返回介绍

solution / 1900-1999 / 1943.Describe the Painting / README_EN

发布于 2024-06-17 01:03:12 字数 7010 浏览 0 评论 0 收藏 0

1943. Describe the Painting

中文文档

Description

There is a long and thin painting that can be represented by a number line. The painting was painted with multiple overlapping segments where each segment was painted with a unique color. You are given a 2D integer array segments, where segments[i] = [starti, endi, colori] represents the half-closed segment [starti, endi) with colori as the color.

The colors in the overlapping segments of the painting were mixed when it was painted. When two or more colors mix, they form a new color that can be represented as a set of mixed colors.

  • For example, if colors 2, 4, and 6 are mixed, then the resulting mixed color is {2,4,6}.

For the sake of simplicity, you should only output the sum of the elements in the set rather than the full set.

You want to describe the painting with the minimum number of non-overlapping half-closed segments of these mixed colors. These segments can be represented by the 2D array painting where painting[j] = [leftj, rightj, mixj] describes a half-closed segment [leftj, rightj) with the mixed color sum of mixj.

  • For example, the painting created with segments = [[1,4,5],[1,7,7]] can be described by painting = [[1,4,12],[4,7,7]] because:
    • [1,4) is colored {5,7} (with a sum of 12) from both the first and second segments.
    • [4,7) is colored {7} from only the second segment.

Return _the 2D array _painting_ describing the finished painting (excluding any parts that are not painted). You may return the segments in any order_.

A half-closed segment [a, b) is the section of the number line between points a and b including point a and not including point b.

 

Example 1:

Input: segments = [[1,4,5],[4,7,7],[1,7,9]]
Output: [[1,4,14],[4,7,16]]
Explanation: The painting can be described as follows:
- [1,4) is colored {5,9} (with a sum of 14) from the first and third segments.
- [4,7) is colored {7,9} (with a sum of 16) from the second and third segments.

Example 2:

Input: segments = [[1,7,9],[6,8,15],[8,10,7]]
Output: [[1,6,9],[6,7,24],[7,8,15],[8,10,7]]
Explanation: The painting can be described as follows:
- [1,6) is colored 9 from the first segment.
- [6,7) is colored {9,15} (with a sum of 24) from the first and second segments.
- [7,8) is colored 15 from the second segment.
- [8,10) is colored 7 from the third segment.

Example 3:

Input: segments = [[1,4,5],[1,4,7],[4,7,1],[4,7,11]]
Output: [[1,4,12],[4,7,12]]
Explanation: The painting can be described as follows:
- [1,4) is colored {5,7} (with a sum of 12) from the first and second segments.
- [4,7) is colored {1,11} (with a sum of 12) from the third and fourth segments.
Note that returning a single segment [1,7) is incorrect because the mixed color sets are different.

 

Constraints:

  • 1 <= segments.length <= 2 * 104
  • segments[i].length == 3
  • 1 <= starti < endi <= 105
  • 1 <= colori <= 109
  • Each colori is distinct.

Solutions

Solution 1

class Solution:
  def splitPainting(self, segments: List[List[int]]) -> List[List[int]]:
    d = defaultdict(int)
    for l, r, c in segments:
      d[l] += c
      d[r] -= c
    s = sorted([[k, v] for k, v in d.items()])
    n = len(s)
    for i in range(1, n):
      s[i][1] += s[i - 1][1]
    return [[s[i][0], s[i + 1][0], s[i][1]] for i in range(n - 1) if s[i][1]]
class Solution {
  public List<List<Long>> splitPainting(int[][] segments) {
    TreeMap<Integer, Long> d = new TreeMap<>();
    for (int[] e : segments) {
      int l = e[0], r = e[1], c = e[2];
      d.put(l, d.getOrDefault(l, 0L) + c);
      d.put(r, d.getOrDefault(r, 0L) - c);
    }
    List<List<Long>> ans = new ArrayList<>();
    long i = 0, j = 0;
    long cur = 0;
    for (Map.Entry<Integer, Long> e : d.entrySet()) {
      if (Objects.equals(e.getKey(), d.firstKey())) {
        i = e.getKey();
      } else {
        j = e.getKey();
        if (cur > 0) {
          ans.add(Arrays.asList(i, j, cur));
        }
        i = j;
      }
      cur += e.getValue();
    }
    return ans;
  }
}
class Solution {
public:
  vector<vector<long long>> splitPainting(vector<vector<int>>& segments) {
    map<int, long long> d;
    for (auto& e : segments) {
      int l = e[0], r = e[1], c = e[2];
      d[l] += c;
      d[r] -= c;
    }
    vector<vector<long long>> ans;
    long long i, j, cur = 0;
    for (auto& it : d) {
      if (it == *d.begin())
        i = it.first;
      else {
        j = it.first;
        if (cur > 0) ans.push_back({i, j, cur});
        i = j;
      }
      cur += it.second;
    }
    return ans;
  }
};

如果你对这篇内容有疑问,欢迎到本站社区发帖提问 参与讨论,获取更多帮助,或者扫码二维码加入 Web 技术交流群。

扫码二维码加入Web技术交流群

发布评论

需要 登录 才能够评论, 你可以免费 注册 一个本站的账号。
列表为空,暂无数据
    我们使用 Cookies 和其他技术来定制您的体验包括您的登录状态等。通过阅读我们的 隐私政策 了解更多相关信息。 单击 接受 或继续使用网站,即表示您同意使用 Cookies 和您的相关数据。
    原文