Two Sum

In this article we will learn about LeetCode First question Two Sum in detail with optimized solution with Example.

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.

Table of Contents

Brute Force Code

Java Example

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};
                }
            }
        }
        return new int[] {};
    }
}

// 2023

C++ Example

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        for (int i = 0; i < nums.size(); i++) {
            for (int j = i + 1; j < nums.size(); j++) {
                if (nums[i] + nums[j] == target) {
                    return {i, j};
                }
            }
        }
        return {};
    }
};
// 2023

Complexity :

Time complexity: O(N²)
Space Complexity: O(1)

Optimized Code

Java Optimized Example

import java.util.HashMap;
import java.util.Map;
 
class Solution {
    public int[] twoSum(int[] nums, int target) {
        Map<Integer, Integer> numToIndex = new HashMap<>();
        for (int i = 0; i < nums.length; i++) {
            if (numToIndex.containsKey(target - nums[i])) {
                return new int[] {numToIndex.get(target - nums[i]), i};
            }
            numToIndex.put(nums[i], i);
        }
        return new int[] {};
    }
}

C++ Optimized Example

#include <unordered_map>
 
class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int, int> mp;
       
        for(int i = 0; i < nums.size(); i++){
            if(mp.find(target - nums[i]) == mp.end())
                mp[nums[i]] = i;
            else
                return {mp[target - nums[i]], i};
        }
 
        return {-1, -1};
    }
};

Complexity :

Time complexity: O(N);
Space Complexity: O(N);

I hope you like this Amazing Two Sum article.🙂