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]

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

**Constraints:**

`2 <= nums.length <= 10`

^{4}`-10`

^{9}<= nums[i] <= 10^{9}`-10`

^{9}<= target <= 10^{9}**Only one valid answer exists.**

**Follow-up: **Can you come up with an algorithm that is less than `O(n`

time complexity?^{2})

## In simple words.

Interview will give us a array with integer number and one target number. We need to return the index of 2 elements those sum will be equal to target number.

**Solution 1**, if your array is sorted:

class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: res = [] for i in range(len(nums)): for c in range(i+1, len(nums)): result = nums[i] + nums[c] if result == target: res.append(i) res.append(c) return res

**Solution **2, if your array is not sorted:

# Here am pring number itself, you can print index of those number as well. def sum_of_two_numbers_with_hash_map(nums, target): #TODO: Working with sorted/unsorted array #TODO: O(n) map = {} for i, v in enumerate(nums): if (target - nums[i]) in map: print([(target - nums[i]), nums[i]]) break else: map[v] = i num = [2,3,5,7,11,6, 13] sum_of_two_numbers_with_left_right_pointer(nums=num, target=12)

Two sum interview question asked by Amazon|Google|Nagarro