From a5e800effe30d6fc011f6456d18e48d20fd04fb3 Mon Sep 17 00:00:00 2001
From: devoalda <linuxes_mojoworld@aleeas.com>
Date: Sat, 18 Feb 2023 08:38:54 +0800
Subject: [PATCH] Added 2 Leetcode Problems

---
 content/en/posts/Leetcode-BinarySearch704.md  | 109 ++++++++++++++++++
 .../en/posts/Leetcode-FirstBadVersion278.md   |  98 ++++++++++++++++
 2 files changed, 207 insertions(+)
 create mode 100644 content/en/posts/Leetcode-BinarySearch704.md
 create mode 100644 content/en/posts/Leetcode-FirstBadVersion278.md

diff --git a/content/en/posts/Leetcode-BinarySearch704.md b/content/en/posts/Leetcode-BinarySearch704.md
new file mode 100644
index 0000000..fcc85cc
--- /dev/null
+++ b/content/en/posts/Leetcode-BinarySearch704.md
@@ -0,0 +1,109 @@
+---
+author: "Devoalda"
+authorEmoji: 🐺
+title: "Leetcode Binary Search (704)"
+date: 2023-02-18T07:39:33+08:00
+description: Leetcode Challenge 704
+draft: false
+hideToc: false
+enableToc: true
+enableTocContent: true
+tocPosition: inner
+tocLevels: ["h1", "h2", "h3"]
+math: true
+libraries: mathjax
+tags:
+  - leetcode
+  - python
+series:
+  -
+categories:
+  - LeetCode
+  - python
+image:
+---
+
+# Introduction
+
+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.
+
+{{< tabs Example_1 Example_2 >}}
+{{< tab >}}
+
+```shell
+Input: nums = [-1,0,3,5,9,12], target = 9
+Output: 4
+Explanation: 9 exists in nums and its index is 4
+```
+
+{{< /tab >}}
+
+{{< tab >}}
+
+```shell
+Input: nums = [-1,0,3,5,9,12], target = 2
+Output: -1
+Explanation: 2 does not exist in nums so return -1
+```
+
+{{< /tab >}}
+{{< /tabs >}}
+
+# Process
+
+This is a normal binary search, splitting each search iteration by $\frac{1}{2}$, I created a new function with `left` and `right` variables to indicate the high and low pointers on the array.
+
+Since the array is already sorted in ascending order, the larger numbers will always be to the right of the mid value.
+The mid index value could be found using the floor division of `left + right`:
+
+```python
+mid = (left + right) // 2
+```
+
+though I realise that using a right shift would make it faster
+
+```python
+mid = (left + right) >> 1
+```
+
+After getting the midpoint, there will be 3 conditions left:
+
+1. `nums[mid] == target`
+2. `nums[mid] < target`
+3. `nums[mid] > target`
+
+For each of this conditions, I'll update the pointers or return the mid value. this is all done in a while loop where `left` needs to be $\le$`right` and `left` should not go past `right` (Value Not Found).
+
+At the end, I'll return `-1` if the value isn't found.
+
+Calling the function in the return statement of the outer function definition, I'll pass `0` and `len(nums) -1` into the binary search function.
+
+This solution takes a Time Complexity of $O(\log n)$ and Space Complexity of $O(1)$ as it only requires the size of the array.
+
+# Solution
+
+```python
+class Solution:
+    def search(self, nums: List[int], target: int) -> int:
+        def bSearch(left, right):
+            while left <= right:
+                mid = (left + right)//2
+                if nums[mid] == target:
+                    return mid
+                elif target < nums[mid]:
+                    right = mid-1
+
+                else:
+                    left = mid + 1
+            return -1
+
+        return bSearch(0, len(nums)-1)
+```
+
+# Afterthoughts
+
+This is my first step into revising for my Data Structures and Algorithms. Binary search is one of the fastest searching algorithms, with $O(\log n)$ Time Complexity. Coupled with a fast Sorting algorithm, Users will be able to sort and search for values quickly and efficiently.
diff --git a/content/en/posts/Leetcode-FirstBadVersion278.md b/content/en/posts/Leetcode-FirstBadVersion278.md
new file mode 100644
index 0000000..f664154
--- /dev/null
+++ b/content/en/posts/Leetcode-FirstBadVersion278.md
@@ -0,0 +1,98 @@
+---
+author: "Devoalda"
+authorEmoji: 🐺
+title: "Leetcode First Bad Version (278)"
+date: 2023-02-18T08:39:33+08:00
+description: Leetcode Challenge 278
+draft: false
+hideToc: false
+enableToc: true
+enableTocContent: true
+tocPosition: inner
+tocLevels: ["h1", "h2", "h3"]
+math: true
+libraries: mathjax
+tags:
+  - leetcode
+  - python
+series:
+  -
+categories:
+  - LeetCode
+  - python
+image:
+---
+
+# Introduction
+
+You are a product manager and currently leading a team to develop a new product. Unfortunately, the latest version of your product fails the quality check. Since each version is developed based on the previous version, all the versions after a bad version are also bad.
+
+Suppose you have `n` versions `[1, 2, ..., n]` and you want to find out the first bad one, which causes all the following ones to be bad.
+
+You are given an API bool `isBadVersion(version)` which returns whether version is bad. Implement a function to find the first bad version. You should minimize the number of calls to the API.
+
+{{< tabs Example_1 Example_2 >}}
+{{< tab >}}
+
+```shell
+Input: n = 5, bad = 4
+Output: 4
+Explanation:
+call isBadVersion(3) -> false
+call isBadVersion(5) -> true
+call isBadVersion(4) -> true
+Then 4 is the first bad version.
+```
+
+{{< /tab >}}
+
+{{< tab >}}
+
+```shell
+Input: nums = [-1,0,3,5,9,12], target = 2
+Output: -1
+Explanation: 2 does not exist in nums so return -1
+```
+
+{{< /tab >}}
+{{< /tabs >}}
+
+# Process
+
+Using a binary search done in [Leetcode Binary Search 704]( {{< ref "posts/Leetcode-BinarySearch704.md" >}}), I was able to get the first bad version.
+
+First, I initialised my `left` variable to `1` as the lower limit was `1` and I couldn't start from 0. I initialised right to be `n` for the number of versions.
+
+In the same while loop, I calculated the mid index again, but this time with right shift. I first checked with the `isBadVersion` API for the midpoint. If it is true, I'll set the `firstIndex` variable to the midpoint, and the right index to `mid - 1` to check the left side.
+
+If the first condition is not met, I'll check the next half by updating `mid + 1` in the while loop.
+
+At the end of the loop, `firstIndex` is returned as the First Bad Version.
+
+This solution takes a Time Complexity of $O(\log n)$ and Space Complexity of $O(1)$
+
+# Solution
+
+```python
+# The isBadVersion API is already defined for you.
+# def isBadVersion(version: int) -> bool:
+
+class Solution:
+    def firstBadVersion(self, n: int) -> int:
+        left = 1
+        right = n
+        firstIndex = -1
+
+        while left <= right:
+            mid = (left + right) >> 1
+            if isBadVersion(mid):
+                firstIndex = mid
+                right = mid - 1
+            else:
+                left = mid + 1
+        return firstIndex
+```
+
+# Afterthoughts
+
+This is almost a duplicate of [Leetcode Binary Search 704]( {{< ref "posts/Leetcode-BinarySearch704.md" >}}), where binary search is used as the most efficient algorithm, splitting the sorted array into half each time through the iteration to search for the values required. I need more practice to be familiar and comfortable with these searching algorithms.