upd

data/leetcode-problems.json

@@ -25823,5 +25823,21 @@
     "id": 3772,
     "link": "https://leetcode.com/problems/maximum-subgraph-score-in-a-tree/",
     "title": "Maximum Subgraph Score in a Tree"
-  }
+  },
+  "797": {"id": 797, "category": "Graph", "title": "All Paths From Source to Target", "difficulty": "Medium", "link": "https://leetcode.com/problems/all-paths-from-source-to-target/", "tags": []},
+  "1008": {"id": 1008, "category": "Tree", "title": "Construct Binary Search Tree from Preorder Traversal", "difficulty": "Medium", "link": "https://leetcode.com/problems/construct-binary-search-tree-from-preorder-traversal/", "tags": []},
+  "1079": {"id": 1079, "category": "Backtracking", "title": "Letter Tile Possibilities", "difficulty": "Medium", "link": "https://leetcode.com/problems/letter-tile-possibilities/", "tags": []},
+  "1261": {"id": 1261, "category": "Tree", "title": "Find Elements in a Contaminated Binary Tree", "difficulty": "Medium", "link": "https://leetcode.com/problems/find-elements-in-a-contaminated-binary-tree/", "tags": []},
+  "1630": {"id": 1630, "category": "Array", "title": "Arithmetic Subarrays", "difficulty": "Medium", "link": "https://leetcode.com/problems/arithmetic-subarrays/", "tags": []},
+  "2110": {"id": 2110, "category": "Array", "title": "Number of Smooth Descent Periods of a Stock", "difficulty": "Medium", "link": "https://leetcode.com/problems/number-of-smooth-descent-periods-of-a-stock/", "tags": []},
+  "237": {"id": 237, "category": "Linked List", "title": "Delete Node in a Linked List", "difficulty": "Medium", "link": "https://leetcode.com/problems/delete-node-in-a-linked-list/", "tags": []},
+  "2482": {"id": 2482, "category": "Matrix", "title": "Difference Between Ones and Zeros in Row and Column", "difficulty": "Medium", "link": "https://leetcode.com/problems/difference-between-ones-and-zeros-in-row-and-column/", "tags": []},
+  "2785": {"id": 2785, "category": "String", "title": "Sort Vowels in a String", "difficulty": "Medium", "link": "https://leetcode.com/problems/sort-vowels-in-a-string/", "tags": []},
+  "1347": {"id": 1347, "category": "String", "title": "Minimum Number of Steps to Make Two Strings Anagram", "difficulty": "Medium", "link": "https://leetcode.com/problems/minimum-number-of-steps-to-make-two-strings-anagram/", "tags": []},
+  "2326": {"id": 2326, "category": "Matrix", "title": "Spiral Matrix IV", "difficulty": "Medium", "link": "https://leetcode.com/problems/spiral-matrix-iv/", "tags": []},
+  "1823": {"id": 1823, "category": "Math", "title": "Find the Winner of the Circular Game", "difficulty": "Medium", "link": "https://leetcode.com/problems/find-the-winner-of-the-circular-game/", "tags": []},
+  "2120": {"id": 2120, "category": "Simulation", "title": "Execution of All Suffix Instructions Staying in a Grid", "difficulty": "Medium", "link": "https://leetcode.com/problems/execution-of-all-suffix-instructions-staying-in-a-grid/", "tags": []},
+  "2221": {"id": 2221, "category": "Array", "title": "Find Triangular Sum of an Array", "difficulty": "Medium", "link": "https://leetcode.com/problems/find-triangular-sum-of-an-array/", "tags": []},
+  "1605": {"id": 1605, "category": "Matrix", "title": "Find Valid Matrix Given Row and Column Sums", "difficulty": "Medium", "link": "https://leetcode.com/problems/find-valid-matrix-given-row-and-column-sums/", "tags": []},
+  "1551": {"id": 1551, "category": "Math", "title": "Minimum Operations to Make Array Equal", "difficulty": "Medium", "link": "https://leetcode.com/problems/minimum-operations-to-make-array-equal/", "tags": []}
 }

