Sri Hari: Batch-5/Neetcode-ALL/Added-articles (#3854)

* Batch-5/Neetcode-ALL/Added-articles

* Batch-5/Neetcode-ALL/Added-articles

* Batch-5/Neetcode-ALL/Added-articles

Author: Srihari2222

Diff for: articles/constrained-subsequence-sum.md

@@ -0,0 +1,387 @@
+## 1. Dynamic Programming (Top-Down)
+
+::tabs-start
+
+```python
+class Solution:
+    def longestObstacleCourseAtEachPosition(self, obstacles: List[int]) -> List[int]:
+        n = len(obstacles)
+        dp = [[-1] * (n + 1) for _ in range(n)]
+
+        def dfs(i, prev):
+            if i < 0:
+                return 0
+            if dp[i][prev] != -1:
+                return dp[i][prev]
+
+            res = dfs(i - 1, prev)
+            if prev == n or obstacles[prev] >= obstacles[i]:
+                res = max(res, 1 + dfs(i - 1, i))
+            dp[i][prev] = res
+            return res
+
+        dfs(n - 1, n)
+        return [1] + [1 + dp[i - 1][i] for i in range(1, n)]
+```
+
+```java
+public class Solution {
+    private int[][] dp;
+
+    public int[] longestObstacleCourseAtEachPosition(int[] obstacles) {
+        int n = obstacles.length;
+        this.dp = new int[n][n + 1];
+        for (int[] row : dp) {
+            Arrays.fill(row, -1);
+        }
+
+        dfs(n - 1, n, obstacles);
+
+        int[] res = new int[n];
+        res[0] = 1;
+        for (int i = 1; i < n; i++) {
+            res[i] = 1 + dp[i - 1][i];
+        }
+        return res;
+    }
+
+    private int dfs(int i, int prev, int[] obstacles) {
+        if (i < 0) {
+            return 0;
+        }
+        if (dp[i][prev] != -1) {
+            return dp[i][prev];
+        }
+
+        int res = dfs(i - 1, prev, obstacles);
+        if (prev == obstacles.length || obstacles[prev] >= obstacles[i]) {
+            res = Math.max(res, 1 + dfs(i - 1, i, obstacles));
+        }
+        return dp[i][prev] = res;
+    }
+}
+```
+
+```cpp
+class Solution {
+public:
+    vector<vector<int>> dp;
+
+    vector<int> longestObstacleCourseAtEachPosition(vector<int>& obstacles) {
+        int n = obstacles.size();
+        this->dp = vector<vector<int>>(n, vector<int>(n + 1, -1));
+
+        dfs(n - 1, n, obstacles);
+
+        vector<int> res(n, 1);
+        for (int i = 1; i < n; i++) {
+            res[i] = 1 + dp[i - 1][i];
+        }
+        return res;
+    }
+
+private:
+    int dfs(int i, int prev, vector<int>& obstacles) {
+        if (i < 0) {
+            return 0;
+        }
+        if (dp[i][prev] != -1) {
+            return dp[i][prev];
+        }
+
+        int res = dfs(i - 1, prev, obstacles);
+        if (prev == obstacles.size() || obstacles[prev] >= obstacles[i]) {
+            res = max(res, 1 + dfs(i - 1, i, obstacles));
+        }
+        return dp[i][prev] = res;
+    }
+};
+```
+
+```javascript
+class Solution {
+    /**
+     * @param {number[]} obstacles
+     * @return {number[]}
+     */
+    longestObstacleCourseAtEachPosition(obstacles) {
+        const n = obstacles.length;
+        const dp = Array.from({ length: n }, () => new Array(n + 1).fill(-1));
+
+        const dfs = (i, prev) => {
+            if (i < 0) {
+                return 0;
+            }
+            if (dp[i][prev] !== -1) {
+                return dp[i][prev];
+            }
+
+            let res = dfs(i - 1, prev);
+            if (prev === n || obstacles[prev] >= obstacles[i]) {
+                res = Math.max(res, 1 + dfs(i - 1, i));
+            }
+            dp[i][prev] = res;
+            return res;
+        };
+
+        dfs(n - 1, n);
+
+        const res = new Array(n).fill(1);
+        for (let i = 1; i < n; i++) {
+            res[i] = 1 + dp[i - 1][i];
+        }
+        return res;
+    }
+}
+```
+
+::tabs-end
+
+### Time & Space Complexity
+
+* Time complexity: $O(n ^ 2)$
+* Space complexity: $O(n ^ 2)$
+
+---
+
+## 2. Dynamic Programming (Binary Search) - I
+
+::tabs-start
+
+```python
+class Solution:
+    def longestObstacleCourseAtEachPosition(self, obstacles: List[int]) -> List[int]:
+        res = []
+        dp = [10**8] * (len(obstacles) + 1)
+
+        for num in obstacles:
+            index = bisect.bisect(dp, num)
+            res.append(index + 1)
+            dp[index] = num
+
+        return res
+```
+
+```java
+public class Solution {
+    public int[] longestObstacleCourseAtEachPosition(int[] obstacles) {
+        int n = obstacles.length;
+        int[] res = new int[n];
+        int[] dp = new int[n + 1];
+        Arrays.fill(dp, (int) 1e8);
+
+        for (int i = 0; i < n; i++) {
+            int index = upperBound(dp, obstacles[i]);
+            res[i] = index + 1;
+            dp[index] = obstacles[i];
+        }
+
+        return res;
+    }
+
+    private int upperBound(int[] dp, int target) {
+        int left = 0, right = dp.length;
+        while (left < right) {
