Given an array of integers nums and an integer target, return the indices of two numbers that add up to the target value.

Example:

Input: nums = [3, 4, 2], target = 6
Output: [1, 2]
Explanation: nums[1] + nums[2] equals 6

Approaches

1. Brute Force Method:

   • Iterate through each element and check every other element for a matching sum
   • Time complexity: O(n²)

2. Hash Map Optimization:

   • Use a hash map to store values and indices as we iterate
   • Check if the complement (target - current value) exists in the map
   • Time complexity: O(n)

Java Implementations

Brute Force Solution:

public int[] findTwoSum(int[] numbers, int sumTarget) {
    for (int first = 0; first < numbers.length; first++) {
        for (int second = first + 1; second < numbers.length; second++) {
            if (numbers[first] + numbers[second] == sumTarget) {
                return new int[]{first, second};
            }
        }
    }
    return new int[0];
}

Optimized Hash Map Solution:

import java.util.HashMap;

public int[] optimizedTwoSum(int[] numbers, int sumTarget) {
    HashMap<Integer, Integer> valueIndexMap = new HashMap<>();
    for (int current = 0; current < numbers.length; current++) {
        int complement = sumTarget - numbers[current];
        if (valueIndexMap.containsKey(complement)) {
            return new int[]{valueIndexMap.get(complement), current};
        }
        valueIndexMap.put(numbers[current], current);
    }
    return new int[0];
}