explanations/1008/en.md

@@ -0,0 +1,52 @@
+## Explanation
+
+### Strategy
+
+**Restate the problem**  
+Given preorder traversal of a BST, reconstruct the tree.
+
+**1.1 Constraints & Complexity**  
+- **Input Size:** up to 100 nodes.  
+- **Time Complexity:** O(n) using a stack to place nodes.  
+- **Space Complexity:** O(n) for the stack/tree nodes.  
+- **Edge Case:** Single-node preorder list.
+
+**1.2 High-level approach**  
+Iterate preorder; use a stack of ancestors. Each value smaller than stack top goes left; otherwise pop until finding parent, then attach right.  
+![BST reconstruction from preorder](https://assets.leetcode.com/static_assets/public/images/LeetCode_logo.png)
+
+**1.3 Brute force vs. optimized strategy**  
+- **Brute Force:** Insert each value via BST insert — still O(n²) worst case (sorted input).  
+- **Optimized:** Monotonic stack to place nodes in O(n).
+
+**1.4 Decomposition**  
+1. Create root from first value; push to stack.  
+2. For each next value:  
+   - If value < stack top, set as left child of top.  
+   - Else pop until stack empty or top > value; last popped is parent; attach as right child.  
+3. Push new node to stack.  
+4. Return root.
+
+### Steps
+
+**2.1 Initialization & Example Setup**  
+Example: `[8,5,1,7,10,12]`; root = 8, stack = [8].
+
+**2.2 Start Checking**  
+Process each value, updating stack and children.
+
+**2.3 Trace Walkthrough**  
+| val | Stack before | Action                          | Child  |
+|-----|--------------|---------------------------------|--------|
+| 5   | [8]          | 5 < 8 → left of 8               | left   |
+| 1   | [8,5]        | 1 < 5 → left of 5               | left   |
+| 7   | [8,5,1]      | pop 1,5 (last popped=5) → right | right  |
+| 10  | [8,7]        | pop 7,8 (last popped=8) → right | right  |
+| 12  | [10]         | pop none → right of 10          | right  |
+
+**2.4 Increment and Loop**  
+Continue until all preorder values are attached.
+
+**2.5 Return Result**  
+Root of the constructed BST.
+

explanations/1079/en.md

@@ -0,0 +1,50 @@
+## Explanation
+
+### Strategy
+
+**Restate the problem**  
+Count all non-empty sequences that can be formed from the multiset of tiles (letters), respecting available counts.
+
+**1.1 Constraints & Complexity**  
+- **Input Size:** `1 <= len(tiles) <= 7`.  
+- **Time Complexity:** O(n! * n) in worst case (backtracking over permutations), acceptable for n <= 7.  
+- **Space Complexity:** O(n) recursion depth + O(1) counts.  
+- **Edge Case:** Single tile → exactly 1 sequence.
+
+**1.2 High-level approach**  
+Use backtracking with frequency counts; at each step, pick a letter with remaining count, decrement, recurse, then restore.  
+![Backtracking over letter counts](https://assets.leetcode.com/static_assets/public/images/LeetCode_logo.png)
+
+**1.3 Brute force vs. optimized strategy**  
+- **Brute Force:** Generate all permutations with duplicates and then deduplicate — costly.  
+- **Optimized:** Use counts to avoid generating identical branches; count as we go.
+
+**1.4 Decomposition**  
+1. Build frequency map of letters.  
+2. DFS: for each letter with count > 0, use it (res++), decrement, recurse, restore.  
+3. Sum all counts encountered.  
+4. Return `res`.
+
+### Steps
+
+**2.1 Initialization & Example Setup**  
+Example: `tiles = "AAB"`; counts: A:2, B:1, `res=0`.
+
+**2.2 Start Checking**  
+Try each available letter, recurse with updated counts.
+
+**2.3 Trace Walkthrough**  
+| Path    | Counts after pick | res increment | Notes        |
+|---------|-------------------|---------------|--------------|
+| A       | A:1,B:1           | +1            | recurse more |
+| AA      | A:0,B:1           | +1            | recurse      |
+| AAB     | A:0,B:0           | +1            | dead end     |
+| AB      | A:1,B:0           | +1            | recurse      |
+| ...     | ...               | ...           | ...          |
+
+**2.4 Increment and Loop**  
+Each pick adds 1 to `res` (for the new sequence) before deeper recursion.
+
+**2.5 Return Result**  
+Final `res = 8` for the example.
+

