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.
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].
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].
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].
numbers
array is sorted. We can use this condition.left
(start) and right
(end)numbers[left]
is the smallest number in the arraynumbers[right]
is the biggest number in the arraynumbers[left]+numbers[right]
check == target
, return the index of left&rightcheck<target
, since it is smaller than the target, move left
pointer to the next index.check>target
, since it is bigger than the target, move right
pointer to the previous index.class Solution:
def twoSum(self, numbers: List[int], target: int) -> List[int]:
left=0
right=len(numbers)-1
while left<right:
check=numbers[left]+numbers[right]
if check==target:
return [left+1,right+1]
elif check<target:
left+=1
else:
right-=1
Time Complexity: O(n)
Space Complexity: O(1)
Runtime: 276 ms
Memory: 14.8 MB
Binary Search
Method?Binary Search
is available and efficient for sorted array, so we can try solving this problem with using Binary search
. References
https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/