Python Exercise 23: Two sum

Question

• Given an array of integers nums and an integer target,

  • return indices of the two numbers such that they add up to target.

• You may assume that each input would have exactly one solution,

  • and you may not use the same element twice.

• You can return the answer in any order.

Example

• Example 1:

Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].

• Example 2:

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

• Example 3:

Input: nums = [3,3], target = 6
Output: [0,1]

Constraints:

• 2 <= nums.length <= 104
• -109 <= nums[i] <= 109
• -109 <= target <= 109
• Only one valid answer exists.

My attempt

First attempt

• The algorithm is work theoretical but too slow, run out of time if last two test case
• algorithm

> find the indices of the two number such that they add up to target
>>find two number if they add up to the target
  for index1, num1 in the nums:
      for index2,num2 in the nums:
          if num2 is not the same number as num1:
              if value1 + value 2 == target:
                  return a list consist of index1 and index2

• code

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        final_list = [[index1,index2] for index1,value1 in enumerate(nums) for index2,value2 in enumerate(nums) if index2 > index1 if value1+value2 ==target]
        return final_list[0]

Second attempt (success pass all test)

• algorithm

> find the indices of the two number such that they add up to target
>>find two number if they add up to the target
  for index1,value1 in nums:
      calculate a second_number that have a sum with value equals to target
      if second is in nums:
          find the indices of the second_number
          if the second_number is not the same number as value1:
              return a list consist of index1 and index2

• code

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        for index1, value1 in enumerate(nums):
            second_number = target - value1
            if second_number in nums:
                second_number_index = nums.index(second_number)
                if second_number_index != index1:
                    return [index1, second_number_index]

Other solution

Claim to be the fastest algorithm with 0(n) complexity

• algorithm

> find the indices of the two number such that they add up to target
 initiate seen as empty dictionary
 for index, value in nums:
     calculate a second_number that have a sum with value equals to target
     if second_number is in seen dictionary:
         return a list of index and the index of second_number
     if second_number is not in seen:
         add the value as key and index as value in the seen dictionary

• code

class Solution:
   def twoSum(self, nums: List[int], target: int) -> List[int]:
       seen = {}
       for index, value in enumerate(nums):
           second_number = target - value

           if remaining in seen:
               return [index, seen[second_number]]

           seen[value] = index

My reflection

This is a good challenge to make me think a method that not just rely on for loop completely. And I learn a new pattern form others to use a dictionary to reduce searching time.

Credit

challenge on leetcode 1
solution on code recipe