Data Stream as Disjoint Intervals

We’ll be using Python’s built-in bisect module for efficient insertion and retrieval of elements.

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from bisect import bisect_left, insort

class SummaryRanges:

    def __init__(self):
        self.data = []

    def addNum(self, value: int) -> None:
        index = bisect_left(self.data, value)
        if index == len(self.data) or self.data[index] != value:
            insort(self.data, value)

    def getIntervals(self) -> List[List[int]]:
        intervals = []
        start = end = None
        for num in self.data:
            if start is None:
                start = end = num
            elif num == end + 1:
                end = num
            else:
                intervals.append([start, end])
                start = end = num

        if start is not None:
            intervals.append([start, end])
        return intervals

This code maintains a sorted list of numbers seen so far. When a new number is added, we insert it into the list using binary search. Then, when we need to retrieve the intervals, we iterate through the sorted list and build the intervals based on consecutive numbers.

This approach ensures that the resulting intervals are sorted by starti, and the intervals are disjoint as required. It leverages Python’s standard library to efficiently manage the sorted list, keeping the complexity reasonable for the given constraints.

The problem “352. Data Stream as Disjoint Intervals” involves handling a data stream such that disjoint intervals can be efficiently retrieved.

This problem can be associated with two other problems:

  1. “295. Find Median from Data Stream” is a simpler problem that shares some similarities with “352. Data Stream as Disjoint Intervals”. Both deal with managing data streams. In “295. Find Median from Data Stream”, we need to find the median of the elements in the stream, which can be viewed as a simpler task compared to managing disjoint intervals.

  2. “218. The Skyline Problem” is a more complex problem and it can be seen as an extended version of “352. Data Stream as Disjoint Intervals”. In “218. The Skyline Problem”, we need to find the outer contour (the skyline) of a city represented by buildings at different heights. This involves managing multiple intervals and merging them, similar to how we handle disjoint intervals in “352. Data Stream as Disjoint Intervals”. But “218. The Skyline Problem” introduces an extra level of complexity as it deals with multiple dimensions (heights of buildings).

So, the order of complexity from simple to more complex would be:

  1. “295. Find Median from Data Stream”
  2. “352. Data Stream as Disjoint Intervals”
  3. “218. The Skyline Problem”.

While these problems have distinct goals, their isomorphic relationship lies in their dealing with streaming data and intervals.

The problem “352. Data Stream as Disjoint Intervals” involves the use of data structures to maintain a set of intervals that can change over time. Here are 10 problems of lesser complexity to prepare:

  1. “Two Sum” (LeetCode Problem #1): This problem involves working with numbers in a data structure to find a pair that meets a certain condition.

  2. “Insert Interval” (LeetCode Problem #57): This problem introduces the idea of inserting an interval into a list of sorted intervals.

  3. “Merge Intervals” (LeetCode Problem #56): Here you practice merging overlapping intervals, which is essential for problem 352.

  4. “Find All Duplicates in an Array” (LeetCode Problem #442): This problem can help you understand how to handle arrays and find specific elements in it.

  5. “Range Sum Query - Immutable” (LeetCode Problem #303): It helps in understanding the handling of cumulative sum which could be beneficial in certain cases.

  6. “Contains Duplicate” (LeetCode Problem #217): This problem can help you practice using a data structure to keep track of elements.

  7. “First Unique Character in a String” (LeetCode Problem #387): This problem also involves tracking elements in a data structure.

  8. “Intersection of Two Arrays” (LeetCode Problem #349): This problem involves working with two sets of numbers, and finding their intersection.

  9. “Summary Ranges” (LeetCode Problem #228): This problem has you convert a list of numbers into a list of intervals, similar to the output of problem 352.

  10. “Meeting Rooms II” (LeetCode Problem #253): This problem requires you to handle a list of intervals and determine how many ‘rooms’ are needed.

These problems will give you the chance to practice with interval manipulation, data structure use, and set manipulation, which are key skills for solving problem 352.

Identifying Problem Isomorphism

Can you help me with finding the isomorphism for this problem?

Which problem does this problem map to the corresponding isomorphic problem on Leetcode ?

How did you identify that this problem is a variation of problem?

