167 Two Sum II

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.

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.

Problem Analysis:

use 2 pointers
there is important assumption that there is a guaranteed solution in the array, so there is less edge cases to handle, eg. no solution found.

Solutions:

class Solution:
    def twoSum(self, numbers: List[int], target: int) -> List[int]:
        # 2 pointers
        # assumption: guaranteed solution in array
        l, r = 0, len(numbers)-1

        while l < r:
            curSum = numbers[l] + numbers[r]
            print(curSum)

            if curSum > target:
                r -=1
            elif curSum < target:
                l +=1
            else:
                return [l+1,r+1]

Walter Teng.

167 Two Sum II

Problem:

Problem Analysis:

Solutions:

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