notes/OJ notes/pages/Binary Search Algorithm.md

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# Binary Search Algorithm
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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/)
- [cppreference, find](https://en.cppreference.com/w/cpp/container/set/find)
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---
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### How to implement Binary search?
#### a: Use cpp's library
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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
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##### 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
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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
// 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;
}
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```