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704 Binary Search 406000777e5f4354a1e4eb947aa1d4c1/704 Binary Search 406000777e5f4354a1e4eb947aa1d4c1.md

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+# 704. Binary Search
+
+Difficulty: easy
+Done: Yes
+Last edited: February 12, 2022 5:53 PM
+Link: https://leetcode.com/problems/binary-search/
+Topic: binary search
+
+# Problem
+
+Given an array of integers `nums` which is sorted in ascending order, and an integer `target`, write a function to search `target` in `nums`. If `target` exists, then return its index. Otherwise, return `-1`.
+
+# Solution
+
+goal of the algorithm is to divide array in half, if our `target` is smaller than current then we take the left side of the array, if `target` is larger than the *current* then we take the right side of the array. Repeat
+
+# Whiteboard
+
+![target = 9](704%20Binary%20Search%20406000777e5f4354a1e4eb947aa1d4c1/Screen_Shot_2022-02-12_at_5.36.32_PM.png)
+
+target = 9
+
+![target = 2 (not existing)](704%20Binary%20Search%20406000777e5f4354a1e4eb947aa1d4c1/Screen_Shot_2022-02-12_at_5.36.49_PM.png)
+
+target = 2 (not existing)
+
+# Code
+
+```python
+class Solution:
+    def search(self, nums: List[int], target: int) -> int:
+        
+        try:
+            i = nums.index(target)
+        except:
+            i = -1
+            
+        return i
+```
+
+```python
+class Solution:
+    def search(self, nums: List[int], target: int) -> int:
+        # iteratively
+        
+        left = 0
+        right = len(nums) - 1
+        while left <= right:
+            middle = left + (right-left) // 2
+            
+            if nums[middle] == target: 
+                return middle
+            
+            if target < nums[middle]: 
+                # use left side
+                right = middle - 1
+            
+            if target > nums[middle]:
+                # use right side
+                left = middle + 1
+                
+        return -1
+							
+```