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1_TwoSum.cpp
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1_TwoSum.cpp
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/*
* Problem: 1
* Name: Two Sum
* Difficulty: Easy
* Topic: Arrays & Hashing
* Link: https://leetcode.com/problems/two-sum/
*/
#include <bits/stdc++.h>
using namespace std;
// Brute-Force Approach
// Time Complexity: O(n^2)
// Space Complexity: O(1)
vector<int> twoSumBruteForce(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 vector<int>();
}
// Two-Pass Hash Table
// Time Complexity: O(n)
// Space Complexity: O(n)
vector<int> twoSumTwoPassHashTable(vector<int>& nums, int target) {
map<int, int> valueIndex;
for (int i = 0; i < nums.size(); i++){
valueIndex[nums[i]] = i;
}
for (int i = 0; i < nums.size(); i++){
int complement = target - nums[i];
if (valueIndex.count(complement) == 1 && valueIndex[complement] != i){
return {i, valueIndex[complement]};
}
}
return vector<int>();
}
// One-Pass Hash Table
// Time Complexity: O(n)
// Space Complexity: O(n)
vector<int> twoSumOnePassHashTable(vector<int>& nums, int target) {
map<int, int> valueIndex;
for (int i = 0; i < nums.size(); i++){
int complement = target - nums[i];
if (valueIndex.count(complement)){
return {i, valueIndex[complement]};
}
valueIndex[nums[i]] = i;
}
return vector<int>();
}