Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

Approach 3: One-pass Hash Table
It turns out we can do it in one-pass. While we iterate and inserting elements into the table, we also look back to check if current element’s complement already exists in the table. If it exists, we have found a solution and return immediately.

class Solution {

  public int[] twoSum(int[] nums, int target) {
    Map<Integer, Integer> hashmap = new HashMap<>();
    for (int i = 0; i < nums.length; i++) {
      int aim = target - nums[i];
      if (hashmap.containsKey(aim)) {
        return new int[]{hashmap.get(aim), i};
      }
      hashmap.put(nums[i], i);
    }
    throw new IllegalArgumentException("No two sum solution");

  }
}

SuccessDetails Runtime: 2 ms, faster than 98.95% of Java online submissions for Two Sum.Memory Usage: 37 MB, less than 99.08% of Java online submissions for Two Sum.Next challenges: