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].
2 <= numbers.length <= 3 * 10^4 -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.
class Solution:
def twoSum(self, numbers: List[int], target: int) -> List[int]:
left, right = 0, len(numbers)-1
while left < right:
sumnums = numbers[left] + numbers [right]
if sumnums == target:
return [left+1, right+1]
elif sumnums < target:
left += 1
else:
right -= 1
return []Initially was a little stumped on how to move the left and right pointers , since target could be found by having included the left or right index element that I passed over.
Time complexity = O(n) Space Complexity = O(1)