vault backup: 2022-06-15 21:50:42
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16
.obsidian/app.sync-conflict-20220615-121016-T44CT3O.json
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.obsidian/app.sync-conflict-20220615-121016-T44CT3O.json
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12
.obsidian/appearance.sync-conflict-20220615-121014-T44CT3O.json
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.obsidian/appearance.sync-conflict-20220615-121014-T44CT3O.json
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.obsidian/community-plugins.sync-conflict-20220615-121016-T44CT3O.json
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.obsidian/community-plugins.sync-conflict-20220615-121016-T44CT3O.json
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[
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"mrj-text-expand"
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]
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129
.obsidian/workspace.sync-conflict-20220615-121017-T44CT3O
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"CS notes/pages/Binary Search Algorithm.md",
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"CS notes/pages/Leetcode Reshape-The-Matrix.md",
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"Index.md",
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"Wiki/Wiki-Index.md",
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"Wiki/Learnings.md",
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"Wiki/Minecraft.md",
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"Wiki/Software Index.md",
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"CS notes/pages/Leetcode Pascal's-Triangle.md"
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]
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}
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# Binary Search Algorithm
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#### 2022-06-13 15:46
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___
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##### Algorithms:
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#algorithm #binary_search
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##### Data structures:
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#DS #array #vector #set #multiset
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##### Difficulty:
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#CS_analysis #difficulty-easy
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##### Related problems:
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```expander
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tag:#coding_problem tag:#binary_search -tag:#template_remove_me
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```
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- [[Leetcode Search-a-2D-Matrix]]
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##### Links:
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- [g4g for manual implementation](https://www.geeksforgeeks.org/binary-search/)
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- [cppreference, find](https://en.cppreference.com/w/cpp/container/set/find)
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___
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### How to implement Binary search?
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#### a: Use cpp's library
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Use cpp's set's [find](https://en.cppreference.com/w/cpp/container/set/find)
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or [equal_range](https://en.cppreference.com/w/cpp/container/multiset/equal_range)
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#### b: Manual
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1. Use a while loop:
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[[Leetcode Search-a-2D-Matrix#Solution]]
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2. Use recursion:
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from g4g:
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```cpp
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// C++ program to implement recursive Binary Search
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#include <bits/stdc++.h>
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using namespace std;
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// A recursive binary search function. It returns
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// location of x in given array arr[l..r] is present,
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// otherwise -1
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int binarySearch(int arr[], int l, int r, int x) {
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if (r >= l) {
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int mid = l + (r - l) / 2;
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// If the element is present at the middle
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// itself
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if (arr[mid] == x)
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return mid;
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// If element is smaller than mid, then
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// it can only be present in left subarray
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if (arr[mid] > x)
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return binarySearch(arr, l, mid - 1, x);
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// Else the element can only be present
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// in right subarray
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return binarySearch(arr, mid + 1, r, x);
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}
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// We reach here when element is not
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// present in array
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return -1;
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}
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int main(void) {
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int arr[] = {2, 3, 4, 10, 40};
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int x = 10;
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int n = sizeof(arr) / sizeof(arr[0]);
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int result = binarySearch(arr, 0, n - 1, x);
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(result == -1) ? cout << "Element is not present in array"
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: cout << "Element is present at index " << result;
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return 0;
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}
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```
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35
CS notes/pages/Leetcode Remove-Linked-List-Elements.md
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35
CS notes/pages/Leetcode Remove-Linked-List-Elements.md
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# Leetcode Remove-Linked-List-Elements
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#### 2022-06-15 21:50
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---
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##### Algorithms:
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#algorithm
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##### Data structures:
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#DS
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##### Difficulty:
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#<CHANGE_ME> #coding_problem #difficulty-easy
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##### Related topics:
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```expander
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tag:#<INSERT_TAG_HERE>
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```
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##### Links:
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- [Link to problem]()
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___
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### Problem
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#### Examples
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```markdown
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```
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#### Constraints
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### Thoughts
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> [!summary]
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> This is a #template_remove_me
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### Solution
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