+            int mid = left + (right - left) / 2;
+            if (dp[mid] > target) {
+                right = mid;
+            } else {
+                left = mid + 1;
+            }
+        }
+        return left;
+    }
+}
+```
+
+```cpp
+class Solution {
+public:
+    vector<int> longestObstacleCourseAtEachPosition(vector<int>& obstacles) {
+        int n = obstacles.size();
+        vector<int> res(n);
+        vector<int> dp(n + 1, 1e8);
+
+        for (int i = 0; i < n; i++) {
+            int index = upper_bound(dp.begin(), dp.end(), obstacles[i]) - dp.begin();
+            res[i] = index + 1;
+            dp[index] = obstacles[i];
+        }
+
+        return res;
+    }
+};
+```
+
+```javascript
+class Solution {
+    /**
+     * @param {number[]} obstacles
+     * @return {number[]}
+     */
+    longestObstacleCourseAtEachPosition(obstacles) {
+        let n = obstacles.length;
+        let res = new Array(n).fill(0);
+        let dp = new Array(n + 1).fill(1e8);
+
+        const upperBound = (dp, target) => {
+            let left = 0, right = dp.length;
+            while (left < right) {
+                let mid = Math.floor((left + right) / 2);
+                if (dp[mid] > target) {
+                    right = mid;
+                } else {
+                    left = mid + 1;
+                }
+            }
+            return left;
+        };
+
+        for (let i = 0; i < n; i++) {
+            let index = upperBound(dp, obstacles[i]);
+            res[i] = index + 1;
+            dp[index] = obstacles[i];
+        }
+
+        return res;
+    }
+}
+```
+
+::tabs-end
+
+### Time & Space Complexity
+
+* Time complexity: $O(n \log n)$
+* Space complexity: $O(n)$
+
+---
+
+## 3. Dynamic Programming (Binary Search) - II
+
+::tabs-start
+
+```python
+class Solution:
+    def longestObstacleCourseAtEachPosition(self, obstacles: List[int]) -> List[int]:
+        res = []
+        dp = []
+
+        for num in obstacles:
+            index = bisect.bisect_right(dp, num)
+            res.append(index + 1)
+
+            if index == len(dp):
+                dp.append(num)
+            else:
+                dp[index] = num
+
+        return res
+```
+
+```java
+public class Solution {
+    public int[] longestObstacleCourseAtEachPosition(int[] obstacles) {
+        int n = obstacles.length;
+        int[] res = new int[n];
+        List<Integer> dp = new ArrayList<>();
+
+        for (int i = 0; i < n; i++) {
+            int index = upperBound(dp, obstacles[i]);
+            res[i] = index + 1;
+
+            if (index == dp.size()) {
+                dp.add(obstacles[i]);
+            } else {
+                dp.set(index, obstacles[i]);
+            }
+        }
+
+        return res;
+    }
+
+    private int upperBound(List<Integer> dp, int target) {
+        int left = 0, right = dp.size();
+        while (left < right) {
+            int mid = left + (right - left) / 2;
+            if (dp.get(mid) > target) {
+                right = mid;
+            } else {
+                left = mid + 1;
+            }
+        }
+        return left;
+    }
+}
+```
+
+```cpp
+class Solution {
+public:
+    vector<int> longestObstacleCourseAtEachPosition(vector<int>& obstacles) {
+        int n = obstacles.size();
+        vector<int> res(n);
+        vector<int> dp;
+
+        for (int i = 0; i < n; i++) {
+            int index = upper_bound(dp.begin(), dp.end(), obstacles[i]) - dp.begin();
+            res[i] = index + 1;
+
+            if (index == dp.size()) {
+                dp.push_back(obstacles[i]);
+            } else {
+                dp[index] = obstacles[i];
+            }
+        }
+
+        return res;
+    }
+};
+```
+
+```javascript
+class Solution {
+    /**
+     * @param {number[]} obstacles
+     * @return {number[]}
+     */
+    longestObstacleCourseAtEachPosition(obstacles) {
+        let n = obstacles.length;
+        let res = new Array(n).fill(0);
+        let dp = [];
+
+        const upperBound = (dp, target) => {
+            let left = 0, right = dp.length;
+            while (left < right) {
+                let mid = Math.floor((left + right) / 2);
+                if (dp[mid] > target) {
+                    right = mid;
+                } else {
+                    left = mid + 1;
+                }
+            }
+            return left;
+        };
+
+        for (let i = 0; i < n; i++) {
+            let index = upperBound(dp, obstacles[i]);
+            res[i] = index + 1;
+
+            if (index === dp.length) {
+                dp.push(obstacles[i]);
+            } else {
+                dp[index] = obstacles[i];
+            }
+        }
+
+        return res;
+    }
+}
+```
+
+::tabs-end
+
+### Time & Space Complexity
+
+* Time complexity: $O(n \log n)$
+* Space complexity: $O(n)$