Leetcode #167
Problem
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, each incremented 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].
Constraints:
2 ⇐ numbers.length ⇐ 3 * 104
-1000 ⇐ numbers[i] ⇐ 1000
numbers is sorted in non-decreasing order.
-1000 ⇐ target ⇐ 1000
The tests are generated such that there is exactly one solution.
My Solution
class Solution:
def twoSum(self, numbers: List[int], target: int) -> List[int]:
length = len(numbers)
a = 0
b = length-1
sum = numbers[a] + numbers[b]
while sum != target:
if sum < target:
a+=1
elif sum > target:
b-=1
sum = numbers[a] + numbers[b]
return [a+1, b+1]i only wrote the while condition to be sum != target because the question specifies that there definitely exists a solution. however, i should be writing something like this instead:
class Solution:
def twoSum(self, numbers: List[int], target: int) -> List[int]:
l, r = 0, len(numbers) - 1
while l < r:
curSum = numbers[l] + numbers[r]
if curSum > target:
r -= 1
elif curSum < target:
l += 1
else:
return [l + 1, r + 1]
return []