Minimum Rectangles to Cover Points

• 1. Sort the points array by x-coordinate. - 2. Create a dp array of size n, where dp[i] is the minimum rectangles to cover points 0 to i. - 3. Iterate i from 0 to n-1. - 4. For each i, iterate j from i down to 0. - 5. If points[i][0] - points[j][0] <= w, update dp[i] with min(dp[i], (j > 0 ? dp[j-1] : 0) + 1). - 6. If the condition fails, break the inner loop as further points won't satisfy it. - 7. The final answer is dp[n-1].

The problem can be solved using dynamic programming after an initial sorting step. The key observation is that the y-coordinates of the points are irrelevant for determining the minimum number of rectangles; only the x-coordinates matter. The width constraint w is the deciding factor for grouping points. The algorithm proceeds as follows:

1. Sort: First, we sort the points array based on the x-coordinates in non-decreasing order. This allows us to process points from left to right.
2. DP State: We define a DP array, dp, where dp[i] stores the minimum number of rectangles required to cover the first i+1 points (i.e., points[0] to points[i]).
3. DP Transition: To calculate dp[i], we consider the i-th point. We must place it in a rectangle. This rectangle can potentially cover previous points as well. We iterate backwards from point i with an index j. As long as points[i][0] - points[j][0] <= w, all points from j to i can be covered by a single new rectangle. The total cost would be 1 (for this new rectangle) plus the cost to cover points up to j-1, which is dp[j-1]. We take the minimum over all valid j. The recurrence relation is: dp[i] = min(dp[j-1] + 1) for all j from 0 to i such that points[i][0] - points[j][0] <= w (with dp[-1] being 0). The final answer is dp[n-1]. Here is the Java implementation:

1import java.util.Arrays;2import java.util.Comparator;3 4class Solution {5    public int minRectanglesToCoverPoints(int[][] points, int w) {6        int n = points.length;7        if (n == 0) {8            return 0;9        }10 11        Arrays.sort(points, Comparator.comparingInt(p -> p[0]));12 13        int[] dp = new int[n];14        // dp[i] = min rectangles to cover points 0...i15 16        for (int i = 0; i < n; i++) {17            // Initialize with the case where point i starts a new rectangle18            // covering only itself, adding to the solution for points 0...i-1.19            dp[i] = (i > 0 ? dp[i-1] : 0) + 1;20 21            // Try to group with previous points22            for (int j = i; j >= 0; j--) {23                if (points[i][0] - points[j][0] <= w) {24                    // Points from j to i can be in one rectangle.25                    // The cost is 1 (for this rectangle) + cost to cover points up to j-1.26                    int prevCost = (j > 0) ? dp[j-1] : 0;27                    dp[i] = Math.min(dp[i], prevCost + 1);28                } else {29                    // Since sorted, no point before j will satisfy the condition30                    break;31                }32            }33        }34        return dp[n-1];35    }36}

• 1. Sort the points array based on the x-coordinate. - 2. If the array is empty, return 0. - 3. Initialize rectangles = 1 and currentRectStartX = points[0][0]. - 4. Iterate through the points starting from the second point (i=1). - 5. If the current point's x-coordinate points[i][0] is greater than currentRectStartX + w, it means the point cannot be covered by the current rectangle. - 6. In that case, increment rectangles and start a new rectangle by updating currentRectStartX to points[i][0]. - 7. After the loop finishes, return the total rectangles count.

This problem has an optimal greedy substructure, which allows for a much more efficient solution than dynamic programming. The core idea remains the same: the problem is one-dimensional, concerning only the x-coordinates of the points. The greedy strategy is as follows:

1. Sort: Sort the points based on their x-coordinates. This is crucial as it allows us to consider points in a left-to-right sweep.
2. Greedy Choice: Iterate through the sorted points. When we encounter the first uncovered point, we must place a new rectangle to cover it. To be maximally efficient (i.e., greedy), we should make this single rectangle cover as many subsequent points as possible.
3. Extend and Cover: We start a new rectangle whose x-range begins at the current point's x-coordinate, let's call it x_start. This rectangle can cover any point p whose x-coordinate p.x is in the range [x_start, x_start + w]. We use this one rectangle to cover the current point and all subsequent points that fall within this range.
4. Iterate: After covering a group of points with one rectangle, we move our consideration to the next point that was not covered and repeat the process until all points are accounted for. This greedy choice is optimal because by extending the rectangle to its maximum possible width, we cover the largest possible set of upcoming points. This choice never prevents us from finding an optimal solution for the remaining points. Here is the Java implementation:

1import java.util.Arrays;2import java.util.Comparator;3 4class Solution {5    public int minRectanglesToCoverPoints(int[][] points, int w) {6        if (points.length == 0) {7            return 0;8        }9 10        // Sort the points based on their x-coordinates11        Arrays.sort(points, Comparator.comparingInt(p -> p[0]));12 13        int rectangles = 1;14        int currentRectStartX = points[0][0];15 16        for (int i = 1; i < points.length; i++) {17            // If the current point's x-coordinate is outside the18            // current rectangle's coverage19            if (points[i][0] > currentRectStartX + w) {20                // We need a new rectangle21                rectangles++;22                // Start the new rectangle from the current point's x-coordinate23                currentRectStartX = points[i][0];24            }25            // Otherwise, the current point is covered by the existing rectangle,26            // so we do nothing and move to the next point.27        }28 29        return rectangles;30    }31}