Add solutions and explanations for problems 2703, 2879, 3512, 2884, 2723, 2627, 3110, 2807, 2894, 2769, 1920, 3760, 3190, 1929, 2888, 300, 322, 373, 380, 399, 427, 433, 502, 530, 637, 909, 3668, 2011, 2942

Author: vikahaze

explanations/1920/en.md

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+## Explanation
+
+### Strategy (The "Why")
+
+**1.1 Constraints & Complexity:**
+- **Input Size:** `1 <= nums.length <= 1000`, `0 <= nums[i] < nums.length`.
+- **Time Complexity:** O(n) where n is the length of nums - we iterate through the array once.
+- **Space Complexity:** O(n) - we create a new array of the same size.
+- **Edge Case:** All elements are distinct and form a valid permutation.
+
+**1.2 High-level approach:**
+The goal is to build an array where each element `ans[i] = nums[nums[i]]`. We iterate through the input array and use each element as an index to access another element.
+
+**1.3 Brute force vs. optimized strategy:**
+- **Brute Force:** Same as optimized - we need to access each element once, which is O(n) time.
+- **Optimized Strategy:** Single pass through the array, building the result array directly.
+
+**1.4 Decomposition:**
+1. Create an empty result array.
+2. Iterate through each index in the input array.
+3. For each index `i`, access `nums[nums[i]]` and append it to the result.
+4. Return the completed result array.
+
+### Steps (The "How")
+
+**2.1 Initialization & Example Setup:**
+Let's use `nums = [0,2,1,5,3,4]`. We initialize `res = []`.
+
+**2.2 Start Checking:**
+We start at index 0 and process each element.
+
+**2.3 Trace Walkthrough:**
+
+| i | nums[i] | nums[nums[i]] | res |
+|---|---------|---------------|-----|
+| 0 | 0 | nums[0] = 0 | [0] |
+| 1 | 2 | nums[2] = 1 | [0,1] |
+| 2 | 1 | nums[1] = 2 | [0,1,2] |
+| 3 | 5 | nums[5] = 4 | [0,1,2,4] |
+| 4 | 3 | nums[3] = 5 | [0,1,2,4,5] |
+| 5 | 4 | nums[4] = 3 | [0,1,2,4,5,3] |
+
+**2.4 Increment and Loop:**
+After processing each index, we move to the next index until we've processed all elements.
+
+**2.5 Return Result:**
+Return `res = [0,1,2,4,5,3]`, which is the built array.
+

explanations/1929/en.md

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+## Explanation
+
+### Strategy (The "Why")
+
+**1.1 Constraints & Complexity:**
+- **Input Size:** `1 <= nums.length <= 1000`.
+- **Time Complexity:** O(n) where n is the length of nums - we create a new array by concatenating.
+- **Space Complexity:** O(n) - we create a new array of size 2n.
+- **Edge Case:** The result array has exactly twice the length of the input.
+
+**1.2 High-level approach:**
+The goal is to create an array that is the concatenation of two copies of the input array. In Python, we can simply use the `+` operator to concatenate lists.
+
+**1.3 Brute force vs. optimized strategy:**
+- **Brute Force:** Same as optimized - list concatenation is already O(n) and optimal.
+- **Optimized Strategy:** Use Python's built-in list concatenation operator.
+
+**1.4 Decomposition:**
+1. Take the input array.
+2. Concatenate it with itself using the `+` operator.
+3. Return the result.
+
+### Steps (The "How")
+
+**2.1 Initialization & Example Setup:**
+Let's use `nums = [1,2,1]`. We want to create `[1,2,1,1,2,1]`.
+
+**2.2 Start Checking:**
+We directly concatenate `nums` with itself: `res = nums + nums`.
+
+**2.3 Trace Walkthrough:**
+
+| Operation | nums | nums + nums | res |
+|-----------|------|-------------|-----|
+| Concatenate | [1,2,1] | [1,2,1] + [1,2,1] | [1,2,1,1,2,1] |
+
+**2.4 Increment and Loop:**
+Not applicable - this is a single operation.
+
+**2.5 Return Result:**
+Return `res = [1,2,1,1,2,1]`, which is the concatenation of two copies of the input array.
+

explanations/2011/en.md

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+## Explanation
+
+### Strategy (The "Why")
+
+**1.1 Constraints & Complexity:**
+- **Input Size:** `1 <= operations.length <= 100`.
+- **Time Complexity:** O(n) where n is the number of operations - we process each operation once.
+- **Space Complexity:** O(1) - we only use a constant amount of space for the result variable.
+- **Edge Case:** All operations could be increments or all could be decrements.
+
+**1.2 High-level approach:**
+The goal is to track the value of variable `X` starting from 0, applying increment or decrement operations. We check each operation string to determine if it's an increment (`++X` or `X++`) or decrement (`--X` or `X--`).
+
+**1.3 Brute force vs. optimized strategy:**
+- **Brute Force:** Same as optimized - we need to check each operation once.
+- **Optimized Strategy:** Single pass through operations, checking the middle character or the operation type.
+
+**1.4 Decomposition:**
+1. Initialize result variable to 0.
+2. Iterate through each operation.
+3. Check if the operation is an increment (contains `++`) or decrement (contains `--`).
+4. Update the result accordingly.
+5. Return the final value.
+
+### Steps (The "How")
+
+**2.1 Initialization & Example Setup:**
+Let's use `operations = ["--X","X++","X++"]`. We initialize `res = 0`.
+
+**2.2 Start Checking:**
+We process each operation in sequence.
+
+**2.3 Trace Walkthrough:**
+
+| Operation | Type | res Before | res After |
+|-----------|------|------------|------------|
+| "--X" | Decrement | 0 | -1 |
+| "X++" | Increment | -1 | 0 |
+| "X++" | Increment | 0 | 1 |
+
+**2.4 Increment and Loop:**
+After processing each operation, we move to the next one until all operations are processed.
+
+**2.5 Return Result:**
+Return `res = 1`, which is the final value after all operations.
+

