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167. Two Sum II - Input Array Is Sorted

Difficulty: Easy

Related Topics: Array, Two Pointers, Binary Search

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].

First Attempt:

Code:

class Solution:
    def twoSum(self, numbers: List[int], target: int) -> List[int]:
        checked = {}
        cnt = 0
        while target-numbers[cnt] not in checked:
            checked[numbers[cnt]] = cnt
            cnt+=1
        return [checked[target - numbers[cnt]]+1,cnt+1]

Feedbacks:

I tried to solve this problem same as two sum problem. But I realized that I need to use two pointers.

Second Attempt:

Code:

class Solution:
    def twoSum(self, numbers: List[int], target: int) -> List[int]:
        left=0
        right=len(numbers)-1

        while left < right:
            sum = numbers[left] + numbers[right]
            if sum == target:
                return [left+1, right+1]
            if sum > target:
                right -= 1
            elif sum < target:
                left += 1

Feedbacks:

Since I need to use two pointers, I set one pointer ‘left’ as 0, another pointer ‘right’ as length of numbers list minus 1. Because the list given is sorted in non-decreasing orders, increasing index will increase numbers and vice versa. So, as I calculate the sum, if sum is larger than target number, I lower the ‘right’ pointer which lowers the sum, and if sum is smaller than target number, I higher the ‘left’ pointer, so sum increases. So, finally, I can get the left and right indicies which sum is equal to the target number.

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