# LeetCode : Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.

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?

### Solution 1 – Single pass: Using enumerate

#### Steps:

1. Create a empty dictionary “`mapping`” for storing the key pairs
2. Enumerate the list (iterating the list with index number)
3. Calculate the difference of target and current value in the enumeration
4. Check if the whether difference is present in the dictionary “mapping”
• If value is present then we get the solution
• Otherwise add the index of the value to dictionary
• For few times in beginning the dictionary will have less values
``````class Solution(object):
def twoSum(self, nums, target):
mapping = {}

for index, val in enumerate(nums):
difference = target - val
if difference in mapping:
return [mapping[difference], index]
else:
mapping[val] = index

s = Solution()
print(s.twoSum([[2,7,11,15]], 9))

#Output
[0,1]

# Space: O(N)
# Time: O(N)``````

### Solution 2

``````def twoSum(self, nums: List[int], target: int) -> List[int]:
for i in range(len(nums)):
for j in range(i+1,len(nums)):
if nums[i]+nums[j]==target:
return[i,j]

print(twoSum([[2,7,11,15]], 9))

#Output
[0,1]``````