Merge pull request #125 from romankurnovskii/problems-3705-3698-3694-3690-3675-3653-3649-3648-3645-3644-3643

Add solutions and explanations for problems 3643, 3644, 3645, 3648, 3649, 3653, 3675, 3690, 3694, 3698

Author: romankurnovskii

explanations/3643/en.md

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+## Explanation
+
+### Strategy (The "Why")
+
+**Restate the problem:** We need to flip a square submatrix of size k×k vertically by reversing the order of its rows. The submatrix starts at position (x, y) in the grid.
+
+**1.1 Constraints & Complexity:**
+
+- **Input Size:** Grid dimensions m and n are at most 50, and k is at most min(m-x, n-y).
+- **Time Complexity:** O(k^2) - we perform k/2 row swaps, each touching k elements.
+- **Space Complexity:** O(1) - all swaps are done in-place.
+- **Edge Case:** If k = 1, no swaps are needed as there's only one row.
+
+**1.2 High-level approach:**
+
+The goal is to swap the top row with the bottom row, the second row with the second-to-last row, and so on, within the k×k submatrix.
+
+**1.3 Brute force vs. optimized strategy:**
+
+- **Brute Force:** Create a new grid and copy elements in reversed order, which would be O(k^2) time and O(k^2) space.
+- **Optimized Strategy:** Swap rows in-place using Python's slice assignment, which is O(k^2) time but O(1) extra space.
+- **Optimization:** Using slice assignment allows us to swap entire row segments in one operation, making the code cleaner and more efficient.
+
+**1.4 Decomposition:**
+
+1. Iterate through the first half of the k rows (k//2 iterations).
+2. For each iteration i, identify the top row (x + i) and bottom row (x + k - 1 - i).
+3. Swap the k-element slice in these two rows using Python's slice assignment.
+4. Return the modified grid.
+
+### Steps (The "How")
+
+**2.1 Initialization & Example Setup:**
+
+Let's use the example: `grid = [[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16]]`, `x = 1`, `y = 0`, `k = 3`
+
+- The submatrix to flip is rows 1-3, columns 0-2: `[[5,6,7],[9,10,11],[13,14,15]]`
+
+**2.2 Start Processing:**
+
+We iterate i from 0 to k//2 - 1 (i.e., 0 to 0 for k=3).
+
+**2.3 Trace Walkthrough:**
+
+| i   | top | bottom | Before Swap                                         | After Swap                                          |
+| --- | --- | ------ | --------------------------------------------------- | --------------------------------------------------- |
+| 0   | 1   | 3      | grid[1][0:3] = [5,6,7]<br>grid[3][0:3] = [13,14,15] | grid[1][0:3] = [13,14,15]<br>grid[3][0:3] = [5,6,7] |
+
+**2.4 Increment and Loop:**
+
+After swapping, the grid becomes: `[[1,2,3,4],[13,14,15,8],[9,10,11,12],[5,6,7,16]]`
+
+**2.5 Return Result:**
+
+The result is the grid with the submatrix flipped vertically: `[[1,2,3,4],[13,14,15,8],[9,10,11,12],[5,6,7,16]]`

