Leetcode 0001. 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.

Easy

Description

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]

• 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(n^2) time complexity?

Solution

class Solution {
    public int[] twoSum(int[] nums, int target) {
        HashMap<Integer, Integer> hashMap = new HashMap<>();
        for (int i = 0; i < nums.length; i++) {
          int complement = target - nums[i];
          if (hashMap.containsKey(complement)) {
            return new int[] {hashMap.get(complement), i};
          }
          hashMap.put(nums[i], i);
        }
        throw new IllegalArgumentException("Not found");
    }
}