Leetcode Day 1: Two Sum Explained

The problem is as follow:

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 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]

Here is how I solved it:

• We want to create a dictionary named index_map to store the integers in nums and their corresponding indices.

index_map = {}

• Then, we will use enumerate to get both index i and value num of each element in nums. For each integer, let's calculate the complement, which is the difference between the target and the current element num.

for i, num in enumerate(nums):
    n = target - num

• Now check the dictionary: if n is in the dictionary, it means we have found the two integers that add up to the target Return the n index and the current index as a list.

if n in index_map: 
    return [index_map[n], i]

• Else, if n is not in the dictionary, add the current element num and index to the dictionary.

index_map[num] = i

Here is the completed solution:

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        index_map = {}
        for i, num in enumerate(nums):
            n = target - num
            if n in index_map: 
                return [index_map[n], i]
            index_map[num] = i