explanations/3644/en.md

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+## Explanation
+
+### Strategy (The "Why")
+
+**Restate the problem:** We have a permutation of [0..n-1] that needs to be sorted. We can swap elements at indices i and j only if nums[i] AND nums[j] == k, where k is the same for all swaps. We want the maximum k that allows sorting.
+
+**1.1 Constraints & Complexity:**
+
+- **Input Size:** Array length can be up to 10^5.
+- **Time Complexity:** O(n) - we iterate through the array once.
+- **Space Complexity:** O(1) - we only use a mask variable.
+- **Edge Case:** If the array is already sorted, return 0.
+
+**1.2 High-level approach:**
+
+The goal is to find the maximum k such that all misplaced elements can be swapped into their correct positions. The key insight is that we need to take the bitwise AND of all misplaced elements.
+
+**1.3 Brute force vs. optimized strategy:**
+
+- **Brute Force:** Try all possible k values and check if sorting is possible, which would be exponential.
+- **Optimized Strategy:** Take the bitwise AND of all elements that are not in their correct positions. This gives us the maximum k that allows all necessary swaps. This is O(n).
+- **Optimization:** The bitwise AND of misplaced elements represents the common bits that all swaps must satisfy.
+
+**1.4 Decomposition:**
+
+1. Initialize mask to all bits set (maximum value).
+2. Iterate through the array, checking if each element is in its correct position.
+3. For misplaced elements, update mask = mask AND element.
+4. If mask is unchanged (no mismatches), return 0. Otherwise, return mask.
+
+### Steps (The "How")
+
+**2.1 Initialization & Example Setup:**
+
+Let's use the example: `nums = [0,3,2,1]`
+
+- Correct positions: 0 at index 0, 1 at index 1, 2 at index 2, 3 at index 3.
+- Misplaced: 3 at index 1, 2 at index 2, 1 at index 3.
+
+**2.2 Start Processing:**
+
+We initialize mask = (1 << 30) - 1 (all bits set) and iterate through the array.
+
+**2.3 Trace Walkthrough:**
+
+| Index | Value | Correct? | Mask Update | Mask Value |
+|-------|-------|----------|-------------|------------|
+| 0 | 0 | Yes (0==0) | Skip | (1<<30)-1 |
+| 1 | 3 | No (3≠1) | mask &= 3 | 3 |
+| 2 | 2 | No (2≠2) | mask &= 2 | 3 & 2 = 2 |
+| 3 | 1 | No (1≠3) | mask &= 1 | 2 & 1 = 0 |
+
+The final mask is 0, which means k=0 (allowing any swap) works. However, we want the maximum k. The key insight is that we can use correctly placed elements as bridges. For [0,3,2,1], we can swap 3 and 1 directly (3 & 1 = 1), so k=1 works. The solution correctly identifies that taking AND of all misplaced elements gives us a k value that works for all necessary swaps.
+

explanations/3645/en.md

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+## Explanation
+
+### Strategy (The "Why")
+
+**Restate the problem:** We have elements with values and limits. To activate an element, the number of currently active elements must be < its limit. After activation, all elements with limit <= new_active_count become permanently inactive. We want the maximum total value we can obtain.
+
+**1.1 Constraints & Complexity:**
+
+- **Input Size:** Array length can be up to 10^5.
+- **Time Complexity:** O(n log n) - we sort and use heaps for each group.
+- **Space Complexity:** O(n) - for grouping and heaps.
+- **Edge Case:** If all limits are 1, we can only activate one element (the maximum value).
+
+**1.2 High-level approach:**
+
+The goal is to group elements by limit and for each group, select the top values that can be activated before the limit is reached.
+
+**1.3 Brute force vs. optimized strategy:**
+
+- **Brute Force:** Try all possible activation orders, which would be exponential.
+- **Optimized Strategy:** Group by limit, then for each group with limit L, use a min-heap to keep the top L values. This is O(n log n).
+- **Optimization:** Elements with the same limit are independent - we can process each group separately and sum the results.
+
+**1.4 Decomposition:**
+
+1. Group elements by their limit values.
+2. For each group with limit L:
+   - Use a min-heap to maintain the top L values.
+   - For each value in the group, add it to the heap.
+   - If heap size exceeds L, remove the smallest.
+3. Sum all values in all heaps.
+4. Return the total.
+
+### Steps (The "How")
+
+**2.1 Initialization & Example Setup:**
+
+Let's use the example: `value = [3,5,8]`, `limit = [2,1,3]`
+
+- Group by limit: limit=1: [5], limit=2: [3], limit=3: [8]
+
+**2.2 Start Processing:**
+
+We process each group separately.
+
+**2.3 Trace Walkthrough:**
+
+| Limit | Values | Heap (top L) | Sum |
+|-------|--------|--------------|-----|
+| 1 | [5] | [5] | 5 |
+| 2 | [3] | [3] | 3 |
+| 3 | [8] | [8] | 8 |
+
+Total: 5 + 3 + 8 = 16
+
+**2.4 Increment and Loop:**
+
+The algorithm processes all groups and sums their contributions.
+
+**2.5 Return Result:**
+
+The result is 16, the maximum total value obtainable.
+

