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Commit 468e051

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feat: add solutions to lc problem: No.1671 (doocs#2118)
No.1671.Minimum Number of Removals to Make Mountain Array
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4 files changed

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‎solution/1600-1699/1671.Minimum Number of Removals to Make Mountain Array/README.md‎

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@@ -60,7 +60,7 @@
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那么最终答案就是 $n - \max(left[i] + right[i] - 1),ドル其中 1ドル \leq i \leq n,ドル并且 $left[i] \gt 1$ 且 $right[i] \gt 1$。
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时间复杂度 $O(n^2),ドル空间复杂度 $O(n)$。其中 $n$ 为数组 `nums` 的长度。
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时间复杂度 $O(n^2),ドル空间复杂度 $O(n)$。其中 $n$ 为数组 $nums$ 的长度。
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<!-- tabs:start -->
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@@ -193,8 +193,8 @@ func minimumMountainRemovals(nums []int) int {
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```ts
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function minimumMountainRemovals(nums: number[]): number {
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const n = nums.length;
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const left = newArray(n).fill(1);
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const right = newArray(n).fill(1);
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const left = Array(n).fill(1);
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const right = Array(n).fill(1);
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for (let i = 1; i < n; ++i) {
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for (let j = 0; j < i; ++j) {
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if (nums[i] > nums[j]) {
@@ -219,6 +219,41 @@ function minimumMountainRemovals(nums: number[]): number {
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}
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```
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### **TypeScript**
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```ts
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impl Solution {
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pub fn minimum_mountain_removals(nums: Vec<i32>) -> i32 {
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let n = nums.len();
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let mut left = vec![1; n];
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let mut right = vec![1; n];
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for i in 1..n {
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for j in 0..i {
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if nums[i] > nums[j] {
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left[i] = left[i].max(left[j] + 1);
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}
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}
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}
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for i in (0..n - 1).rev() {
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for j in i + 1..n {
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if nums[i] > nums[j] {
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right[i] = right[i].max(right[j] + 1);
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}
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}
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}
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let mut ans = 0;
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for i in 0..n {
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if left[i] > 1 && right[i] > 1 {
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ans = ans.max(left[i] + right[i] - 1);
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}
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}
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(n as i32) - ans
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}
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}
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```
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### **...**
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```

‎solution/1600-1699/1671.Minimum Number of Removals to Make Mountain Array/README_EN.md‎

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## Solutions
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**Solution 1: Dynamic Programming**
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This problem can be transformed into finding the longest increasing subsequence and the longest decreasing subsequence.
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We define $left[i]$ as the length of the longest increasing subsequence ending with $nums[i],ドル and define $right[i]$ as the length of the longest decreasing subsequence starting with $nums[i]$.
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Then the final answer is $n - \max(left[i] + right[i] - 1),ドル where 1ドル \leq i \leq n,ドル and $left[i] \gt 1$ and $right[i] \gt 1$.
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The time complexity is $O(n^2),ドル and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.
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<!-- tabs:start -->
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### **Python3**
@@ -173,8 +183,8 @@ func minimumMountainRemovals(nums []int) int {
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```ts
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function minimumMountainRemovals(nums: number[]): number {
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const n = nums.length;
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const left = newArray(n).fill(1);
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const right = newArray(n).fill(1);
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const left = Array(n).fill(1);
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const right = Array(n).fill(1);
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for (let i = 1; i < n; ++i) {
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for (let j = 0; j < i; ++j) {
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if (nums[i] > nums[j]) {
@@ -199,6 +209,41 @@ function minimumMountainRemovals(nums: number[]): number {
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}
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```
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### **TypeScript**
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```ts
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impl Solution {
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pub fn minimum_mountain_removals(nums: Vec<i32>) -> i32 {
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let n = nums.len();
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let mut left = vec![1; n];
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let mut right = vec![1; n];
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for i in 1..n {
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for j in 0..i {
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if nums[i] > nums[j] {
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left[i] = left[i].max(left[j] + 1);
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}
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}
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}
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for i in (0..n - 1).rev() {
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for j in i + 1..n {
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if nums[i] > nums[j] {
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right[i] = right[i].max(right[j] + 1);
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}
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}
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}
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let mut ans = 0;
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for i in 0..n {
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if left[i] > 1 && right[i] > 1 {
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ans = ans.max(left[i] + right[i] - 1);
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}
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}
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(n as i32) - ans
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}
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}
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```
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### **...**
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```
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impl Solution {
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pub fn minimum_mountain_removals(nums: Vec<i32>) -> i32 {
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let n = nums.len();
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let mut left = vec![1; n];
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let mut right = vec![1; n];
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for i in 1..n {
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for j in 0..i {
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if nums[i] > nums[j] {
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left[i] = left[i].max(left[j] + 1);
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}
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}
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}
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for i in (0..n - 1).rev() {
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for j in i + 1..n {
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if nums[i] > nums[j] {
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right[i] = right[i].max(right[j] + 1);
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}
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}
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}
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let mut ans = 0;
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for i in 0..n {
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if left[i] > 1 && right[i] > 1 {
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ans = ans.max(left[i] + right[i] - 1);
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}
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}
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(n as i32) - ans
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}
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}

‎solution/1600-1699/1671.Minimum Number of Removals to Make Mountain Array/Solution.ts‎

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function minimumMountainRemovals(nums: number[]): number {
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const n = nums.length;
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const left = newArray(n).fill(1);
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const right = newArray(n).fill(1);
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const left = Array(n).fill(1);
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const right = Array(n).fill(1);
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for (let i = 1; i < n; ++i) {
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for (let j = 0; j < i; ++j) {
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if (nums[i] > nums[j]) {

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