1. Two Sum

Problem Statement:

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:

TwoSum.java

import java.util.HashMap;
import java.util.Map;

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

  public static int[] twoSum2(int nums[], int target) {
    Map<Integer, Integer> numMap = new HashMap<>();
    for (int i = 0; i < nums.length; i++) {
        int complement = target - nums[i];
        if (numMap.containsKey(complement)) {
            return new int[] { numMap.get(complement), i };
        }
        numMap.put(nums[i], i);
    }
    return new int[]{};
  }
}

TwoSum.ts

function twoSum(nums: number[], target: number): number[] {
    const numMap = new Map()

    for (let i = 0; i < nums.length; i++) {
        const complement = target - nums[i];
        if (numMap.has(complement)) {
            return [numMap.get(complement), i]
        }
        numMap.set(nums[i], i)
    }
    return []
};

function twoSum2(nums: number[], target: number): number[] {
    const memo = {}
    nums.forEach((item, index) => memo[item] = index);

    for (let i = 0; i < nums.length; i++) {
        let x = target - nums[i]
        if (memo[x] && memo[x] !== i) {
            return [i, memo[x]]
        }
    }

    return []
};

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