Tow Sum

1. Two Sum

Easy

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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]

 

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?

Python solution using HashMap

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

        for i, n in enumerate(nums):
            difference = target - n
            if difference in new_hash_map:
                return(i, new_hash_map[difference])
            else:
                new_hash_map[n] = i

Time complexity O(N)

Java Solution

class Solution {
    public boolean containsDuplicate(int[] nums) {
        Set<Integer> numbers = new HashSet<Integer>();
        for (int num : nums) {
            if (numbers.contains(num)) return true;
            numbers.add(num);
        }

        return false;
    }
}

Time complexity O(N)