# Leetcode ## [Two Sum](https://leetcode.com/problems/two-sum/)
DifficultyEasy
Topicsarrrays, hashmap
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. **Example 1:** ``` Input: nums = [2,7,11,15], target = 9 Output: [0,1] Explanation: Because nums[0] + nums[1] == 9, we return [0, 1]. ``` **Example 2:** ``` Input: nums = [3,2,4], target = 6 Output: [1,2] ``` **Example 3:** ``` Input: nums = [3,3], target = 6 Output: [0,1] ``` **Constraints:** * `2 <= nums.length <= 104` * `109 <= nums[i] <= 109` * `109 <= target <= 109` * **Only one valid answer exists.** **Follow-up:** Can you come up with an algorithm that is less than ``` O(n2) ``` time complexity? #### **Solution: Bruteforce** ```python class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: result = set() for x in range(0,len(nums)): for y in range(0,len(nums)): if x == y: continue if nums[x] + nums[y] == target: result.add(x) result.add(y) return result ```
Solution 2: Use hashmap: ```python class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: result = [] hashmap = {} for x in range(0, len(nums)): minus_data = target - nums[x] if minus_data in hashmap: result.append(hashmap[minus_data]) result.append(x) hashmap[nums[x]] = x return result ```
{% embed url="" %} ## [Two Sum II - Input Array Is Sorted](https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/)
Difficultymedium
Topicsarrrays, two pointers
Given a **1-indexed** array of integers `numbers` that is already ***sorted in non-decreasing order***, find two numbers such that they add up to a specific `target` number. Let these two numbers be `numbers[index1]` and `numbers[index2]` where `1 <= index1 < index2 <= numbers.length`. Return *the indices of the two numbers,* `index1` *and* `index2`*, **added by one** as an integer array* `[index1, index2]` *of length 2.* The tests are generated so that there is **exactly one solution**. You **may not** use the same element twice. Your solution must use only constant extra space. **Example 1:** 
Input: numbers = [2,7,11,15], target = 9
Output: [1,2]
Explanation: The sum of 2 and 7 is 9. 
Therefore, index1 = 1, index2 = 2. We return [1, 2].
**Example 2:** 
Input: numbers = [2,3,4], target = 6
Output: [1,3]
Explanation: The sum of 2 and 4 is 6. Therefore index1 = 1, index2 = 3. 
We return [1, 3].
**Example 3:** 
Input: numbers = [-1,0], target = -1
Output: [1,2]
Explanation: The sum of -1 and 0 is -1. Therefore index1 = 1, index2 = 2. 
We return [1, 2].
**Constraints:** * `2 <= numbers.length <= 3 * 104` * `-1000 <= numbers[i] <= 1000` * `numbers` is sorted in **non-decreasing order**. * `-1000 <= target <= 1000` * The tests are generated so that there is **exactly one solution**. #### Solution ``` // Some code ``` #### Additional Resources {% embed url="" %} ## Contains Duplicate #### Additional Resources {% embed url="" %} ## Min Stack Design a stack that supports push, pop, top, and retrieving the minimum element in constant time. Implement the `MinStack` class: * `MinStack()` initializes the stack object. * `void push(int val)` pushes the element `val` onto the stack. * `void pop()` removes the element on the top of the stack. * `int top()` gets the top element of the stack. * `int getMin()` retrieves the minimum element in the stack. You must implement a solution with `O(1)` time complexity for each function. **Example 1:** ``` Input ["MinStack","push","push","push","getMin","pop","top","getMin"] [[],[-2],[0],[-3],[],[],[],[]] Output [null,null,null,null,-3,null,0,-2] Explanation MinStack minStack = new MinStack(); minStack.push(-2); minStack.push(0); minStack.push(-3); minStack.getMin(); // return -3 minStack.pop(); minStack.top(); // return 0 minStack.getMin(); // return -2 ``` **Constraints:** * `231 <= val <= 231 - 1` * Methods `pop`, `top` and `getMin` operations will always be called on **non-empty** stacks. * At most `3 * 104` calls will be made to `push`, `pop`, `top`, and `getMin`. #### Solution ```python class MinStack(object): def __init__(self): self.stack = [] self.min_stack = [] def push(self, val): self.stack.append(val) val = min(val, self.min_stack[-1] if self.min_stack else val) self.min_stack.append(val) def pop(self): self.stack.pop() self.min_stack.pop() def top(self): return self.stack[-1] if self.stack else None def getMin(self): return self.min_stack[-1] if self.min_stack else None # Your MinStack object will be instantiated and called as such: # obj = MinStack() # obj.push(val) # obj.pop() # param_3 = obj.top() # param_4 = obj.getMin() ``` #### Additional Resources {% embed url="" %} ## Valid Parentheses ## Merge Two Sorted List ## Daily Temperatures --- # Agent Instructions: Querying This Documentation If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question. Perform an HTTP GET request on the current page URL with the `ask` query parameter: ``` GET https://docs.hackerspot.net/coding-practices/leetcode.md?ask= ``` The question should be specific, self-contained, and written in natural language. The response will contain a direct answer to the question and relevant excerpts and sources from the documentation. Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.