Easiest Solution 99% beats
Intuition\nThe initial thought behind solving this problem is to find the minimum number of rectangles required to cover all given points, assuming each rectang
### Intuition\nThe initial thought behind solving this problem is to find the minimum number of rectangles required to cover all given points, assuming each rectangle has a width `w` and extends vertically to cover all points horizontally aligned with it. The intuition is to sort the x-coordinates of the points and then iterate through them to determine the minimum number of vertical ranges needed to cover all points, considering the width constraint.\n\n### Approach\n1. **Collect X-coordinates**: First, collect all the x-coordinates of the points into a vector. This step assumes that the points are aligned horizontally, and each x-coordinate represents a potential starting point for a covering rectangle.\n   \n2. **Sort X-coordinates**: Sort the collected x-coordinates. Sorting allows us to process the points from left to right, which is crucial for determining the minimum number of vertical ranges needed to cover all points.\n\n3. **Determine Minimum Number of Rectangles**:\n   - Initialize a variable `minima` to hold the maximum x-coordinate encountered so far and set it to the smallest x-coordinate after sorting. This variable represents the bottom edge of the current rectangle being considered.\n   - Initialize a counter `count` to 1, representing the minimum number of rectangles needed to cover the points.\n   - Iterate through the sorted x-coordinates. For each coordinate, check if adding `w` to the current `minima` exceeds the next x-coordinate. If it does, increment `minima` to the current x-coordinate and increment `count` by 1. This indicates that a new rectangle is needed to cover the points up to the current x-coordinate.\n   - Continue iterating until all x-coordinates have been processed.\n\n### Complexity Analysis\n- **Time Complexity**: The time complexity of this algorithm is dominated by the sorting operation, which requires \\(O(n \\log n)\\) time, where `n` is the number of points. Additionally, the linear iteration through the sorted coordinates contributes \\(O(n)\\) time complexity. Therefore, the overall time complexity is \\(O(n \\log n)\\).\n\n- **Space Complexity**: The space complexity arises from storing the x-coordinates in a vector. Since the vector stores `n` integers, the space complexity is \\(O(n)\\).\n\n### Code Explanation\nThe provided code snippet implements the described approach. It starts by collecting and sorting the x-coordinates of the points. Then, it iterates through the sorted coordinates to determine the minimum number of rectangles needed to cover all points, considering the width constraint. The algorithm efficiently finds the minimum number of vertical ranges by dynamically adjusting the bottom edge of each covering rectangle based on the width and the positions of the points.\n\nCitations:\n\n\n# Code\n```\nclass Solution {\npublic:\n    int minRectanglesToCoverPoints(vector<vector<int>>& points, int w) {\n        vector<int> direcX;\n        for (auto i = 0; i < points.size(); i++) {\n            direcX.push_back(points[i][0]);\n        }\n        sort(direcX.begin(), direcX.end());\n        int minima = direcX[0];\n        int size = direcX.size();\n        int i = 0;\n        int count = 1;\n        while (i < size) {\n            if ((minima + w) < direcX[i]) {\n                minima=direcX[i];\n                count++;\n            }\n            i++;\n        }\n        return count;\n    }\n};\n```
['Array', 'Greedy', 'Sorting', 'C++']