Two Sum

Problem

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 <= 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(n2) time complexity?

My Solution

Solving the problem is easy once you realize that you don’t need to sum every pair in the array, but rather for each element in the array calculate the difference to the target value and if the difference is present in the array, jackpot, now you have two numbers that sum to the target.

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        prevs = {}
        for i, num in enumerate(nums):
            if (num in prevs):
                return [prevs[num], i]

            prevs[target - num] = i

        return []

var twoSum = function(nums, target) {
  const prevs = {};
  for (i=0;i<nums.length;i++) {
    const num = nums[i];
    if (prevs[num] !== undefined) {
      return [prevs[num], i];
    }

    prevs[target - num] = i;
  };

  return [];
};