TwoSumII

Problem

https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/

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[index:sub:`1`] and numbers[index:sub:`2`] where 1 <= index:sub:`1`< index:sub:`2`<= numbers.length.

Return the indices of the two numbers *``index``:sub:```1``` *and index:sub:`2`, each incremented by one, as an integer array [index:sub:`1`, index:sub:`2`] 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 * 10:sup:`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.

Solution

Set two pointers i and j to the beginning and end of numbers. If twosum = numbers[i] + numbers[j] is less than target, we can increase twosum by using a larger numbers[i] by incrementing i. If twosum is greater than target, we can decrease twosum by using a smaller numbers[j] by decrementing j. Stop when twosum == target. This algorithm can find the twosum in \(O(n \\log n)\) time and no extra space.

Pattern

Array, Two Pointers, Binary Search

Code

from typing import List


def twoSum(numbers: List[int], target: int) -> List[int]:
    """Find two numbers in ``numbers`` that add up to ``target`` and return
    their one offset indicies.
    """
    i = 0
    j = len(numbers) - 1
    while i < j:
        twosum = numbers[i] + numbers[j]
        if twosum == target:
            break
        elif twosum < target:
            i += 1
        else:
            j -= 1
    return [i + 1, j + 1]

Test

>>> from TwoSumII import twoSum
>>> twoSum([2, 7, 11, 15], 9)
[1, 2]
>>> twoSum([2, 3, 4], 6)
[1, 3]
>>> twoSum([-1, 0], -1)
[1, 2]

TwoSumII.twoSum(numbers: List[int], target: int) → List[int]: Find two numbers in numbers that add up to target and return their one offset indicies.