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+# Leetcode Binary-Search
+
+#### 2022-07-09 09:34
+
+> ##### Algorithms:
+> #algorithm #binary_search 
+> ##### Data structures:
+> #DS #array 
+> ##### Difficulty:
+> #coding_problem #difficulty-easy
+> ##### Additional tags:
+> #leetcode #CS_list_need_practicing 
+> ##### Revisions:
+> N/A
+
+##### Related topics:
+```expander
+tag:#binary_search
+```
+ 
+ 
+
+##### Links:
+- [Link to problem](https://leetcode.com/problems/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`.
+
+You must write an algorithm with `O(log n)` runtime complexity.
+
+#### Examples
+
+**Example 1:**
+
+**Input:** nums = [-1,0,3,5,9,12], target = 9
+**Output:** 4
+**Explanation:** 9 exists in nums and its index is 4
+
+**Example 2:**
+
+**Input:** nums = [-1,0,3,5,9,12], target = 2
+**Output:** -1
+**Explanation:** 2 does not exist in nums so return -1
+
+#### Constraints
+
+-   `1 <= nums.length <= 104`
+-   `-104 < nums[i], target < 104`
+-   All the integers in `nums` are **unique**.
+-   `nums` is sorted in ascending order.
+
+### Thoughts
+
+> [!summary]
+> This is a #binary_search problem.
+
+Key takeout:
+```
+int r = nums.size() - 1;
+```
+make sure r is never OOB (l == r && r = array.size())
+
+### Solution
+
+==#TODO: write in recursion==
+
+iteration
+```cpp
+class Solution {
+public:
+  int search(vector<int> &nums, int target) {
+    // Why - 1?
+    // Make sure mid is never OOB (l == r && r = size)
+    int r = nums.size() - 1;
+    int l = 0;
+    int mid;
+    int val;
+
+    do {
+      // l + ( r - l ) / 2 or (l + r) / 2 Both ok
+      mid = (l + r) / 2;
+      val = nums[mid];
+      if (val == target) {
+        return mid;
+      } else if (val < target) {
+        l = mid + 1;
+      } else {
+        r = mid - 1;
+      }
+    } while (l <= r);
+
+    return -1;
+  }
+};
+```
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