This approach uses a simple brute force algorithm, which consists of checking each pair of numbers to see if they add up to the target. Although this is straightforward, it is not the most efficient method. We loop through each number in the array using two nested loops, effectively trying all combinations of two numbers.
Time Complexity: O(n^2)
Space Complexity: O(1)
1import java.util.*;
2
3public class Main {
4 public static int[] twoSum(int[] nums, int target) {
5 for (int i = 0; i < nums.length; i++) {
6 for (int j = i + 1; j < nums.length; j++) {
7 if (nums[i] + nums[j] == target) {
8 return new int[] { i, j };
9 }
10 }
11 }
12 return new int[] {};
13 }
14
15 public static void main(String[] args) {
16 int[] nums = { 2, 7, 11, 15 };
17 int target = 9;
18 int[] result = twoSum(nums, target);
19 System.out.println("[" + result[0] + ", " + result[1] + "]");
20 }
21}In the Java solution, a nested for loop is utilized to find the two integers that sum up to the target. Indices are returned once the condition is met. The program outputs the result as an array of two integers.
This efficient approach utilizes a hash map to store the difference between the target and the current element (called 'complement') as the key and the element's index as the value. As we iterate through the array, we check if the current element is a key in the hash map, which indicates that its complement was seen earlier, thus allowing us to retrieve the correct indices quickly.
Time Complexity: O(n)
Space Complexity: O(n)
1using System;
2using System.Collections.Generic;
3
4class Program {
5 static int[] TwoSum(int[] nums, int target) {
6 Dictionary<int, int> map = new Dictionary<int, int>();
7 for (int i = 0; i < nums.Length; i++) {
8 int complement = target - nums[i];
9 if (map.ContainsKey(complement)) {
10 return new int[] { map[complement], i };
11 }
12 map[nums[i]] = i;
13 }
14 return new int[] {};
15 }
16
17 static void Main() {
18 int[] nums = { 2, 7, 11, 15 };
19 int target = 9;
20 int[] result = TwoSum(nums, target);
21 Console.WriteLine($"[{result[0]}, {result[1]}]");
22 }
23}The C# solution uses a dictionary to track each number and its index. As we iterate through the array, we calculate each number's complement and check if it exists in the dictionary, thereby providing an efficient solution.