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from sortedcontainers import SortedList

class SummaryRanges:
    def __init__(self):
        self.sl = SortedList()

    def addNum(self, val: int) -> None:
        n = len(self.sl)
        i = self.sl.bisect_left([val, val])

        if i > 0 and self.sl[i-1][0] <= val <= self.sl[i-1][1] or i < n and self.sl[i][0] <= val <= self.sl[i][1]:
            pass

        elif 0 < i < n and self.sl[i-1][1] == val-1 and self.sl[i][0] == val+1:
            new = [self.sl[i-1][0], self.sl[i][1]]
            del self.sl[i]
            del self.sl[i-1]
            self.sl.add(new)

        elif 0 < i and self.sl[i-1][1] == val-1:
            new = [self.sl[i-1][0], val]
            del self.sl[i-1]
            self.sl.add(new)

        elif i < n and self.sl[i][0] == val+1:
            new = [val, self.sl[i][1]]
            del self.sl[i]
            self.sl.add(new)
        else:
            self.sl.add([val, val])

    def getIntervals(self) -> List[List[int]]:
        return self.sl

Data Stream as Disjoint Intervals - This problem involves handling a data stream and summarizing it into disjoint intervals. This is an Interval Summarization Problem.

Language Agnostic Coding Drills

This problem deals with managing a list of intervals, where intervals are added and retrieved dynamically. This requires an understanding of list manipulations, interval handling, and binary search for efficient operations. The key tasks performed by this code are as follows:

  1. Initial Setup: This is the part where you are initializing your data structure or any other preliminary setup. In this case, an object of the SortedList class is being initialized. This SortedList is used to keep the intervals sorted so that binary search can be performed on the list.

  2. List Operations: Add an element to the SortedList and delete an element from the SortedList.

  3. Interval Handling: Understand how to handle and manipulate intervals. This includes identifying whether a number falls within an existing interval, whether a number extends an existing interval, or whether a number bridges two existing intervals.

  4. Binary Search: It is used to find the possible position of the inserted value in the sorted list. This helps in placing the new interval correctly in the SortedList, thus maintaining the sorted order of intervals.

  5. Conditions and Branching: There are multiple if-elif-else conditions here that control the flow of the logic based on the position of the number in the SortedList.

Now, breaking down the problem-solving approach step by step:

  1. Initialize an object of the SortedList class.

  2. Whenever a new number is added, use binary search to find its possible position in the sorted list.

  3. Check various conditions:

    • If the number is already included in an interval, do nothing.
    • If the number bridges two intervals, merge those two intervals.
    • If the number extends an existing interval, extend that interval.
    • If none of the above conditions are met, add a new interval consisting of just this number.

  4. For retrieving intervals, return the current state of the SortedList as it maintains all the current intervals.

Remember, the order of the conditions is important. The first one to match should be the one executed. If the conditions are rearranged, the code may not function correctly.

Targeted Drills in Python

  1. Initial Setup:

    • Drill 1: Practise initializing and using data structures. Specifically, you can practise initializing an object of the SortedList class.
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    from sortedcontainers import SortedList

sl = SortedList()

  2. List Operations:

    • Drill 2: Practise adding and deleting elements from a list. Use SortedList for this purpose.
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    sl = SortedList()

# Add elements
sl.add(5)
sl.add(3)
sl.add(9)

# Delete elements
sl.remove(3)

  3. Interval Handling:

    • Drill 3: Write a function to check if a number is within a certain interval.
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    def in_interval(number, interval):
    return interval[0] <= number <= interval[1]
    • Drill 4: Write a function to check if a number can extend a certain interval.
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    def can_extend_interval(number, interval):
    return number == interval[0] - 1 or number == interval[1] + 1

  4. Binary Search:

    • Drill 5: Use the SortedList’s bisect_left function to practise binary search. This function can be used to find the possible position of an element in a sorted list.
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    sl = SortedList([1, 2, 3, 4, 5, 6])
position = sl.bisect_left(4)

print(position)  # Output: 3

  5. Conditions and Branching:

    • Drill 6: Practise writing multiple conditions and branching in your code. For this, you can create a basic calculator where user input determines which operation to perform.
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    def basic_calculator(a, b, operation):
    if operation == 'add':
        return a + b
    elif operation == 'subtract':
        return a - b
    elif operation == 'multiply':
        return a * b
    elif operation == 'divide':
        return a / b
    else:
        return "Invalid operation!"

  6. Check if number is in Interval

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    def check_interval(number, interval):
    return interval[0] <= number <= interval[1]

print(check_interval(3, [1, 5]))  # Output: True
print(check_interval(6, [1, 5]))  # Output: False

  7. Perform Interval Merge

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    def merge_intervals(interval1, interval2):
    return [min(interval1[0], interval2[0]), max(interval1[1], interval2[1])]

print(merge_intervals([1, 5], [4, 8]))  # Output: [1, 8]

Each of these drills targets a specific operation or concept used in the code. You can extend these drills and make them more complex as you get comfortable with the operations. For example, for the interval merge, you can later deal with cases where intervals do not overlap.