How do you solve Summary Ranges on LeetCode?

Summary Ranges (LeetCode #228) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The input array is sorted and contains unique integers, which simplifies the logic for detecting ranges.

What is the brute force solution for Summary Ranges?

The brute force approach for Summary Ranges is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking each number in the array and determining if it continues a range or starts a new one. This method is straightforward but inefficient, as it may require multiple passes through the data. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Summary Ranges?

Summary Ranges uses the following concepts: Array. The recommended patterns to study are: Array, Two Pointers.

What are the interview tips for Summary Ranges?

Clearly explain your thought process as you write code. Consider edge cases and test your solution with them. Use meaningful variable names to enhance code readability.

What are common mistakes when solving Summary Ranges?

Not handling edge cases like an empty array. Confusing the start and end of ranges, especially when adding to the results.

#228

Summary Ranges

Easy

ArrayArrayTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a single pass through the array to group numbers into ranges. This approach is efficient because it only requires one iteration, making it faster and more suitable for larger datasets.

⚙️

Algorithm

4 steps

1Step 1: Initialize an empty list to store ranges and set the start of the first range.
2Step 2: Iterate through the array, checking if the current number is consecutive to the previous number.
3Step 3: If it is not consecutive, finalize the current range and start a new one.
4Step 4: After the loop, finalize the last range.

solution.py20 lines

1# Full working Python code
2
3def summaryRanges(nums):
4    if not nums:
5        return []
6    ranges = []
7    start = nums[0]
8    for i in range(1, len(nums)):
9        if nums[i] != nums[i - 1] + 1:
10            if start == nums[i - 1]:
11                ranges.append(str(start))
12            else:
13                ranges.append(f'{start}->{nums[i - 1]}')
14            start = nums[i]
15    if start == nums[-1]:
16        ranges.append(str(start))
17    else:
18        ranges.append(f'{start}->{nums[-1]}')
19    return ranges
20

ℹ

Complexity note: The time complexity is O(n) because we only make a single pass through the array, checking each number once. The space complexity is O(n) for storing the resulting ranges.

1The input array is sorted and contains unique integers, which simplifies the logic for detecting ranges.
2Each range can be identified by checking if the current number is consecutive to the previous one.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.