Problem

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

Solution

Brute-Force

Brute-Force runs in $O(n^2)$ :

class Solution {
    public int[] twoSum(int[] nums, int target) {
        // O(n^2)
        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");
    }
}

Twice Hash

Hash twice: First takes $O(n)$ to build HashMap. Second to search the complement $O(n)$ times in $O(1)$ :

import java.util.*;

class Solution {
    public int[] twoSum(int[] nums, int target) {
        Map<Integer, Integer> val_loc = new HashMap<Integer, Integer>();
        
        // O(n)
        for (int i=0; i<nums.length; i++) {
            val_loc.put(nums[i], i);
        }
        
        // O(n)
        for (int i=0; i<nums.length; i++) {
            // O(1)
            Integer j = val_loc.get(target - nums[i]);
            
            if (j != null) {
                if (j != i) {
                    return (i < j) ? (new int[] {i, j}) : (new int[] {j, i});
                }
            }
        }
        
        throw new IllegalArgumentException("No solution");
    }
}

Once Hash

Search the existed elements in table while insert new pairs to it. Thus loop one time in $O(n)$ :

import java.util.*;

class Solution {
    public int[] twoSum(int[] nums, int target) {
        Map<Integer, Integer> val_loc = new HashMap<Integer, Integer>();
        
        // O(n)
        for (int i=0; i<nums.length; i++) {
            // O(1)
            Integer j = val_loc.get(target - nums[i]);
            
            if (j != null) {
                if (j != i) {
                    return (i < j) ? (new int[] {i, j}) : (new int[] {j, i});
                }
            }
            
            val_loc.put(nums[i], i);
        }
                
        throw new IllegalArgumentException("No solution");
    }   
}