Problem

Solution

Intuition

This problem asks us to keep track of continuous ranges within an array that contains unique increaing elements. This problem seems easy to solve as a two pointer problem keeping track of the start and the end of each range and adding it to the result as necessary.

Approach

We utilize two pointers to keep track of the current contiguous range. We increment the right pointer with the two following conditions:

1. If the current number is not $1$ more than the previous number, it is not part of a continuous range and we add the range to our result. If there is more than $1$ number in the range, we create a string representation of a range of numbers from L to R - 1 otherwise just add L as a string. L moves to R and we increment R.
2. If the current number that R points to is $1$ more than the number at R - 1, then we increment R.

Complexity

• Time complexity: $O(n)$ as we iterate though the array once doing a constant time operation on each step.
• Space complexity: $O(n)$ to store the result.

Code

class Solution:
    def summaryRanges(self, nums: List[int]) -> List[str]:
        if len(nums) == 0:
            return []
            
        res = []

        L = 0
        R = 1

        while R < len(nums):
            if nums[R] != nums[R - 1] + 1:
                if R - L == 1:
                    res.append(str(nums[L]))
                else:
                    res.append(str(nums[L]) + "->" + str(nums[R - 1]))
                L = R
                R += 1
            else:
                R += 1
        
        if R - L == 1:
            res.append(str(nums[L]))
        else:
            res.append(str(nums[L]) + "->" + str(nums[R - 1]))
    
        return res