花花酱 LeetCode 882. Random Point in Non-overlapping Rectangles

zxi — Sun, 29 Jul 2018 00:45:20 +0000

Problem

Given a list of non-overlapping axis-aligned rectangles rects, write a function pick which randomly and uniformily picks an integer point in the space covered by the rectangles.

Note:

An integer point is a point that has integer coordinates.
A point on the perimeter of a rectangle is included in the space covered by the rectangles.
ith rectangle = rects[i] = [x1,y1,x2,y2], where [x1, y1] are the integer coordinates of the bottom-left corner, and [x2, y2] are the integer coordinates of the top-right corner.
length and width of each rectangle does not exceed 2000.
1 <= rects.length <= 100
pick return a point as an array of integer coordinates [p_x, p_y]
pick is called at most 10000 times.

Example 1:

Input: 
["Solution","pick","pick","pick"]
[[[[1,1,5,5]]],[],[],[]]
Output: 
[null,[4,1],[4,1],[3,3]]

Example 2:

Input: 
["Solution","pick","pick","pick","pick","pick"]
[[[[-2,-2,-1,-1],[1,0,3,0]]],[],[],[],[],[]]
Output: 
[null,[-1,-2],[2,0],[-2,-1],[3,0],[-2,-2]]

Explanation of Input Syntax:

The input is two lists: the subroutines called and their arguments. Solution‘s constructor has one argument, the array of rectangles rects. pick has no arguments. Arguments are always wrapped with a list, even if there aren’t any.

Solution: Binary Search

Same as LeetCode 880. Random Pick with Weight

Use area of the rectangles as weights.

Time complexity: Init: O(n) Pick: O(logn)

Space complexity: O(n)

// Author: Huahua
// Running time: 92 ms
class Solution {
public:
  Solution(vector> rects): rects_(std::move(rects)) {
    sums_ = vector(rects_.size());
    for (int i = 0; i < rects_.size(); ++i) {
      sums_[i] = (rects_[i][2] - rects_[i][0] + 1) * (rects_[i][3] - rects_[i][1] + 1);
      if (i > 0) sums_[i] += sums_[i - 1];
    }
  }

  vector pick() {
    const int s = nextInt(sums_.back()) + 1;
    int index = lower_bound(sums_.begin(), sums_.end(), s) - sums_.begin();
    const auto& rect = rects_[index];
    return {rect[0] + nextInt(rect[2] - rect[0] + 1),
            rect[1] + nextInt(rect[3] - rect[1] + 1)};
  }
    
private:  
  vector sums_;
  vector> rects_;
  
  // Returns a random int in [0, n - 1]
  int nextInt(int n) {
    return rand() / static_cast(RAND_MAX) * n;
  }
};

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花花酱 LeetCode 882. Random Point in Non-overlapping Rectangles

Problem

Solution: Binary Search

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