Leetcode #2655: Find Maximal Uncovered Ranges
In this guide, we solve Leetcode #2655 Find Maximal Uncovered Ranges 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
You are given an integer n which is the length of a 0-indexed array nums, and a 0-indexed 2D-array ranges, which is a list of sub-ranges of nums (sub-ranges may overlap). Each row ranges[i] has exactly 2 cells: ranges[i][0], which shows the start of the ith range (inclusive) ranges[i][1], which shows the end of the ith range (inclusive) These ranges cover some cells of nums and leave some cells uncovered.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Array, Sorting
Intuition
Sorting reveals structure that is hard to see in the original order.
Once sorted, a linear scan is usually enough to compute the answer.
Approach
Sort the data and sweep through it while maintaining a small state.
This keeps the logic straightforward and reliable.
Steps:
- Sort the data.
- Scan in order while maintaining state.
- Update the best answer.
Example
Input: n = 10, ranges = [[3,5],[7,8]]
Output: [[0,2],[6,6],[9,9]]
Explanation: The ranges (3, 5) and (7, 8) are covered, so if we simplify the array nums to a binary array where 0 shows an uncovered cell and 1 shows a covered cell, the array becomes [0,0,0,1,1,1,0,1,1,0] in which we can observe that the ranges (0, 2), (6, 6) and (9, 9) aren't covered.
Python Solution
class Solution:
def findMaximalUncoveredRanges(
self, n: int, ranges: List[List[int]]
) -> List[List[int]]:
ranges.sort()
last = -1
ans = []
for l, r in ranges:
if last + 1 < l:
ans.append([last + 1, l - 1])
last = max(last, r)
if last + 1 < n:
ans.append([last + 1, n - 1])
return ans
Complexity
The time complexity is , and the space complexity is , where is the length of the array . The space complexity is , where is the length of the array .
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.