# 1. Two Sum

## 14 Dec 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.

Example 1:

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

Follow-up: Can you come up with an algorithm that is less than `O(n2) `time complexity?

class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
i=0
j=0
length = len(nums)
length2 = len(nums)
# print(length)
for i in range(length):
# print(i)
length2 = length-i-1
for j in range(length2):
# print(j)
# print(length2)
if((nums[i]+nums[i+j+1])==target):
return [i, i+j+1]
``````class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
dictionary = {}
for i, value in enumerate(nums):
print(i)
print(value)
remain = target - nums[i]
print(remain)
if remain in dictionary:
return [dictionary[remain], i]
else:
dictionary[value] = i

print('---------------')
``````