explanations/2627/en.md

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+## Explanation
+
+### Strategy (The "Why")
+
+**1.1 Constraints & Complexity:**
+- **Input Size:** `0 <= t <= 1000`, `1 <= calls.length <= 10`.
+- **Time Complexity:** O(1) per call - setTimeout/clearTimeout are O(1).
+- **Space Complexity:** O(1) - we store one timeout ID.
+- **Edge Case:** Multiple rapid calls, only the last one executes.
+
+**1.2 High-level approach:**
+The goal is to implement a debounce function that delays execution and cancels previous pending executions. We use setTimeout to delay, and clearTimeout to cancel previous timeouts.
+
+**1.3 Brute force vs. optimized strategy:**
+- **Brute Force:** Same as optimized - debounce pattern is standard.
+- **Optimized Strategy:** Store timeout ID, clear on new call, set new timeout.
+
+**1.4 Decomposition:**
+1. Maintain a timeout ID variable.
+2. When function is called, clear any existing timeout.
+3. Set a new timeout to execute the function after delay `t`.
+4. Return the debounced function.
+
+### Steps (The "How")
+
+**2.1 Initialization & Example Setup:**
+Let's use `t = 50`, calls at times 50 and 75. We want the function to execute only once at time 125.
+
+**2.2 Start Checking:**
+We track timeout IDs and clear/set them appropriately.
+
+**2.3 Trace Walkthrough:**
+
+| Time | Call | Action | Timeout ID | Execution Time |
+|------|------|--------|------------|----------------|
+| 50 | dlog(1) | Set timeout | id1 | 100 (cancelled) |
+| 75 | dlog(2) | Clear id1, set timeout | id2 | 125 |
+
+**2.4 Increment and Loop:**
+Not applicable - this handles individual calls.
+
+**2.5 Return Result:**
+Function executes at time 125 with input 2.
+

explanations/2703/en.md

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-# Problem 2703: Return Length of Arguments Passed
-
-**Difficulty:** Easy  
-**LeetCode Link:** https://leetcode.com/problems/return-length-of-arguments-passed/
-
 ## Explanation
 
 ### Strategy (The "Why")
 
-**Restate the problem:** We need to write a function that counts how many arguments are passed to it. In Python, when arguments are passed as a list, we simply return the length of that list.
-
 **1.1 Constraints & Complexity:**
-
-- **Input Size:** We have at most 100 arguments.
-- **Time Complexity:** $O(1)$ - accessing the length of a list is a constant-time operation.
-- **Space Complexity:** $O(1)$ - we only return an integer, no extra space is used.
-- **Edge Case:** If no arguments are passed (empty list), we return 0.
+- **Input Size:** The function receives a variable number of arguments, where `0 <= args.length <= 100`.
+- **Time Complexity:** O(1) - We simply return the length of the arguments, which is a constant-time operation.
+- **Space Complexity:** O(1) - No additional space is needed beyond the input.
+- **Edge Case:** When no arguments are passed, the function should return 0.
 
 **1.2 High-level approach:**
-
-The goal is to return the number of arguments passed to the function. Since the arguments are provided as a list, we simply return the length of that list.
+The goal is to count how many arguments were passed to the function. In Python, we can use the `*args` syntax to accept a variable number of arguments, and the built-in `len()` function to count them.
 
 **1.3 Brute force vs. optimized strategy:**
-
-- **Brute Force:** Iterate through the arguments and count them manually. This is $O(n)$ time.
-- **Optimized Strategy:** Use the built-in `len()` function which is $O(1)$ time.
-- **Optimization:** Using `len()` is both simpler and more efficient than manual counting.
+- **Brute Force:** Not applicable - this is already the simplest possible solution.
+- **Optimized Strategy:** Directly return the length of the arguments tuple, which is O(1) time complexity.
 
 **1.4 Decomposition:**
-
-1. Receive the arguments as a list parameter.
-2. Return the length of the list using the built-in `len()` function.
+1. Accept variable arguments using `*args` syntax.
+2. Calculate the length of the arguments.
+3. Return the count.
 
 ### Steps (The "How")
 
 **2.1 Initialization & Example Setup:**
+Let's say we call `argumentsLength(5)`. The function receives one argument, so `args = (5,)` (a tuple with one element).
 
