95 lines
2 KiB
Markdown
95 lines
2 KiB
Markdown
# 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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#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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##### 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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##### Tips:
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> [!tip] Why `mid = l + (r - l) / 2`, not `mid = (l + r) / 2`
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> Avoids integer overflow when l and r is big
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> [!tip] Why `r = array.size() - 1`
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> Avoids OOB when (l == r && r == array.size())
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> This happens if 1 is not subtracted
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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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