explanations/1261/en.md

@@ -0,0 +1,48 @@
+## Explanation
+
+### Strategy
+
+**Restate the problem**  
+Recover a contaminated binary tree where original root was 0 and children follow `left = 2*x+1`, `right = 2*x+2`. Support queries to check if a target value exists.
+
+**1.1 Constraints & Complexity**  
+- **Input Size:** Up to `1e4` nodes, height <= 20.  
+- **Time Complexity:** O(n) to recover; O(1) average for `find` via set lookup.  
+- **Space Complexity:** O(n) to store recovered values.  
+- **Edge Case:** Single-node tree.
+
+**1.2 High-level approach**  
+DFS from root, assign values by the given formulas, store all in a hash set for O(1) membership.  
+![Tree recovery with value formulas](https://assets.leetcode.com/static_assets/public/images/LeetCode_logo.png)
+
+**1.3 Brute force vs. optimized strategy**  
+- **Brute Force:** Recover on every `find`, re-walking the tree — repeated O(n).  
+- **Optimized:** Recover once, store in a set — O(n) build, O(1) queries.
+
+**1.4 Decomposition**  
+1. DFS from root with value parameter.  
+2. Assign `val`, insert into set.  
+3. Recurse to children with `2*val+1` and `2*val+2`.  
+4. `find` checks membership in the set.
+
+### Steps
+
+**2.1 Initialization & Example Setup**  
+Start at root with value 0, empty set.
+
+**2.2 Start Processing**  
+DFS visits each node, computing and storing values.
+
+**2.3 Trace Walkthrough**  
+| Node | Assigned val | Action        |
+|------|--------------|---------------|
+| root | 0            | add to set    |
+| left | 1            | add to set    |
+| right| 2            | add to set    |
+
+**2.4 Increment and Loop**  
+Continue recursively; each node’s children follow the formula.
+
+**2.5 Return Result**  
+`find(target)` returns `target in set`.
+

explanations/1347/en.md

@@ -0,0 +1,53 @@
+## Explanation
+
+### Strategy
+
+**Restate the problem**  
+Given two equal-length lowercase strings `s` and `t`, find the minimum number of character replacements in `t` to make it an anagram of `s`.
+
+**1.1 Constraints & Complexity**  
+- **Input Size:** up to 5×10^4 characters.  
+- **Time Complexity:** O(n) to count frequencies.  
+- **Space Complexity:** O(1) since alphabet size is fixed (26).  
+- **Edge Case:** Already anagrams → 0 steps.
+
+**1.2 High-level approach**  
+Count letters in both strings; for each character, if `s` needs more than `t` has, that deficit contributes to the answer.  
+![Frequency gap guiding replacements](https://assets.leetcode.com/static_assets/public/images/LeetCode_logo.png)
+
+**1.3 Brute force vs. optimized strategy**  
+- **Brute Force:** Try all replacement combinations — exponential.  
+- **Optimized:** Frequency difference — O(n), straightforward.
+
+**1.4 Decomposition**  
+1. Count frequencies of `s` and `t`.  
+2. For each letter, compute `max(0, count_s - count_t)` and sum.  
+3. That sum is the number of replacements needed in `t`.  
+4. Return the sum.
+
+### Steps
+
+**2.1 Initialization & Example Setup**  
+Example: `s="leetcode"`, `t="practice"`.
+
+**2.2 Start Checking**  
+Build `count_s`, `count_t`.
+
+**2.3 Trace Walkthrough**  
+| Char | count_s | count_t | deficit (if s needs more) |
+|------|---------|---------|----------------------------|
+| e    | 3       | 1       | +2                         |
+| l    | 1       | 0       | +1                         |
+| t    | 1       | 1       | 0                          |
+| c    | 1       | 2       | 0                          |
+| o    | 1       | 0       | +1                         |
+| d    | 1       | 0       | +1                         |
+| ...  | ...     | ...     | ...                        |
+Total = 5.
+
+**2.4 Increment and Loop**  
+Sum deficits over all 26 letters.
+
+**2.5 Return Result**  
+Return the total replacements (5 in the example).
+

explanations/1551/en.md

@@ -0,0 +1,46 @@
+## Explanation
+
+### Strategy
+
+**Restate the problem**  
+Array is `[1,3,5,...,2n-1]`; in one operation, decrement one element and increment another. Find min operations to make all elements equal.
+
+**1.1 Constraints & Complexity**  
+- **Input Size:** `1 <= n <= 1e4`.  
+- **Time Complexity:** O(n) to sum needed moves.  
+- **Space Complexity:** O(1).  
+- **Edge Case:** n=1 → 0 operations.
+
+**1.2 High-level approach**  
+Target value is `n` (the average). Pair symmetric elements; each pair needs `(target - current_lower)` moves. Sum over first half.  
+![Balancing around the center value](https://assets.leetcode.com/static_assets/public/images/LeetCode_logo.png)
+
+**1.3 Brute force vs. optimized strategy**  
+- **Brute Force:** Simulate operations — inefficient.  
+- **Optimized:** Direct formula using symmetry — O(n).
+
+**1.4 Decomposition**  
+1. Target = n.  
+2. For i in [0 .. n/2 - 1], current = 2*i+1.  
+3. Add `target - current` to result.  
+4. Return result.
+
+### Steps
+
+**2.1 Initialization & Example Setup**  
+Example: n=3; array [1,3,5], target=3.
+
+**2.2 Start Checking**  
+Only i=0 in first half: current=1 → need 2 moves.
+
+**2.3 Trace Walkthrough**  
+| i | current | target | add to res |
+|---|---------|--------|------------|
+| 0 | 1       | 3      | 2          |
+
+**2.4 Increment and Loop**  
+Loop over first half (integer division).
+
+**2.5 Return Result**  
+Return accumulated moves (2 for n=3).
+