-Let's use the example input: `args = [5]`.
-
-- The function receives a list containing one element: `[5]`
-- We need to return the count of arguments, which is 1
-
-**2.2 Start Processing:**
-
-The function receives `args` as a parameter, which is already a list of all arguments.
+**2.2 Start Checking:**
+We initialize `res = 0`, then immediately calculate `res = len(args)`.
 
 **2.3 Trace Walkthrough:**
 
-| Input | `len(args)` | Output |
-|-------|-------------|--------|
-| `[5]` | 1           | 1      |
-| `[{}, null, "3"]` | 3 | 3      |
-| `[]` | 0 | 0      |
+| Call | args | len(args) | res |
+|------|------|-----------|-----|
+| `argumentsLength(5)` | `(5,)` | 1 | 1 |
+| `argumentsLength({}, null, "3")` | `({}, None, "3")` | 3 | 3 |
+| `argumentsLength()` | `()` | 0 | 0 |
 
-**2.4 Return Result:**
-
-After calculating `len(args)`, we return the result directly.
+**2.4 Increment and Loop:**
+Not applicable - this is a single operation.
 
 **2.5 Return Result:**
-
-The function returns the length of the `args` list, which represents the count of arguments passed to the function.
-
+Return the value stored in `res`, which represents the count of arguments passed to the function.

explanations/2723/en.md

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+## Explanation
+
+### Strategy (The "Why")
+
+**1.1 Constraints & Complexity:**
+- **Input Size:** Two promises that resolve with numbers.
+- **Time Complexity:** O(1) - we wait for both promises and add their values.
+- **Space Complexity:** O(1) - constant space.
+- **Edge Case:** Promises might resolve at different times.
+
+**1.2 High-level approach:**
+The goal is to return a new promise that resolves with the sum of two input promises. We use `Promise.all()` to wait for both promises, then add their resolved values.
+
+**1.3 Brute force vs. optimized strategy:**
+- **Brute Force:** Same as optimized - Promise.all() is the standard approach.
+- **Optimized Strategy:** Use async/await with Promise.all() for clean code.
+
+**1.4 Decomposition:**
+1. Wait for both promises to resolve using Promise.all().
+2. Extract the resolved values.
+3. Add them together.
+4. Return the sum.
+
+### Steps (The "How")
+
+**2.1 Initialization & Example Setup:**
+Let's use `promise1` resolves to 2, `promise2` resolves to 5. We want to return a promise that resolves to 7.
+
+**2.2 Start Checking:**
+We await both promises and add their values.
+
+**2.3 Trace Walkthrough:**
+
+| Step | Operation | Result |
+|------|-----------|--------|
+| 1 | await Promise.all([promise1, promise2]) | [2, 5] |
+| 2 | val1 = 2, val2 = 5 | - |
+| 3 | res = 2 + 5 | 7 |
+
+**2.4 Increment and Loop:**
+Not applicable - this is a single async operation.
+
+**2.5 Return Result:**
+Return a promise that resolves to `res = 7`.
+

explanations/2769/en.md

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+## Explanation
+
+### Strategy (The "Why")
+
+**1.1 Constraints & Complexity:**
+- **Input Size:** `1 <= num, t <= 50`.
+- **Time Complexity:** O(1) - We calculate the answer directly using a mathematical formula.
+- **Space Complexity:** O(1) - Only constant space is used.
+- **Edge Case:** When `t = 0`, the maximum achievable number equals `num`.
+
+**1.2 High-level approach:**
+The goal is to find the maximum value of `x` such that after at most `t` operations (where each operation decreases `x` by 1 and increases `num` by 1), `x` becomes equal to `num`. To maximize `x`, we should always decrease `x` and increase `num` in each operation.
+
+**1.3 Brute force vs. optimized strategy:**
+- **Brute Force:** Try all possible values of `x` and simulate the operations - O(t) time.
+- **Optimized Strategy:** Use mathematical reasoning: after `t` operations, `x - t = num + t`, so `x = num + 2*t` - O(1) time.
+
+**1.4 Decomposition:**
+1. Understand that to maximize `x`, we should decrease `x` and increase `num` in each operation.
+2. After `t` operations: `x` becomes `x - t` and `num` becomes `num + t`.
+3. For them to be equal: `x - t = num + t`.
+4. Solve for `x`: `x = num + 2*t`.
+
+### Steps (The "How")
+
+**2.1 Initialization & Example Setup:**
+Let's use `num = 4, t = 1`. We want to find the maximum `x` such that after 1 operation, `x` equals `num`.
+
+**2.2 Start Checking:**
+We apply the formula: `x = num + 2*t = 4 + 2*1 = 6`.
+
+**2.3 Trace Walkthrough:**
+
+| Operation | x | num | Condition |
+|-----------|---|-----|-----------|
+| Initial | 6 | 4 | - |
+| After 1 op | 6-1=5 | 4+1=5 | 5 == 5 ✓ |
+
+**2.4 Increment and Loop:**
+Not applicable - this is a direct calculation.
+
+**2.5 Return Result:**
+Return `res = 6`, which is the maximum achievable value of `x`.
+