TwoSum.java

package leetcode;

/*
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 <= 104
-109 <= nums[i] <= 109
-109 <= target <= 109
Only one valid answer exists.
 */


public class TwoSum {

    public int[] twoSum(int[] nums, int target) {
        for (int i = 0; i < nums.length - 1; 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 found");
    }

    public static void main(String[] args) {
        TwoSum solution = new TwoSum();

        // Test case 1
        int[] nums1 = {2, 7, 11, 15};
        int target1 = 9;
        int[] result1 = solution.twoSum(nums1, target1);
        System.out.println("Indices: [" + result1[0] + ", " + result1[1] + "]");

        // Test case 2
        int[] nums2 = {3, 2, 4};
        int target2 = 6;
        int[] result2 = solution.twoSum(nums2, target2);
        System.out.println("Indices: [" + result2[0] + ", " + result2[1] + "]");

        // Test case 3
        int[] nums3 = {3, 3};
        int target3 = 6;
        int[] result3 = solution.twoSum(nums3, target3);
        System.out.println("Indices: [" + result3[0] + ", " + result3[1] + "]");
    }
}