[📝LeetCode #1] Two Sum

🎀 The Problem

Given an array of integers nums and an integer target, return the 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:

Input: nums = [2,7,11,15], target = 9
Output: [0,1]

👩‍💻 My Answer

class Solution {
    public int[] twoSum(int[] nums, int target) {
        int index = 0;

        while(index < nums.length){
            for (int i = index+1; i < nums.length; i++) {
                if (nums[i] == target - nums[index]) {
                    int[] output = new int[2];
                    output[0] = index;
                    output[1] = i;
                    return output;
                }
            }
            index++;
        }

        return null;
    }
}

Pro & Con

✖️ Runtime & Memory
✖️ Brute Force Code (nested loop)

💋 Ideal Answer

Approach - "Hash Table"

I read the solution on LeetCode and found that using a hash table is the fastest one for this.

New Code

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

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

        return null;
    }
}

💡 What I Learned

How I construct int[] in return line return new int[] {map.get(target - nums[i]),i};

shine @shine7on