1. Two Sum

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.

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

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

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

Hashmap

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

To find an O(n) solution, we need to go through num once and for each num in the array, check if the difference between target and num exists in num.

nums = [2,7,11,15], target = 9

# iteration 1:
num = 2, target = 9
difference = 9 - 2 = 7

7 exists

To check if the number (7 in the example) exists, we need to make a hashmap. The hashmap will map each number in nums with its index in num: { number : index }.

The same value/index cannot be used twice. For example 3 ([3, 3], 6), the result is [0, 1], but not [0, 0].

nums = [2,7,11,15], target = 9
hashmap = {}

# iteration 1:
num = 2, target = 9
difference = 9 - 2 = 7

7 not in hashmap, so add 2 to it
hashmap = { 2 : 0 }

# iteration 2
num = 7, target = 9
difference = 9 - 7 = 2

2 is in the hashmap with index 0 (no need to add 7)
we are currently on index 1

return [0, 1]

Another example:

nums = [3,2,4], target = 6
hashmap = {}

# iteration 1
num = 3, target = 6
difference = 3

3 not in hashmap, so add 3
hashmap = { 3 : 0 }

# iteration 2
num = 2, target = 6
difference = 4

4 not in hashmap, so add 2
hashmap = { 3 : 0, 2 : 1 }

# iteration 3
num = 4, target = 6
difference = 2

2 is in hashmap with value 1
index of num (4) is 2

return [1, 2]

Python

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        # number : index
        hashmap = {}

        for i in range(len(nums)):
            num = nums[i]

            diff = target - num

            if diff in hashmap:
                return [i, hashmap[diff]]
            else:
                hashmap[num] = i