# Binary Search Algorithm

#### 2022-06-13 15:46

---

##### Algorithms:

#algorithm #binary_search

##### Data structures:

#array #vector #set #multiset

##### Difficulty:

#CS_analysis #difficulty-easy

##### Related problems:

##### Links:

- [g4g for manual implementation](https://www.geeksforgeeks.org/binary-search/)
- [cppreference, find](https://en.cppreference.com/w/cpp/container/set/find)

---

### How to implement Binary search?

#### a: Use cpp's library

Use cpp's set's [find](https://en.cppreference.com/w/cpp/container/set/find)
or [equal_range](https://en.cppreference.com/w/cpp/container/multiset/equal_range)

#### b: Manual

##### Tips:

> [!tip] Why `mid = l + (r - l) / 2`, not `mid = (l + r) / 2`
> Avoids integer overflow when l and r is big

> [!tip] Why `r = array.size() - 1`
> Avoids OOB when (l == r && r == array.size())
> This happens if 1 is not subtracted

1. Use a while loop:
   [[Leetcode Search-a-2D-Matrix#Solution]]

2. Use recursion:
   from g4g:

```cpp
// C++ program to implement recursive Binary Search
#include <bits/stdc++.h>
using namespace std;

// A recursive binary search function. It returns
// location of x in given array arr[l..r] is present,
// otherwise -1
int binarySearch(int arr[], int l, int r, int x) {
  if (r >= l) {
    int mid = l + (r - l) / 2;

    // If the element is present at the middle
    // itself
    if (arr[mid] == x)
      return mid;

    // If element is smaller than mid, then
    // it can only be present in left subarray
    if (arr[mid] > x)
      return binarySearch(arr, l, mid - 1, x);

    // Else the element can only be present
    // in right subarray
    return binarySearch(arr, mid + 1, r, x);
  }

  // We reach here when element is not
  // present in array
  return -1;
}

int main(void) {
  int arr[] = {2, 3, 4, 10, 40};
  int x = 10;
  int n = sizeof(arr) / sizeof(arr[0]);
  int result = binarySearch(arr, 0, n - 1, x);
  (result == -1) ? cout << "Element is not present in array"
                 : cout << "Element is present at index " << result;
  return 0;
}

```
