LeetCode 1: Two Sum in Java

Given an integer array nums and an integer target, return the indexes of two distinct elements whose sum equals the target.

Brute force

class Solution {
    public int[] twoSum(int[] nums, int target) {
        for (int i = 0; i < nums.length; i++) {
            for (int j = i + 1; j < nums.length; j++) {
                if (nums[i] + nums[j] == target) {
                    return new int[]{i, j};
                }
            }
        }
        throw new IllegalArgumentException("No solution");
    }
}

Time complexity is O(n²) and extra space is O(1).

One-pass HashMap

For each value, calculate the required complement and check whether it has already been seen:

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

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

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

        throw new IllegalArgumentException("No solution");
    }
}

Average time complexity is O(n) and space complexity is O(n). Checking before inserting prevents the current element from matching itself and correctly handles inputs such as [3, 3] with target 6.