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Commit3a66f08

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Sri Hari: Batch-5/Neetcode-ALL/Added-articles (neetcode-gh#3854)
* Batch-5/Neetcode-ALL/Added-articles* Batch-5/Neetcode-ALL/Added-articles* Batch-5/Neetcode-ALL/Added-articles
1 parent9e68fa1 commit3a66f08

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‎articles/constrained-subsequence-sum.md

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‎articles/find-the-kth-largest-integer-in-the-array.md

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##1. Dynamic Programming (Top-Down)
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::tabs-start
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```python
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classSolution:
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deflongestObstacleCourseAtEachPosition(self,obstacles: List[int]) -> List[int]:
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n=len(obstacles)
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dp= [[-1]* (n+1)for _inrange(n)]
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defdfs(i,prev):
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if i<0:
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return0
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if dp[i][prev]!=-1:
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return dp[i][prev]
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res= dfs(i-1, prev)
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if prev== nor obstacles[prev]>= obstacles[i]:
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res=max(res,1+ dfs(i-1, i))
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dp[i][prev]= res
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return res
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dfs(n-1, n)
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return [1]+ [1+ dp[i-1][i]for iinrange(1, n)]
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```
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```java
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publicclassSolution {
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privateint[][] dp;
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publicint[]longestObstacleCourseAtEachPosition(int[]obstacles) {
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int n= obstacles.length;
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this.dp=newint[n][n+1];
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for (int[] row: dp) {
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Arrays.fill(row,-1);
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}
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dfs(n-1, n, obstacles);
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int[] res=newint[n];
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res[0]=1;
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for (int i=1; i< n; i++) {
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res[i]=1+ dp[i-1][i];
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}
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return res;
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}
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privateintdfs(inti,intprev,int[]obstacles) {
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if (i<0) {
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return0;
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}
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if (dp[i][prev]!=-1) {
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return dp[i][prev];
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}
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int res= dfs(i-1, prev, obstacles);
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if (prev== obstacles.length|| obstacles[prev]>= obstacles[i]) {
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res=Math.max(res,1+ dfs(i-1, i, obstacles));
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}
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return dp[i][prev]= res;
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}
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}
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```
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```cpp
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classSolution {
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public:
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vector<vector<int>> dp;
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vector<int> longestObstacleCourseAtEachPosition(vector<int>& obstacles) {
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int n = obstacles.size();
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this->dp = vector<vector<int>>(n, vector<int>(n + 1, -1));
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dfs(n - 1, n, obstacles);
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vector<int> res(n, 1);
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for (int i = 1; i < n; i++) {
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res[i] = 1 + dp[i - 1][i];
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}
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return res;
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}
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private:
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intdfs(int i, int prev, vector<int>& obstacles) {
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if (i < 0) {
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return 0;
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}
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if (dp[i][prev] != -1) {
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return dp[i][prev];
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}
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int res = dfs(i - 1, prev, obstacles);
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if (prev == obstacles.size() || obstacles[prev] >= obstacles[i]) {
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res = max(res, 1 + dfs(i - 1, i, obstacles));
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}
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return dp[i][prev] = res;
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}
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};
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```
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```javascript
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class Solution {
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/**
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* @param {number[]} obstacles
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* @return {number[]}
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*/
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longestObstacleCourseAtEachPosition(obstacles) {
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const n = obstacles.length;
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const dp = Array.from({ length: n }, () => new Array(n + 1).fill(-1));
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const dfs = (i, prev) => {
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if (i < 0) {
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return 0;
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}
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if (dp[i][prev] !== -1) {
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return dp[i][prev];
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}
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let res = dfs(i - 1, prev);
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if (prev === n || obstacles[prev] >= obstacles[i]) {
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res = Math.max(res, 1 + dfs(i - 1, i));
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}
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dp[i][prev] = res;
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return res;
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};
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dfs(n - 1, n);
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const res = new Array(n).fill(1);
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for (let i = 1; i < n; i++) {
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res[i] = 1 + dp[i - 1][i];
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}
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return res;
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}
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}
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```
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::tabs-end
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###Time & Space Complexity
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* Time complexity: $O(n ^ 2)$
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* Space complexity: $O(n ^ 2)$
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---
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##2. Dynamic Programming (Binary Search) - I
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::tabs-start
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```python
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classSolution:
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deflongestObstacleCourseAtEachPosition(self,obstacles: List[int]) -> List[int]:
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res= []
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dp= [10**8]* (len(obstacles)+1)
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for numin obstacles:
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index= bisect.bisect(dp, num)
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res.append(index+1)
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dp[index]= num
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return res
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```
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```java
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publicclassSolution {
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publicint[]longestObstacleCourseAtEachPosition(int[]obstacles) {
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int n= obstacles.length;
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int[] res=newint[n];
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int[] dp=newint[n+1];
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Arrays.fill(dp, (int)1e8);
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for (int i=0; i< n; i++) {
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int index= upperBound(dp, obstacles[i]);
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res[i]= index+1;
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dp[index]= obstacles[i];
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}
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return res;
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}
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privateintupperBound(int[]dp,inttarget) {
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int left=0, right= dp.length;
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while (left< right) {
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int mid= left+ (right- left)/2;
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if (dp[mid]> target) {
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right= mid;
