[Leetcode]167. Two Sum II - Input Array Is Sorted

limelimejiwon·2022년 3월 11일

📄 Description

Given a 1-indexed array of integers numbers that is already sorted in non-decreasing order, find two numbers such that they add up to a specific target number. Let these two numbers be numbers[index1] and numbers[index2] where 1 <= index1 < index2 <= numbers.length.

Return the indices of the two numbers, index1 and index2, added by one as an integer array [index1, index2] _of length 2.

The tests are generated such that there is exactly one solution. You may not use the same element twice.

Your solution must use only constant extra space.

Example 1:

Input: numbers = [2,7,11,15], target = 9
Output: [1,2]
Explanation: The sum of 2 and 7 is 9. Therefore, index1 = 1, index2 = 2. We return [1, 2].

Example 2:

Input: numbers = [2,3,4], target = 6
Output: [1,3]
Explanation: The sum of 2 and 4 is 6. Therefore index1 = 1, index2 = 3. We return [1, 3].

Example 3:

Input: numbers = [-1,0], target = -1
Output: [1,2]
Explanation: The sum of -1 and 0 is -1. Therefore index1 = 1, index2 = 2. We return [1, 2].

🔨 My Solution

  • numbers array is sorted. We can use this condition.
  • use two pointers for left(start) and right(end)
    - numbers[left] is the smallest number in the array
    • numbers[right] is the biggest number in the array
  • check numbers[left]+numbers[right]
    (1) if check == target, return the index of left&right
    (2) if check<target, since it is smaller than the target, move left pointer to the next index.
    (3) if check>target, since it is bigger than the target, move right pointer to the previous index.

💻 My Submission

class Solution:
    def twoSum(self, numbers: List[int], target: int) -> List[int]:
        while left<right:
            if check==target:
                return [left+1,right+1]
            elif check<target:

🔎 Complexity

Time Complexity: O(n)
Space Complexity: O(1)

Runtime: 276 ms
Memory: 14.8 MB

❓ How Can I improve it?

  • using Binary Search Method?
    - Since Binary Search is available and efficient for sorted array, so we can try solving this problem with using Binary search.


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