Two Sum II - Input Array Is Sorted

The key idea

Because the array is already sorted, you do not need a hash map. Put one pointer at each end: if the pair sum is too big, move the right pointer left to shrink it; if too small, move the left pointer right to grow it. Each move discards exactly the values that can never be part of the answer, so a single pass in O(1) extra space finds the unique pair.

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, 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.

Constraints

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

Examples

Input: numbers = [2,7,11,15], target = 9 Output: [1,2]

Input: numbers = [2,3,4], target = 6 Output: [1,3]

Input: numbers = [-1,0], target = -1 Output: [1,2]

Complexity

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

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