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Commit e720d2b

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feat: add solutions to lc problem: No.1866 (doocs#1981)
No.1866.Number of Ways to Rearrange Sticks With K Sticks Visible
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‎solution/1800-1899/1866.Number of Ways to Rearrange Sticks With K Sticks Visible/README.md‎

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@@ -210,6 +210,39 @@ func rearrangeSticks(n int, k int) int {
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}
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```
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### **TypeScript**
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```ts
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function rearrangeSticks(n: number, k: number): number {
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const mod = 10 ** 9 + 7;
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const f: number[][] = Array.from({ length: n + 1 }, () =>
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Array.from({ length: k + 1 }, () => 0),
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);
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f[0][0] = 1;
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for (let i = 1; i <= n; ++i) {
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for (let j = 1; j <= k; ++j) {
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f[i][j] = (f[i - 1][j - 1] + (i - 1) * f[i - 1][j]) % mod;
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}
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}
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return f[n][k];
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}
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```
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```ts
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function rearrangeSticks(n: number, k: number): number {
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const mod = 10 ** 9 + 7;
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const f: number[] = Array(n + 1).fill(0);
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f[0] = 1;
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for (let i = 1; i <= n; ++i) {
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for (let j = k; j; --j) {
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f[j] = (f[j] * (i - 1) + f[j - 1]) % mod;
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}
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f[0] = 0;
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}
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return f[k];
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}
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```
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### **...**
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```

‎solution/1800-1899/1866.Number of Ways to Rearrange Sticks With K Sticks Visible/README_EN.md‎

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## Solutions
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**Solution 1: Dynamic Programming**
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We define $f[i][j]$ to represent the number of permutations of length $i$ in which exactly $j$ sticks can be seen. Initially, $f[0][0]=1$ and the rest $f[i][j]=0$. The answer is $f[n][k]$.
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Consider whether the last stick can be seen. If it can be seen, it must be the longest. Then there are $i - 1$ sticks in front of it, and exactly $j - 1$ sticks can be seen, which is $f[i - 1][j - 1]$. If the last stick cannot be seen, it can be any one except the longest stick. Then there are $i - 1$ sticks in front of it, and exactly $j$ sticks can be seen, which is $f[i - 1][j] \times (i - 1)$.
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Therefore, the state transition equation is:
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$$
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f[i][j] = f[i - 1][j - 1] + f[i - 1][j] \times (i - 1)
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$$
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The final answer is $f[n][k]$.
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We notice that $f[i][j]$ is only related to $f[i - 1][j - 1]$ and $f[i - 1][j],ドル so we can use a one-dimensional array to optimize the space complexity.
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The time complexity is $O(n \times k),ドル and the space complexity is $O(k)$. Here, $n$ and $k$ are the two integers given in the problem.
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<!-- tabs:start -->
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### **Python3**
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}
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```
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### **TypeScript**
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```ts
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function rearrangeSticks(n: number, k: number): number {
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const mod = 10 ** 9 + 7;
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const f: number[][] = Array.from({ length: n + 1 }, () =>
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Array.from({ length: k + 1 }, () => 0),
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);
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f[0][0] = 1;
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for (let i = 1; i <= n; ++i) {
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for (let j = 1; j <= k; ++j) {
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f[i][j] = (f[i - 1][j - 1] + (i - 1) * f[i - 1][j]) % mod;
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}
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}
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return f[n][k];
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}
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```
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```ts
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function rearrangeSticks(n: number, k: number): number {
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const mod = 10 ** 9 + 7;
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const f: number[] = Array(n + 1).fill(0);
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f[0] = 1;
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for (let i = 1; i <= n; ++i) {
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for (let j = k; j; --j) {
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f[j] = (f[j] * (i - 1) + f[j - 1]) % mod;
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}
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f[0] = 0;
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}
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return f[k];
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}
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```
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### **...**
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```
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function rearrangeSticks(n: number, k: number): number {
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const mod = 10 ** 9 + 7;
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const f: number[] = Array(n + 1).fill(0);
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f[0] = 1;
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for (let i = 1; i <= n; ++i) {
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for (let j = k; j; --j) {
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f[j] = (f[j] * (i - 1) + f[j - 1]) % mod;
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}
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f[0] = 0;
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}
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return f[k];
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}

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