0001. Two Sum

Easy | Hash Table | 28 ms (93.73%), 13.4 MB (79.11%)

❓ Problem Statement

Source: LeetCode - Two Sum GitHub: Solution / Performance

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.

✍🏻 Constraints & Example

Constraints:

• 2 <= nums.length <= 10^4

• -10^9 <= nums[i] <= 10^9

• -10^9 <= target <= 10^9

• Only one valid answer exists.

Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Output: Because nums[0] + nums[1] == 9, we return [0, 1].

Input: nums = [3,2,4], target = 6
Output: [1,2]

Input: nums = [3,3], target = 6
Output: [0,1]

💡 Ideas

Each number in the list of integers has its own "remain" value by subtracting its value from the target number.

Store the remains as key with the index as the value in the hash table.

So that we could iterate each element in the list of integers by the index, and then check whether the value of remain exists.

🤖 Python3

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        hashTable = dict()

        # ==================================================
        #  Array + Hash Table                              =
        # ==================================================
        # time  : O(n)
        # space : O(n)
        
        for i in range(len(nums)):
            remain = target - nums[i]
            
            if remain in hashTable:
                return [i, hashTable[remain]]
            
            # We could overwrite since there is only one solution
            hashTable[nums[i]] = i

🤖 Java

class Solution {
    /**  
     * @time  : O(n)
     * @space : O(n)
     */
    
    public int[] twoSum(int[] nums, int target) {
        Map<Integer, Integer> map = new HashMap<>();
        
        for(int i=0 ; i<nums.length ; i++) {
            int remain = target - nums[i];
            
            if(map.containsKey(remain)) {
                return new int[] {i, map.get(remain)};
            }
            
            map.put(nums[i], i);
        }
        
        return null;
    }
}