#228

Easy

auto_awesomeArray-driven solution strategy

LeetCode Problem Workspace

Summary Ranges

Summary Ranges involves converting a sorted array of unique integers into a minimal list of range strings.

Problem Statement

You are given a sorted array of unique integers, nums. Your task is to convert this array into a sorted list of range strings that cover all the elements in the array.

A range is defined as a continuous sequence of integers, and for each sequence, you should return it in the format 'start->end' if the range has more than one element, or just 'start' if it consists of a single element.

Examples

Example 1

Input: nums = [0,1,2,4,5,7]

Output: ["0->2","4->5","7"]

The ranges are: [0,2] --> "0->2" [4,5] --> "4->5" [7,7] --> "7"

Example 2

Input: nums = [0,2,3,4,6,8,9]

Output: ["0","2->4","6","8->9"]

The ranges are: [0,0] --> "0" [2,4] --> "2->4" [6,6] --> "6" [8,9] --> "8->9"

Constraints

0 <= nums.length <= 20
-231 <= nums[i] <= 231 - 1
All the values of nums are unique.
nums is sorted in ascending order.

Solution Approach

Iterate through the array

You can iterate through the given array and identify consecutive numbers. For each group of consecutive numbers, create a range. If the numbers are not consecutive, close the previous range and start a new one.

Handle edge cases

Make sure to account for edge cases such as when the array is empty or has only one element. Also, ensure that the output is correctly formatted when there's a single number in a range.

Optimize for time complexity

While a simple iteration will work, you should consider how the solution scales with increasing array sizes. Optimizing the approach for time and space complexity is essential, given the constraint of up to 20 elements.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the iteration over the array, which is O(n) where n is the number of elements in nums. The space complexity is O(n) because we need to store the output list of ranges, which could be as large as the input array in the worst case.

What Interviewers Usually Probe

Checks the candidate's ability to optimize array traversal.
Evaluates how the candidate handles edge cases in a sorted array.
Assesses the candidate's approach to time and space complexity trade-offs.

Common Pitfalls or Variants

Common pitfalls

Failing to handle single-element ranges correctly.
Not optimizing the solution when scaling the array size.
Misinterpreting the problem as needing a brute-force solution instead of an efficient array-driven approach.

Follow-up variants

Allow duplicates in the array and handle them.
Modify the problem to deal with unsorted input arrays.
Implement the problem with additional constraints like a maximum range size.

FAQ

What is the time complexity of the Summary Ranges problem?

The time complexity is O(n) as you iterate through the array once, where n is the number of elements in nums.

How do I handle edge cases in the Summary Ranges problem?

Edge cases include when the array is empty, has a single element, or has no consecutive numbers. Ensure you cover these cases in your solution.

Can I solve the Summary Ranges problem without sorting the array?

Yes, the problem guarantees that the input array is already sorted, so no additional sorting is necessary.

What is the best way to format ranges in the Summary Ranges problem?

Ranges should be formatted as 'start->end' for sequences with multiple numbers, and 'start' for single elements.

How can GhostInterview help me prepare for the Summary Ranges problem?

GhostInterview guides you through breaking down the array-driven approach and ensures you understand how to optimize both time and space complexities.

terminal

Solution

Solution 1: Two Pointers

We can use two pointers $i$ and $j$ to find the left and right endpoints of each interval.

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class Solution:
    def summaryRanges(self, nums: List[int]) -> List[str]:
        def f(i: int, j: int) -> str:
            return str(nums[i]) if i == j else f'{nums[i]}->{nums[j]}'

        i = 0
        n = len(nums)
        ans = []
        while i < n:
            j = i
            while j + 1 < n and nums[j + 1] == nums[j] + 1:
                j += 1
            ans.append(f(i, j))
            i = j + 1
        return ans

Continue Practicing

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