O(n^2)$。其中 $n$ 为房子的数量。
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}
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```
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#### TypeScript
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```ts
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function minDistance(houses: number[], k: number): number {
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houses.sort((a, b) => a - b);
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const n = houses.length;
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const g: number[][] = Array.from({ length: n }, () => Array(n).fill(0));
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for (let i = n - 2; i >= 0; i--) {
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for (let j = i + 1; j < n; j++) {
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g[i][j] = g[i + 1][j - 1] + houses[j] - houses[i];
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}
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}
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const inf = Number.POSITIVE_INFINITY;
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const f: number[][] = Array.from({ length: n }, () => Array(k + 1).fill(inf));
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for (let i = 0; i < n; i++) {
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f[i][1] = g[0][i];
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}
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for (let j = 2; j <= k; j++) {
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for (let i = j - 1; i < n; i++) {
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for (let p = i - 1; p >= 0; p--) {
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f[i][j] = Math.min(f[i][j], f[p][j - 1] + g[p + 1][i]);
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}
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}
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}
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return f[n - 1][k];
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->

‎solution/1400-1499/1478.Allocate Mailboxes/README_EN.md‎

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<strong>Input:</strong> houses = [1,4,8,10,20], k = 3
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<strong>Output:</strong> 5
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<strong>Explanation:</strong> Allocate mailboxes in position 3, 9 and 20.
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Minimum total distance from each houses to nearest mailboxes is |3-1| + |4-3| + |9-8| + |10-9| + |20-20| = 5
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Minimum total distance from each houses to nearest mailboxes is |3-1| + |4-3| + |9-8| + |10-9| + |20-20| = 5
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<!-- solution:start -->
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### Solution 1
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### Solution 1: Dynamic Programming
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We define $f[i][j]$ to represent the minimum total distance between the houses and their nearest mailbox, when placing $j$ mailboxes among the first $i+1$ houses. Initially, $f[i][j] = \infty,ドル and the final answer will be $f[n-1][k]$.
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We can iterate over the last house $p$ controlled by the $j-1$-th mailbox, i.e., 0ドル \leq p \leq i-1$. The $j$-th mailbox will control the houses in the range $[p+1, \dots, i]$. Let $g[i][j]$ denote the minimum total distance when placing a mailbox for the houses in the range $[i, \dots, j]$. The state transition equation is:
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$$
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f[i][j] = \min_{0 \leq p \leq i-1} \{f[p][j-1] + g[p+1][i]\}
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$$
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where $g[i][j]$ is computed as follows:
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$$
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g[i][j] = g[i + 1][j - 1] + \textit{houses}[j] - \textit{houses}[i]
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$$
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The time complexity is $O(n^2 \times k),ドル and the space complexity is $O(n^2),ドル where $n$ is the number of houses.
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<!-- tabs:start -->
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}
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```
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#### TypeScript
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```ts
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function minDistance(houses: number[], k: number): number {
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houses.sort((a, b) => a - b);
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const n = houses.length;
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const g: number[][] = Array.from({ length: n }, () => Array(n).fill(0));
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for (let i = n - 2; i >= 0; i--) {
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for (let j = i + 1; j < n; j++) {
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g[i][j] = g[i + 1][j - 1] + houses[j] - houses[i];
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}
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}
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const inf = Number.POSITIVE_INFINITY;
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const f: number[][] = Array.from({ length: n }, () => Array(k + 1).fill(inf));
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for (let i = 0; i < n; i++) {
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f[i][1] = g[0][i];
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}
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for (let j = 2; j <= k; j++) {
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for (let i = j - 1; i < n; i++) {
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for (let p = i - 1; p >= 0; p--) {
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f[i][j] = Math.min(f[i][j], f[p][j - 1] + g[p + 1][i]);
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}
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}
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}
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return f[n - 1][k];
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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 minDistance(houses: number[], k: number): number {
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houses.sort((a, b) => a - b);
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const n = houses.length;
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const g: number[][] = Array.from({ length: n }, () => Array(n).fill(0));
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for (let i = n - 2; i >= 0; i--) {
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for (let j = i + 1; j < n; j++) {
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g[i][j] = g[i + 1][j - 1] + houses[j] - houses[i];
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}
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}
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const inf = Number.POSITIVE_INFINITY;
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const f: number[][] = Array.from({ length: n }, () => Array(k + 1).fill(inf));
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for (let i = 0; i < n; i++) {
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f[i][1] = g[0][i];
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}
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for (let j = 2; j <= k; j++) {
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for (let i = j - 1; i < n; i++) {
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for (let p = i - 1; p >= 0; p--) {
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f[i][j] = Math.min(f[i][j], f[p][j - 1] + g[p + 1][i]);
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}
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}
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}
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return f[n - 1][k];
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}

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