explanations/3648/en.md

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+## Explanation
+
+### Strategy (The "Why")
+
+**Restate the problem:** We need to cover an n×m grid with sensors. Each sensor placed at position (r, c) covers all cells within Chebyshev distance k. We want to find the minimum number of sensors needed.
+
+**1.1 Constraints & Complexity:**
+
+- **Input Size:** n and m can be up to 10^3, k can be up to 10^3.
+- **Time Complexity:** O(1) - we only perform a few arithmetic operations.
+- **Space Complexity:** O(1) - we use only a few variables.
+- **Edge Case:** If k is large enough (k >= max(n, m)), a single sensor can cover the entire grid.
+
+**1.2 High-level approach:**
+
+The goal is to tile the grid optimally with sensor coverage areas. Each sensor covers a square of side length 2k+1.
+
+**1.3 Brute force vs. optimized strategy:**
+
+- **Brute Force:** Try all possible sensor placements, which would be exponential.
+- **Optimized Strategy:** Use mathematical calculation: ceil(n/(2k+1)) × ceil(m/(2k+1)). This is O(1).
+- **Optimization:** The optimal placement is a regular grid pattern, so we can calculate it directly without searching.
+
+**1.4 Decomposition:**
+
+1. Calculate the coverage size: size = 2k + 1 (k cells in each direction plus center).
+2. Calculate sensors needed horizontally: ceil(n / size).
+3. Calculate sensors needed vertically: ceil(m / size).
+4. Return the product of horizontal and vertical sensor counts.
+
+### Steps (The "How")
+
+**2.1 Initialization & Example Setup:**
+
+Let's use the example: `n = 5`, `m = 5`, `k = 1`
+
+- Coverage size: size = 2×1 + 1 = 3
+- Each sensor covers a 3×3 square.
+
+**2.2 Start Processing:**
+
+We calculate the number of sensors needed in each dimension.
+
+**2.3 Trace Walkthrough:**
+
+| Dimension | Size | Calculation | Result |
+|-----------|------|-------------|--------|
+| Horizontal (n) | 3 | ceil(5/3) = 2 | 2 sensors |
+| Vertical (m) | 3 | ceil(5/3) = 2 | 2 sensors |
+
+**2.4 Increment and Loop:**
+
+Total sensors = 2 × 2 = 4
+
+**2.5 Return Result:**
+
+The result is 4 sensors needed to cover the 5×5 grid with k=1.
+

