TwoSum.java

package Lc1_TwoSum;

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

/**
 * 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 <= 10^4
 * -10^9 <= nums[i] <= 10^9
 * -10^9 <= target <= 10^9
 * Only one valid answer exists.
 */
public class TwoSum {

    /**
     * Solution 1: Brute Force
     * Time complexity : O(n^2)
     * Space complexity : O(1)
     */
    public int[] solution1(int[] nums, int target) {
        int len = nums.length;
        for (int i = 0; i < len - 1; i++) {
            for (int j = i + 1; j < len; j++) {
                if (nums[i] == target - nums[j]) {
                    return new int[]{i, j};
                }
            }
        }
        return new int[]{};
    }

    /**
     * Solution 2: Two-pass Hash Table
     * Time complexity : O(n)
     * Space complexity : O(n)
     */
    public int[] solution2(int[] nums, int target) {
        Map<Integer, Integer> numMap = new HashMap<>();
        int len = nums.length;
        for (int i = 0; i < len; i++) {
            numMap.put(nums[i], i);
        }
        for (int i = 0; i < len; i++) {
            int complement = target - nums[i];
            if (numMap.containsKey(complement) && numMap.get(complement) != i) {
                return new int[]{i, numMap.get(complement)};
            }
        }
        return new int[]{};
    }

    /**
     * Solution 3: One-pass Hash Table
     * Time complexity : O(n)
     * Space complexity : O(n)
     */
    public int[] solution3(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[]{};
    }
}