|
1 | 1 | --- |
2 | 2 | title: "Refinaid" |
3 | | -publishedAt: '2024-06-25' |
| 3 | +publishedAt: 2024-08-27 |
4 | 4 | category: Web Development |
5 | 5 | tags: |
6 | 6 | - Go |
7 | 7 | - Python |
8 | 8 | - Leetcode |
9 | 9 | - Array |
10 | 10 | - Hash Table |
11 | | -summary: 'Given an array of integers `nums` and an integer `target`, *return indices of the two numbers such that they add up to `target`*.' |
| 11 | +summary: |
12 | 12 | banner: /images/projects/refinaid.png |
13 | | -alt: "LeetCode 0001. Two Sum - Hash Map Solution | Go, Python, C++" |
14 | | -mathjax: true |
| 13 | +alt: |
15 | 14 | --- |
16 | 15 |
|
17 | | -Link 👉🏻 [1. Two Sum](https://leetcode.com/problems/two-sum/) |
18 | | - |
19 | | - |
20 | | - |
21 | | -### Description |
22 | | - |
23 | | - |
24 | | -Given an array of integers `nums` and an integer `target`, *return indices of the two numbers such that they add up to `target`*. |
25 | | - |
26 | | -You may assume that each input would have exactly one solution, and you may not use the same element twice. |
27 | | - |
28 | | -You can return the answer in any order. |
29 | | - |
30 | | -**Example 1:** |
31 | | -- Input: `nums = [2,7,11,15], target = 9` |
32 | | -- Output: `[0,1]` |
33 | | -- Explanation: `Because nums[0] + nums[1] == 9, we return [0, 1].` |
34 | | - |
35 | | -**Example 2:** |
36 | | -- Input: `nums = [3,2,4], target = 6` |
37 | | -- Output: `[1,2]` |
38 | | - |
39 | | -**Example 3:** |
40 | | -- Input: `nums = [3,3], target = 6` |
41 | | -- Output: `[0,1]` |
42 | | - |
43 | | -**Constraints:** |
44 | | - |
45 | | -- <code>2 <= nums.length <= 10<sup>4</sup></code> |
46 | | -- <code>-10<sup>9</sup> <= nums[i] <= 10<sup>9</sup></code> |
47 | | -- <code>-10<sup>9</sup> <= target <= 10<sup>9</sup></code> |
48 | | -- <code>Only one valid answer exists.</code> |
49 | | - |
50 | | -**Follow-up:** Can you come up with an algorithm that is less than <code>O(n<sup>2</sup>)</code> time complexity? |
51 | | - |
52 | | -### Intuition |
53 | | - |
54 | | -Use HashMap to keep numbers and their indices we found. |
55 | | - |
56 | | - |
57 | | -> Create a Hash Table in `GO` with `make` function <sup>[Golang Maps](https://www.geeksforgeeks.org/golang-maps/)</sup>, and use `map[key] = value` to set the value of the key. |
58 | | -> |
59 | | -> ```go |
60 | | -> numMap := make(map[int]int) |
61 | | -> ``` |
62 | | -
|
63 | | -> **How to Work with maps?** <sup>[Go maps in action#Working with maps](https://go.dev/blog/maps)</sup> |
64 | | -> |
65 | | -> A two-value assignment tests for the existence of a key: |
66 | | -> |
67 | | -> ```go |
68 | | -> i, ok := m["route"] |
69 | | -> ``` |
70 | | -> |
71 | | -> In this statement, the first `value (i)` is assigned the value stored under the key "route". If that key doesn't exist, i is the value type's zero `value (0)`. The second `value (ok)` is a `bool` that is true if the key exists in the map, and false if not. |
72 | | -> |
73 | | -> To test for a key without retrieving the value, use an underscore in place of the first value: |
74 | | -> |
75 | | -> ```go |
76 | | -> _, ok := m["route"] |
77 | | -> ``` |
78 | | -> |
79 | | -> To iterate over the contents of a map, use the range keyword: |
80 | | -> |
81 | | -> ```go |
82 | | -> for key, value := range m { |
83 | | -> fmt.Println("Key:", key, "Value:", value) |
84 | | -> } |
85 | | -> ``` |
86 | | -
|
87 | | -### Approach |
88 | | -
|
89 | | -1. Traverse the `nums` array and store the difference between the `target` and the current `number` as the `key` and the `index` as the `value` in the HashMap. |
90 | | -2. If the current `number` is already in the HashMap, return the `index` of the current `number` and the `index` stored in the HashMap. |
91 | | -3. We still need to return an empty array if there is no solution. |
92 | | -
|
93 | | -
|
94 | | -### Complexity |
95 | | -- Time complexity: $O(n)$ |
96 | | -
|
97 | | -- Space complexity: $O(n)$ |
98 | | -
|
99 | | -### Code |
100 | | -
|
101 | | -```go |
102 | | -// Go Solution |
103 | | -func twoSum(nums []int, target int) []int { |
104 | | - hashMap := make(map[int]int) |
105 | | -
|
106 | | - for i, num := range nums { |
107 | | - if j, ok := hashMap[num]; ok { |
108 | | - return []int{j, i} |
109 | | - } |
110 | | - hashMap[target - num] = i |
111 | | - } |
112 | | -
|
113 | | - return []int{} |
114 | | -} |
115 | | -``` |
116 | | -
|
117 | | -```python |
118 | | -# Python Solution |
119 | | -class Solution: |
120 | | - def twoSum(self, nums: List[int], target: int) -> List[int]: |
121 | | - hMap = {} |
122 | | - |
123 | | - for i in range(len(nums)): |
124 | | - if nums[i] in hMap: |
125 | | - return [hMap[nums[i]], i] |
126 | | - else: |
127 | | - hMap[target - nums[i]] = i |
128 | | - |
129 | | - return [] |
130 | | -``` |
131 | | - |
132 | | -```cpp |
133 | | -// C++ Solution |
134 | | -class Solution { |
135 | | -public: |
136 | | - vector<int> twoSum(vector<int>& nums, int target) { |
137 | | - unordered_map<int, int> hashMap; |
138 | | - |
139 | | - for (int i = 0; i < nums.size(); ++i) { |
140 | | - int num = nums[i]; |
141 | | - if (hashMap.find(num) != hashMap.end()) { |
142 | | - return {hashMap[num], i}; |
143 | | - } |
144 | | - hashMap[target - num] = i; |
145 | | - } |
146 | | - |
147 | | - return {}; |
148 | | - } |
149 | | -}; |
150 | | -``` |
151 | | - |
152 | | - |
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