Complete Binary Search 2

Problem1.py

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+"""
+(https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/)
+
+Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value.
+
+Your algorithm's runtime complexity must be in the order of O(log n).
+
+If the target is not found in the array, return [-1, -1].
+"""
+class Solution(object):
+    def searchRange(self, nums, target):
+        """
+        :type nums: List[int]
+        :type target: int
+        :rtype: List[int]
+        """
+        def findFirst(nums, target):
+            low, high = 0, len(nums) - 1
+            first = -1
+            while low <= high:
+                mid = (low + high) // 2
+                if nums[mid] == target:
+                    first = mid
+                    high = mid - 1  # keep going left
+                elif nums[mid] < target:
+                    low = mid + 1
+                else:
+                    high = mid - 1
+            return first
+
+        def findLast(nums, target):
+            low, high = 0, len(nums) - 1
+            last = -1
+            while low <= high:
+                mid = (low + high) // 2
+                if nums[mid] == target:
+                    last = mid
+                    low = mid + 1  # keep going right
+                elif nums[mid] < target:
+                    low = mid + 1
+                else:
+                    high = mid - 1
+            return last
+
+        first = findFirst(nums, target)
+        last = findLast(nums, target)
+        return [first, last]

Problem2.py

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+
+"""
+## Problem 2: (https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/)
+
+Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
+
+(i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).
+
+Find the minimum element.
+
+You may assume no duplicate exists in the array.
+
+Example 1:
+Input: [3,4,5,1,2]
+Output: 1
+
+Example 2:
+Input: [4,5,6,7,0,1,2]
+Output: 0
+"""
+
+class Solution(object):
+    def findMin(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: int
+        """
+        low, high = 0, len(nums) - 1
+
+        while low < high:
+            mid = (low + high)//2
+            if nums[mid] > nums[high]:
+                low = mid  + 1
+            else:
+                high = mid 
+
+        return nums[low]

Problem3.py

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+"""
+## Problem 3: (https://leetcode.com/problems/find-peak-element/)
+A peak element is an element that is greater than its neighbors.
+
+Given an input array nums, where nums[i] ≠ nums[i+1], find a peak element and return its index.
+
+The array may contain multiple peaks, in that case return the index to any one of the peaks is fine.
+
+You may imagine that nums[-1] = nums[n] = -∞.
+
+Example 1:
+
+Input: nums = [1,2,3,1]
+Output: 2
+Explanation: 3 is a peak element and your function should return the index number 2.
+Example 2:
+
+Input: nums = [1,2,1,3,5,6,4]
+Output: 1 or 5 
+Explanation: Your function can return either index number 1 where the peak element is 2, 
+
+             or index number 5 where the peak element is 6.
+Note:
+
+"""
+
+class Solution(object):
+    def findPeakElement(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: int
+        """
+        low, high = 0, len(nums)-1
+        while(low<high):
+            mid = (low + high)//2
+            if nums[mid] < nums[mid+1]:
+                low = mid + 1
+            else:
+                high = mid
+        return low