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}else {
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left= mid+1;
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}
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}
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return left;
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}
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}
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```
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```cpp
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classSolution {
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public:
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vector<int> longestObstacleCourseAtEachPosition(vector<int>& obstacles) {
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int n = obstacles.size();
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vector<int> res(n);
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vector<int> dp(n + 1, 1e8);
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for (int i = 0; i < n; i++) {
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int index = upper_bound(dp.begin(), dp.end(), obstacles[i]) - dp.begin();
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res[i] = index + 1;
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dp[index] = obstacles[i];
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}
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return res;
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}
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};
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```
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```javascript
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classSolution {
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/**
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*@param{number[]}obstacles
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*@return{number[]}
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*/
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longestObstacleCourseAtEachPosition(obstacles) {
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let n=obstacles.length;
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let res=newArray(n).fill(0);
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let dp=newArray(n+1).fill(1e8);
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constupperBound= (dp,target)=> {
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let left=0, right=dp.length;
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while (left< right) {
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let mid=Math.floor((left+ right)/2);
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if (dp[mid]> target) {
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right= mid;
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}else {
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left= mid+1;
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}
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}
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return left;
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};
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for (let i=0; i< n; i++) {
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let index=upperBound(dp, obstacles[i]);
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res[i]= index+1;
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dp[index]= obstacles[i];
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}
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return res;
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}
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}
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```
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::tabs-end
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###Time & Space Complexity
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* Time complexity: $O(n \log n)$
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* Space complexity: $O(n)$
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---
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##3. Dynamic Programming (Binary Search) - II
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::tabs-start
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```python
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classSolution:
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deflongestObstacleCourseAtEachPosition(self,obstacles: List[int]) -> List[int]:
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res= []
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dp= []
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for numin obstacles:
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index= bisect.bisect_right(dp, num)
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res.append(index+1)
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if index==len(dp):
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dp.append(num)
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else:
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dp[index]= num
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return res
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```
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```java
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publicclassSolution {
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publicint[]longestObstacleCourseAtEachPosition(int[]obstacles) {
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int n= obstacles.length;
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int[] res=newint[n];
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List<Integer> dp=newArrayList<>();
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for (int i=0; i< n; i++) {
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int index= upperBound(dp, obstacles[i]);
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res[i]= index+1;
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if (index== dp.size()) {
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dp.add(obstacles[i]);
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}else {
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dp.set(index, obstacles[i]);
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}
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}
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return res;
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}
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privateintupperBound(List<Integer>dp,inttarget) {
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int left=0, right= dp.size();
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while (left< right) {
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int mid= left+ (right- left)/2;
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if (dp.get(mid)> target) {
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right= mid;
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}else {
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left= mid+1;
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}
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}
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return left;
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}
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}
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```
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```cpp
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classSolution {
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public:
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vector<int> longestObstacleCourseAtEachPosition(vector<int>& obstacles) {
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int n = obstacles.size();
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vector<int> res(n);
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vector<int> dp;
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for (int i = 0; i < n; i++) {
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int index = upper_bound(dp.begin(), dp.end(), obstacles[i]) - dp.begin();
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res[i] = index + 1;
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if (index == dp.size()) {
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dp.push_back(obstacles[i]);
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} else {
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dp[index] = obstacles[i];
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}
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}
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return res;
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}
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};
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```
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```javascript
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classSolution {
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/**
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*@param{number[]}obstacles
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*@return{number[]}
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*/
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longestObstacleCourseAtEachPosition(obstacles) {
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let n=obstacles.length;
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let res=newArray(n).fill(0);
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let dp= [];
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constupperBound= (dp,target)=> {
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let left=0, right=dp.length;
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while (left< right) {
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let mid=Math.floor((left+ right)/2);
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if (dp[mid]> target) {
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right= mid;
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}else {
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left= mid+1;
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}
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}
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return left;
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};
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for (let i=0; i< n; i++) {
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let index=upperBound(dp, obstacles[i]);
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res[i]= index+1;
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if (index===dp.length) {
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dp.push(obstacles[i]);
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}else {
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dp[index]= obstacles[i];
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}
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}
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return res;
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}
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}
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```
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::tabs-end
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###Time & Space Complexity
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* Time complexity: $O(n \log n)$
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* Space complexity: $O(n)$

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