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‎Problemset/0352-Data Stream as Disjoint Intervals/README.md

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# 352. Data Stream as Disjoint Intervals
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Given a data stream input of non-negative integers <code>a<sub>1</sub>, a<sub>2</sub>, ..., a<sub>n</sub></code>, summarize the numbers seen so far as a list of disjoint intervals.
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Implement the `SummaryRanges` class:
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* `SummaryRanges()` Initializes the object with an empty stream.
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* `void addNum(int value)` Adds the integer `value` to the stream.
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* `int[][] getIntervals()` Returns a summary of the integers in the stream currently as a list of disjoint intervals <code>[start<sub>i</sub>, end<sub>i</sub>]</code>. The answer should be sorted by <code>start<sub>i</sub></code>.
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#### Example 1:
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<pre>
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<strong>Input:</strong>
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["SummaryRanges", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals"]
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[[], [1], [], [3], [], [7], [], [2], [], [6], []]
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<strong>Output:</strong>
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[null, null, [[1, 1]], null, [[1, 1], [3, 3]], null, [[1, 1], [3, 3], [7, 7]], null, [[1, 3], [7, 7]], null, [[1, 3], [6, 7]]]
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<strong>Explanation:</strong>
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SummaryRanges summaryRanges = new SummaryRanges();
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summaryRanges.addNum(1); // arr = [1]
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summaryRanges.getIntervals(); // return [[1, 1]]
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summaryRanges.addNum(3); // arr = [1, 3]
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summaryRanges.getIntervals(); // return [[1, 1], [3, 3]]
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summaryRanges.addNum(7); // arr = [1, 3, 7]
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summaryRanges.getIntervals(); // return [[1, 1], [3, 3], [7, 7]]
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summaryRanges.addNum(2); // arr = [1, 2, 3, 7]
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summaryRanges.getIntervals(); // return [[1, 3], [7, 7]]
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summaryRanges.addNum(6); // arr = [1, 2, 3, 6, 7]
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summaryRanges.getIntervals(); // return [[1, 3], [6, 7]]
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</pre>
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#### Constraints:
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* <code>0 <= value <= 10<sup>4</sup></code>
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* At most <code>3 * 10<sup>4</sup></code> calls will be made to `addNum` and `getIntervals`.
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* At most <code>10<sup>2</sup></code> calls will be made to `getIntervals`.
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**Follow up:** What if there are lots of merges and the number of disjoint intervals is small compared to the size of the data stream?
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## Solutions (Python)
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### 1. Solution
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```Python
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from sortedcontainers import SortedList
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class SummaryRanges:
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def __init__(self):
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self.intervals = SortedList()
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self.nums = set()
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def addNum(self, value: int) -> None:
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if value in self.nums:
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return
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i = self.intervals.bisect_left([value, value])
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self.nums.add(value)
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if value - 1 in self.nums and value + 1 in self.nums:
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x, _ = self.intervals.pop(i - 1)
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_, y = self.intervals.pop(i - 1)
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self.intervals.add([x, y])
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elif value - 1 in self.nums:
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x, _ = self.intervals.pop(i - 1)
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self.intervals.add([x, value])
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elif value + 1 in self.nums:
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_, y = self.intervals.pop(i)
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self.intervals.add([value, y])
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else:
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self.intervals.add([value, value])
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def getIntervals(self) -> List[List[int]]:
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return list(self.intervals)
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# Your SummaryRanges object will be instantiated and called as such:
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# obj = SummaryRanges()
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# obj.addNum(value)
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# param_2 = obj.getIntervals()
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```

