[LeetCode/Python] 228. Summary Ranges

📌 Problem

[LeetCode] 228. Summary Ranges

💡 Try & Error

Initial Attempt

• Append ranges only when a break in consecutiveness occurs
  (연속이 끊기는 지점에서만 range append)
• Handle last index using conditional checks inside the loop
  (루프 내부에서 i == len(nums) - 1 예외 처리 추가)
• Result generation dependent on loop execution
  (마지막 range를 루프 안에서 처리하는 구조)

Issue

• For input size 1 (e.g. nums = [-1]), loop never executes
  (nums = [-1] 같은 길이 1 배열에서 for-loop 미실행)
• No range appended, resulting in empty output
  (append가 한 번도 발생하지 않아 빈 배열 반환)
• Loop-dependent logic causing edge case failure
  (루프 의존적인 결과 생성 구조 문제)

Key Insight

• Separation of "range termination" and "finalization" required
  (연속 구간 문제의 핵심은 "마지막 구간 처리")
• Intermediate ranges handled inside the loop
  (끊기는 지점은 루프에서 처리)
• Final range appended after loop completion
  (마지막 구간은 루프 종료 후 무조건 한 번 append)
• Single-element input handled naturally
  (길이 1 배열 자연스럽게 커버 가능)

📌 Solution

Solution 1: Upgraded Version

Idea

• Preserve original for-loop structure
• Append range when consecutiveness breaks
• Defer last range handling until after loop
• Ensure coverage for len(nums) == 1

Code

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

        results = []
        start, end = nums[0], nums[0]
        for i in range(1, len(nums)):
            if nums[i] != nums[i - 1] + 1:
                end = nums[i-1]
                if end == start:
                    results.append(f"{start}")
                else:
                    results.append(f"{start}->{end}")
                start = nums[i]

                if i == len(nums)-1:
                    results.append(f"{start}")
            else:
                if i == len(nums)-1:
                    results.append(f"{start}->{nums[i]}")
            
            print(f"index {i} -> start = {start}, end = {end}, results = {results}")

        return results

Complexity

• Time Complexity: $O(n)$
• Space Complexity: $O(n)$ (output 포함)

⭐️ Solution 2: Optimal Version (Two Pointers)

Idea

• Single-pass two-pointer traversal
• Expand range while numbers remain consecutive
• Emit range immediately after expansion ends
• Minimal branching and clean termination logic

Code

class Solution:
    def summaryRanges(self, nums: List[int]) -> List[str]:
        if not nums:
            return []
        
        results = []
        start = nums[0]

        for i in range(1, len(nums)):
            if nums[i] != nums[i - 1] + 1:
                end = nums[i - 1]
                if start == end:
                    results.append(str(start))
                else:
                    results.append(f"{start}->{end}")
                start = nums[i]

        end = nums[-1]
        if start == end:
            results.append(str(start))
        else:
            results.append(f"{start}->{end}")

        return results

Complexity

• Time Complexity: $O(n)$
• Space Complexity: $O(1)$ extra (output 제외)