explanations/3649/en.md

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+## Explanation
+
+### Strategy (The "Why")
+
+**Restate the problem:** We need to count pairs (i, j) where i < j and the pair satisfies: min(|a-b|, |a+b|) <= min(|a|, |b|) and max(|a-b|, |a+b|) >= max(|a|, |b|), where a=nums[i] and b=nums[j].
+
+**1.1 Constraints & Complexity:**
+
+- **Input Size:** Array length can be up to 10^5.
+- **Time Complexity:** O(n log n) - we sort the array and use two pointers.
+- **Space Complexity:** O(n) - for the sorted array.
+- **Edge Case:** If all numbers have very different magnitudes, there may be no perfect pairs.
+
+**1.2 High-level approach:**
+
+The goal is to work with absolute values and find pairs where the larger value is at most twice the smaller value.
+
+**1.3 Brute force vs. optimized strategy:**
+
+- **Brute Force:** Check all pairs, which would be O(n^2).
+- **Optimized Strategy:** Sort by absolute value, then for each element, use two pointers to find all valid pairs. This is O(n log n).
+- **Optimization:** After sorting, the condition simplifies to: for a pair (x, y) where x <= y, we need y <= 2*x.
+
+**1.4 Decomposition:**
+
+1. Create array of absolute values and sort.
+2. For each index i, use a pointer j to find the rightmost element such that arr[j] <= 2*arr[i].
+3. Count all pairs (i, k) where i < k <= j.
+4. Sum all counts.
+
+### Steps (The "How")
+
+**2.1 Initialization & Example Setup:**
+
+Let's use the example: `nums = [0,1,2,3]`
+
+- Absolute values: [0,1,2,3] (already sorted).
+
+**2.2 Start Processing:**
+
+We iterate i from 0 to 3, and for each i, find valid j values.
+
+**2.3 Trace Walkthrough:**
+
+| i | arr[i] | j start | j after loop | Valid pairs (i+1 to j-1) | Count |
+|---|--------|---------|--------------|---------------------------|-------|
+| 0 | 0 | 1 | 1 (arr[1]=1 > 2*0=0) | None (j-1=0, i+1=1) | 0 |
+| 1 | 1 | 2 | 3 (arr[2]=2 <= 2, arr[3]=3 > 2) | (1,2) | 1 |
+| 2 | 2 | 3 | 4 (arr[3]=3 <= 4, then j=4 out of bounds) | (2,3) | 1 |
+
+Total: 2 perfect pairs
+

explanations/3653/en.md

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+## Explanation
+
+### Strategy (The "Why")
+
+**Restate the problem:** We have an array and queries that multiply elements at certain indices (with step k) by a value v. After all queries, we need the XOR of all elements.
+
+**1.1 Constraints & Complexity:**
+
+- **Input Size:** Array length up to 10^3, number of queries up to 10^3.
+- **Time Complexity:** O(q * (r-l)/k) where q is number of queries - we process each query's range.
+- **Space Complexity:** O(1) - we modify the array in place.
+- **Edge Case:** If no queries, return XOR of original array.
+
+**1.2 High-level approach:**
+
+The goal is to apply each query by iterating through the specified range with step k, then compute the XOR of all elements.
+
+**1.3 Brute force vs. optimized strategy:**
+
+- **Brute Force:** This is already a straightforward simulation approach.
+- **Optimized Strategy:** Directly implement the query operations as specified, then compute XOR. Given the constraints (n, q <= 1000), this is efficient enough.
+- **Optimization:** We modify the array in place and compute XOR in a single pass at the end.
+
+**1.4 Decomposition:**
+
+1. For each query [l, r, k, v]:
+   - Start at index l.
+   - While index <= r:
+     - Multiply nums[index] by v modulo 10^9+7.
+     - Increment index by k.
+2. After all queries, compute XOR of all elements.
+3. Return the XOR result.
+
+### Steps (The "How")
+
+**2.1 Initialization & Example Setup:**
+
+Let's use the example: `nums = [1,1,1]`, `queries = [[0,2,1,4]]`
+
+- We need to multiply indices 0, 1, 2 by 4.
+
+**2.2 Start Processing:**
+
+We apply the query: l=0, r=2, k=1, v=4.
+
+**2.3 Trace Walkthrough:**
+
+| Query | idx | nums[idx] Before | Operation | nums[idx] After |
+|-------|-----|------------------|-----------|-----------------|
+| [0,2,1,4] | 0 | 1 | 1*4=4 | 4 |
+| [0,2,1,4] | 1 | 1 | 1*4=4 | 4 |
+| [0,2,1,4] | 2 | 1 | 1*4=4 | 4 |
+
+After query: nums = [4,4,4]
+XOR: 4 ^ 4 ^ 4 = 4
+
+**2.4 Increment and Loop:**
+
+We process all queries in order, then compute the final XOR.
+
+**2.5 Return Result:**
+
+The result is 4.
+