Two Sum Leetcode Optimal Solution | C++ | Java | Python

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.

Solutions

✅Time: O(n)
✅Space: O(n)

C++

class Solution {
 public:
  vector<int> twoSum(vector<int>& nums, int target) {
    unordered_map<int, int> numToIndex;

    for (int i = 0; i < nums.size(); ++i) {
      if (numToIndex.count(target - nums[i]))
        return {numToIndex[target - nums[i]], i};
      numToIndex[nums[i]] = i;
    }

    throw;
  }
};

Java

class Solution {
  public int[] twoSum(int[] nums, int target) {
    Map<Integer, Integer> numToIndex = new HashMap<>();

    for (int i = 0; i < nums.length; ++i) {
      if (numToIndex.containsKey(target - nums[i]))
        return new int[] {numToIndex.get(target - nums[i]), i};
      numToIndex.put(nums[i], i);
    }

    throw new IllegalArgumentException();
  }
}

Python

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

    for i, num in enumerate(nums):
      if target - num in numToIndex:
        return numToIndex[target - num], i
      numToIndex[num] = i

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