‎Problemset/0352-Data Stream as Disjoint Intervals/README_CN.md

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# 352. 将数据流变为多个不相交区间
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给你一个由非负整数 <code>a<sub>1</sub>, a<sub>2</sub>, ..., a<sub>n</sub></code> 组成的数据流输入，请你将到目前为止看到的数字总结为不相交的区间列表。
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实现 `SummaryRanges` 类：
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* `SummaryRanges()` 使用一个空数据流初始化对象。
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* `void addNum(int val)` 向数据流中加入整数 `val` 。
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* `int[][] getIntervals()` 以不相交区间 <code>[start<sub>i</sub>, end<sub>i</sub>]</code> 的列表形式返回对数据流中整数的总结。
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#### 示例 1:
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<pre>
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<strong>输入:</strong>
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["SummaryRanges", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals"]
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[[], [1], [], [3], [], [7], [], [2], [], [6], []]
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<strong>输出:</strong>
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[null, null, [[1, 1]], null, [[1, 1], [3, 3]], null, [[1, 1], [3, 3], [7, 7]], null, [[1, 3], [7, 7]], null, [[1, 3], [6, 7]]]
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<strong>解释:</strong>
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SummaryRanges summaryRanges = new SummaryRanges();
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summaryRanges.addNum(1); // arr = [1]
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summaryRanges.getIntervals(); // 返回 [[1, 1]]
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summaryRanges.addNum(3); // arr = [1, 3]
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summaryRanges.getIntervals(); // 返回 [[1, 1], [3, 3]]
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summaryRanges.addNum(7); // arr = [1, 3, 7]
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summaryRanges.getIntervals(); // 返回 [[1, 1], [3, 3], [7, 7]]
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summaryRanges.addNum(2); // arr = [1, 2, 3, 7]
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summaryRanges.getIntervals(); // 返回 [[1, 3], [7, 7]]
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summaryRanges.addNum(6); // arr = [1, 2, 3, 6, 7]
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summaryRanges.getIntervals(); // 返回 [[1, 3], [6, 7]]
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</pre>
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#### 提示:
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* <code>0 <= value <= 10<sup>4</sup></code>
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* 最多调用 `addNum` 和 `getIntervals` 方法 <code>3 * 10<sup>4</sup></code> 次
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**进阶：**如果存在大量合并，并且与数据流的大小相比，不相交区间的数量很小，该怎么办?
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## 题解 (Python)
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### 1. 题解
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```Python
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from sortedcontainers import SortedList
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class SummaryRanges:
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def __init__(self):
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self.intervals = SortedList()
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self.nums = set()
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def addNum(self, value: int) -> None:
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if value in self.nums:
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return
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i = self.intervals.bisect_left([value, value])
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self.nums.add(value)
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if value - 1 in self.nums and value + 1 in self.nums:
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x, _ = self.intervals.pop(i - 1)
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_, y = self.intervals.pop(i - 1)
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self.intervals.add([x, y])
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elif value - 1 in self.nums:
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x, _ = self.intervals.pop(i - 1)
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self.intervals.add([x, value])
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elif value + 1 in self.nums:
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_, y = self.intervals.pop(i)
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self.intervals.add([value, y])
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else:
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self.intervals.add([value, value])
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def getIntervals(self) -> List[List[int]]:
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return list(self.intervals)
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# Your SummaryRanges object will be instantiated and called as such:
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# obj = SummaryRanges()
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# obj.addNum(value)
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# param_2 = obj.getIntervals()
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```

‎Problemset/0352-Data Stream as Disjoint Intervals/Solution.py

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from sortedcontainers import SortedList
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class SummaryRanges:
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def __init__(self):
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self.intervals = SortedList()
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self.nums = set()
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def addNum(self, value: int) -> None:
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if value in self.nums:
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return
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i = self.intervals.bisect_left([value, value])
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self.nums.add(value)
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if value - 1 in self.nums and value + 1 in self.nums:
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x, _ = self.intervals.pop(i - 1)
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_, y = self.intervals.pop(i - 1)
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self.intervals.add([x, y])
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elif value - 1 in self.nums:
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x, _ = self.intervals.pop(i - 1)
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self.intervals.add([x, value])
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elif value + 1 in self.nums:
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_, y = self.intervals.pop(i)
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self.intervals.add([value, y])
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else:
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self.intervals.add([value, value])
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def getIntervals(self) -> List[List[int]]:
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return list(self.intervals)
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# Your SummaryRanges object will be instantiated and called as such:
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# obj = SummaryRanges()
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# obj.addNum(value)
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# param_2 = obj.getIntervals()

‎README.md

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[347][347l] |[Top K Frequent Elements][347] |![rs]
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[349][349l] |[Intersection of Two Arrays][349] |![rs]
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[350][350l] |[Intersection of Two Arrays II][350] |![rs]
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[352][352l] |[Data Stream as Disjoint Intervals][352] |![py]
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[357][357l] |[Count Numbers with Unique Digits][357] |![rs]
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[365][365l] |[Water and Jug Problem][365] |![rb]
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[367][367l] |[Valid Perfect Square][367] |![rs]
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[347]:Problemset/0347-Top%20K%20Frequent%20Elements/README.md#347-top-k-frequent-elements
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[349]:Problemset/0349-Intersection%20of%20Two%20Arrays/README.md#349-intersection-of-two-arrays
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[350]:Problemset/0350-Intersection%20of%20Two%20Arrays%20II/README.md#350-intersection-of-two-arrays-ii
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[352]:Problemset/0352-Data%20Stream%20as%20Disjoint%20Intervals/README.md#352-data-stream-as-disjoint-intervals
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[357]:Problemset/0357-Count%20Numbers%20with%20Unique%20Digits/README.md#357-count-numbers-with-unique-digits
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[365]:Problemset/0365-Water%20and%20Jug%20Problem/README.md#365-water-and-jug-problem
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[367]:Problemset/0367-Valid%20Perfect%20Square/README.md#367-valid-perfect-square
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[347l]:https://leetcode.com/problems/top-k-frequent-elements/
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[349l]:https://leetcode.com/problems/intersection-of-two-arrays/
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[350l]:https://leetcode.com/problems/intersection-of-two-arrays-ii/
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[352l]:https://leetcode.com/problems/data-stream-as-disjoint-intervals/
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[357l]:https://leetcode.com/problems/count-numbers-with-unique-digits/
27822785
[365l]:https://leetcode.com/problems/water-and-jug-problem/
27832786
[367l]:https://leetcode.com/problems/valid-perfect-square/

‎README_CN.md

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[347][347l] |[前 K 个高频元素][347] |![rs]
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[349][349l] |[两个数组的交集][349] |![rs]
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[350][350l] |[两个数组的交集 II][350] |![rs]
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[352][352l] |[将数据流变为多个不相交区间][352] |![py]
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[357][357l] |[计算各个位数不同的数字个数][357] |![rs]
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[365][365l] |[水壶问题][365] |![rb]
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[367][367l] |[有效的完全平方数][367] |![rs]
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[347]:Problemset/0347-Top%20K%20Frequent%20Elements/README_CN.md#347-前-k-个高频元素
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[349]:Problemset/0349-Intersection%20of%20Two%20Arrays/README_CN.md#349-两个数组的交集
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[350]:Problemset/0350-Intersection%20of%20Two%20Arrays%20II/README_CN.md#350-两个数组的交集-ii
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[352]:Problemset/0352-Data%20Stream%20as%20Disjoint%20Intervals/README_CN.md#352-将数据流变为多个不相交区间
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[357]:Problemset/0357-Count%20Numbers%20with%20Unique%20Digits/README_CN.md#357-计算各个位数不同的数字个数
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[365]:Problemset/0365-Water%20and%20Jug%20Problem/README_CN.md#365-水壶问题
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[367]:Problemset/0367-Valid%20Perfect%20Square/README_CN.md#367-有效的完全平方数
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[347l]:https://leetcode.cn/problems/top-k-frequent-elements/
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[349l]:https://leetcode.cn/problems/intersection-of-two-arrays/
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[350l]:https://leetcode.cn/problems/intersection-of-two-arrays-ii/
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[352l]:https://leetcode.cn/problems/data-stream-as-disjoint-intervals/
27812784
[357l]:https://leetcode.cn/problems/count-numbers-with-unique-digits/
27822785
[365l]:https://leetcode.cn/problems/water-and-jug-problem/
27832786
[367l]:https://leetcode.cn/problems/valid-perfect-square/

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