LeetCode Algorithm 0001 - Two Sum (Easy)

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LeetCode Algorithm 0001 - Two Sum (Easy)

Problem Link: https://leetcode.com/problems/two-sum/description/

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Description

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].

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Solution C++

#pragma once

#include "pch.h"

// Problem: https://leetcode.com/problems/two-sum/description/

namespace P1TwoSum
{
    class Solution
    {
        public:
        vector<int> twoSum(vector<int>& nums, int target)
        {
#if false // two loop, time O(n^2), space O(1)
            for (size_t i = 0; i < nums.size(); i++)
            {
                for (size_t j = i + 1; j < nums.size(); j++)
                {
                    if (nums[i] + nums[j] == target)
                    {
                        return vector<int>{i, value2Index[complement]};
                    }
                }
            }

            logic_error e = logic_error("No two sum solution");
            throw e;
#endif

#if false // Two-pass Hash Table, time O(n), space O(n)

            unordered_map<int, int> value2Index;

            // add each element's value and its index to the table
            for (size_t i = 0; i < nums.size(); i++)
            {
                value2Index[nums[i]] = i;
            }

            // check if each element's complement
            for (int i = 0; i < nums.size(); i++)
            {
                int complement = target - nums[i];
                if (value2Index.count(complement) && value2Index[complement] != i)
                {
                    return vector<int>{i, value2Index[complement]};
                }
            }

            logic_error e = logic_error("No two sum solution");
            throw e;
#endif

#if true // One-pass Hash Table, time O(n), space O(n)

            unordered_map<int, int> value2Index;

            for (int i = 0; i < nums.size(); i++)
            {
                int complement = target - nums[i];
                if (value2Index.count(complement))
                {
                    return vector<int>{i, value2Index[complement]};
                }
                value2Index[nums[i]] = i;
            }

            logic_error e = logic_error("No two sum solution");
            throw e;
#endif

        }
    };
}

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