feat(792-binary-search): add py3 solution

0792_binary-search/README.md

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+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.
+
+**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.

0792_binary-search/python3/solution.py

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+class Solution:
+    def search(self, nums: List[int], target: int) -> int:
+        start = 0
+        end = len(nums) - 1
+        
+        while start <= end:
+            mid = (start + end) // 2
+            
+            if nums[mid] == target:
+                return mid
+            elif target < nums[mid]:
+                end = mid - 1
+            else:
+                start = mid + 1
+        
+        return -1