1. 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.

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.

Follow-up: Can you come up with an algorithm that is less than O(n²) time complexity?

Solution 1:

Brute force in Java:

class Solution {
    public static int[] twoSum(int[] nums, int target) {
        int[] result = new int[2];
        for (int i = 0; i < nums.length; i++) {
            for (int j = i + 1; j < nums.length; j++) {
                if (nums[i] + nums[j] == target) {
                    result[0] = i;
                    result[1] = j;
                }
            }
        }
        return result;
    }

    public static void main(String[] args) {
        int[] nums = {2, 7, 11, 15};
        int target = 9;

        int[] result = twoSum(nums, target);
        for (int i = 0; i < result.length; i++)
            System.out.print(result[i] + " ");
        System.out.println();
    }
}

Solution 2

Hash table solution

• Time complexity: O(n)
• Space complexity: O(n)

import time
from typing import List

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

        for i in range(len(nums)):
            res = target - nums[i]
            if res in cache_dict:
                return [cache_dict[res], i]
            else:
                cache_dict[nums[i]] = i
            


if __name__ == '__main__':
    s = Solution()
    start = time.time()
    print(s.twoSum([2,7,11,15], 9))
    print()
    print(s.twoSum([3,2,4], 6))
    print()
    print(s.twoSum([3, 3], 6))
    print()
    print("cost time: " + str(time.time() - start))

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