TwoSum.java

package arrays.easy;

import java.util.Arrays;
import java.util.HashMap;

import static java.util.Arrays.binarySearch;

/**
 * Problem 1 in Leetcode: https://leetcode.com/problems/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]
 *
 * Example 2:
 * Input: nums = [3,2,4], target = 6
 * Output: [1,2]
 *
 * Example 3:
 * Input: nums = [3,3], target = 6
 * Output: [0,1]
 */

public class TwoSum {
    public static void main(String[] args) {
        int[] nums = {3, 3};
        int target = 6;
        System.out.println("Brute Force Approach: " + Arrays.toString(bruteForceApproach(nums, target)));
        System.out.println("HashMap Approach: " + Arrays.toString(hashmapApproach(nums, target)));
    }

    private static int[] bruteForceApproach(int[] nums, int target) {
        int n = nums.length;
        for (int i = 0; i < n - 1; i++) {
            for (int j = i + 1; j < n; j++) {
                if (nums[i] + nums[j] == target) {
                    return new int[]{i, j};
                }
            }
        }
        return new int[]{-1, -1};
    }

    private static int[] hashmapApproach(int[] nums, int target) {
        int n = nums.length;
        HashMap<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < n; i++) {
            int difference = target - nums[i];
            if (map.containsKey(difference)) {
                return new int[]{map.get(difference), i};
            }
            map.put(nums[i], i);
        }
        return new int[]{-1, -1};
    }
}