228.Summary_Ranges(Easy).md

228. Summary Ranges (Easy)

Date and Time: Sep 6, 2024, 15:10 (EST)

Link: https://leetcode.com/problems/summary-ranges/

Question:

You are given a sorted unique integer array nums.

A range [a,b] is the set of all integers from a to b (inclusive).

Return the smallest sorted list of ranges that cover all the numbers in the array exactly. That is, each element of nums is covered by exactly one of the ranges, and there is no integer x such that x is in one of the ranges but not in nums.

Each range [a,b] in the list should be output as:

• "a->b" if a != b

• "a" if a == b

Example 1:

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

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

Explanation: 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"]

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

Constraints:

• 0 <= nums.length <= 20

• -2^31 <= nums[i] <= 2^31 - 1

• All the values of nums are unique.

• nums is sorted in ascending order.

Walk-through:

1. First we check if nums is empty or not.

2. We use start to keep track of the first element of the continue interval, if nums[i] != nums[i-1]+1, we know nums[i] is not part of the previous interval, so we can add this interval [start, nums[i-1]] into res. However, we have two cases to check, if start == nums[i-1], the previous value should be added to res by itself. Otherwise, we add the interval str(start) + "->" + str(nums[i-1]). After that, we update start = nums[i].

3. We handle the last element in nums, we repeat to check if start == nums[-1], we add the last element by itself to res[], otherwise, we add str(start) + "->" + str(nums[i-1]).

Python Solution:

class Solution:
    def summaryRanges(self, nums: List[int]) -> List[str]:
        if not nums:
            return []

        res = []
        start = nums[0]
        for i in range(1, len(nums)):
            if nums[i] != nums[i-1]+1:
                if start == nums[i-1]:
                    res.append(str(start))
                else:
                    res.append(str(start) + "->" + str(nums[i-1]))
                start = nums[i]
        
        if start == nums[-1]:
            res.append(str(start))
        else:
            res.append(str(start) + "->" + str(nums[-1]))
        return res

Time Complexity: $O(n)$, n is length of nums.
Space Complexity: $O(n)$