Leetcode #352: Data Stream as Disjoint Intervals
In this guide, we solve Leetcode #352 Data Stream as Disjoint Intervals in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given a data stream input of non-negative integers a1, a2, ..., an, summarize the numbers seen so far as a list of disjoint intervals. Implement the SummaryRanges class: SummaryRanges() Initializes the object with an empty stream.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Union Find, Design, Hash Table, Binary Search, Data Stream, Ordered Set
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input
["SummaryRanges", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals"]
[[], [1], [], [3], [], [7], [], [2], [], [6], []]
Output
[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]]]
Explanation
SummaryRanges summaryRanges = new SummaryRanges();
summaryRanges.addNum(1); // arr = [1]
summaryRanges.getIntervals(); // return [[1, 1]]
summaryRanges.addNum(3); // arr = [1, 3]
summaryRanges.getIntervals(); // return [[1, 1], [3, 3]]
summaryRanges.addNum(7); // arr = [1, 3, 7]
summaryRanges.getIntervals(); // return [[1, 1], [3, 3], [7, 7]]
summaryRanges.addNum(2); // arr = [1, 2, 3, 7]
summaryRanges.getIntervals(); // return [[1, 3], [7, 7]]
summaryRanges.addNum(6); // arr = [1, 2, 3, 6, 7]
summaryRanges.getIntervals(); // return [[1, 3], [6, 7]]
Python Solution
class SummaryRanges:
def __init__(self):
self.mp = SortedDict()
def addNum(self, val: int) -> None:
n = len(self.mp)
ridx = self.mp.bisect_right(val)
lidx = n if ridx == 0 else ridx - 1
keys = self.mp.keys()
values = self.mp.values()
if (
lidx != n
and ridx != n
and values[lidx][1] + 1 == val
and values[ridx][0] - 1 == val
):
self.mp[keys[lidx]][1] = self.mp[keys[ridx]][1]
self.mp.pop(keys[ridx])
elif lidx != n and val <= values[lidx][1] + 1:
self.mp[keys[lidx]][1] = max(val, self.mp[keys[lidx]][1])
elif ridx != n and val >= values[ridx][0] - 1:
self.mp[keys[ridx]][0] = min(val, self.mp[keys[ridx]][0])
else:
self.mp[val] = [val, val]
def getIntervals(self) -> List[List[int]]:
return list(self.mp.values())
# # Your SummaryRanges object will be instantiated and called as such:
# # obj = SummaryRanges()
# # obj.addNum(val)
# # param_2 = obj.getIntervals()
Complexity
The time complexity is O(n). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.