explanations/1605/en.md

@@ -0,0 +1,51 @@
+## Explanation
+
+### Strategy
+
+**Restate the problem**  
+Construct a non-negative matrix that matches given row sums and column sums.
+
+**1.1 Constraints & Complexity**  
+- **Input Size:** up to 500 rows/cols.  
+- **Time Complexity:** O(m*n) greedy fill.  
+- **Space Complexity:** O(m*n) for the output matrix.  
+- **Edge Case:** Trivial 1x1 matrix equals the shared sum.
+
+**1.2 High-level approach**  
+Greedy: at cell (i,j), place `min(rowSum[i], colSum[j])`, then decrement both sums; proceed row by row.  
+![Greedy fill consuming row/col budgets](https://assets.leetcode.com/static_assets/public/images/LeetCode_logo.png)
+
+**1.3 Brute force vs. optimized strategy**  
+- **Brute Force:** Search all matrices satisfying sums — exponential.  
+- **Optimized:** Greedy works because any feasible solution can allocate the minimal remaining budget without violating totals.
+
+**1.4 Decomposition**  
+1. Initialize `res` with zeros.  
+2. For each cell `(i,j)`:  
+   - `val = min(rowSum[i], colSum[j])`  
+   - Set `res[i][j] = val`, update rowSum/colSum.  
+3. Continue until all cells processed.  
+4. Return `res`.
+
+### Steps
+
+**2.1 Initialization & Example Setup**  
+Example: `rowSum=[3,8]`, `colSum=[4,7]`, res all zeros.
+
+**2.2 Start Checking**  
+At (0,0): min(3,4)=3 → res[0][0]=3; rowSum=[0,8], colSum=[1,7].
+
+**2.3 Trace Walkthrough**  
+| Cell | rowSum before | colSum before | placed | rowSum after | colSum after |
+|------|---------------|---------------|--------|--------------|--------------|
+| (0,0)| 3             | 4             | 3      | 0            | 1            |
+| (0,1)| 0             | 7             | 0      | 0            | 7            |
+| (1,0)| 8             | 1             | 1      | 7            | 0            |
+| (1,1)| 7             | 7             | 7      | 0            | 0            |
+
+**2.4 Increment and Loop**  
+Proceed left-to-right, top-to-bottom.
+
+**2.5 Return Result**  
+Matrix satisfies all row/column sums.
+

explanations/1630/en.md

@@ -0,0 +1,47 @@
+## Explanation
+
+### Strategy
+
+**Restate the problem**  
+For each query `[l_i, r_i]`, decide if the subarray can be rearranged into an arithmetic progression.
+
+**1.1 Constraints & Complexity**  
+- **Input Size:** `n, m <= 500`.  
+- **Time Complexity:** O(m * k log k), where k is subarray length (sort each subarray).  
+- **Space Complexity:** O(k) per query for the sorted copy.  
+- **Edge Case:** Subarray of length 2 is always arithmetic.
+
+**1.2 High-level approach**  
+Extract subarray, sort it, ensure consecutive differences are equal.  
+![Sorted subarray check](https://assets.leetcode.com/static_assets/public/images/LeetCode_logo.png)
+
+**1.3 Brute force vs. optimized strategy**  
+- **Brute Force:** Try all permutations — factorial blowup.  
+- **Optimized:** Sorting then one pass diff check — O(k log k).
+
+**1.4 Decomposition**  
+1. For each query, copy subarray `nums[l:r+1]`.  
+2. Sort it.  
+3. Compute common diff = `sorted[1]-sorted[0]`.  
+4. Verify all consecutive diffs match; if any differ, mark false; else true.
+
+### Steps
+
+**2.1 Initialization & Example Setup**  
+Example: `nums=[4,6,5,9,3,7]`, query `[0,2]` → subarray `[4,6,5]`.
+
+**2.2 Start Checking**  
+Sort → `[4,5,6]`, diff = 1.
+
+**2.3 Trace Walkthrough**  
+| Subarray (sorted) | Step | Diff check          | OK? |
+|-------------------|------|----------------------|-----|
+| [4,5,6]           | 4→5  | 1 == diff            | ✓   |
+|                   | 5→6  | 1 == diff            | ✓   |
+
+**2.4 Increment and Loop**  
+Repeat for each query independently.
+
+**2.5 Return Result**  
+Append boolean for each query into `res`.
+