Binary search -2 completed

Find Minimum in Rotated Sorted Array.py

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+# Check if the first elememt in the array < last elemet in the array. If it is, return the 
+# first element. Else, take the mid and see if array[mid-1]>array[mid]. if it is, return mid. Else,
+#  check which side of the array is sorted. If the left side is sorted, low = mid+1 else high = mid -1
+
+#  Time complexity: O(logn)
+#  Space complexity: O(1)
+
+
+class Solution(object):
+    def findMin(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: int
+        """
+        low = 0
+        high = len(nums)-1
+
+        while low<=high:
+            if nums[low]<nums[high]:
+                return nums[low]        
+            mid = (low + high)//2
+            if mid>0 and nums[mid-1]>nums[mid]:
+                return nums[mid]
+            else:
+                if nums[low]<=nums[mid]:
+                    low = mid+1
+                else:
+                    high=mid-1
+        return nums[mid]
+
+
+

Find Peak Element.py

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+# If there is only one element, return its index. If there are 2 elements, retuen the index of the 
+# grater one. Else, calcualte mid. Check if array[mid-1]<array[mid] and array[mid]>array[mid+1] to see 
+# if its the peek element. If true, return mid. Else, check if array[mid+1]>array[mid], if yes, low = mid+1
+# else high=mid -1
+
+# Time complexity: O(logn)
+# Space complexity: O(1)
+
+
+class Solution(object):
+    def findPeakElement(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: int
+        """
+        n = len(nums)
+        if n==1:
+            return 0
+        if n==2:
+            return 0 if nums[0]>nums[1] else 1
+
+        low=0
+        high = n-1
+        while low<=high:
+            mid = (low+high)//2
+            if (mid==0 or nums[mid]>nums[mid-1]) and (mid==n-1 or nums[mid]>nums[mid+1] ):
+                return mid
+            elif mid<n-1 and nums[mid]<nums[mid+1]:
+                low=mid+1
+            elif  mid+1<n and nums[mid]>nums[mid+1]:
+                high=mid-1
+

First and last positioned Element in Sorted Array.py

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+# For first index - Take mid, if array[mid] is equal to target, check if the element before it is leeser than the 
+# current element. If it is, then return mid. Else, high=mid-1. If the elememt < target, low =mid+1
+#  else high=mid-1
+# For last index - Take mid, if array[mid] is equal to target, check if the element after it is grater than the 
+# current element. If it is, then return mid. Else, low=mid+1. If the elememt < target, low =mid+1
+#  else high=mid-1
+
+# Time complexity: O(logn)
+# Space complexity: O(1)
+
+class Solution(object):
+    def searchRange(self, nums, target):
+        """
+        :type nums: List[int]
+        :type target: int
+        :rtype: List[int]
+        """
+        low = 0
+        high = len(nums)-1
+        first_ind=-1
+        while low<=high:
+            mid = (low+high)//2
+            if nums[mid] == target:
+                if mid == 0 or nums[mid-1]<nums[mid]:
+                    first_ind=mid
+                    break
+                else:
+                    high=mid-1
+            else:
+                if nums[mid] < target:
+                    low = mid+1
+                else:
+                    high = mid-1
+
+        low = 0
+        high = len(nums)-1
+        last_ind =-1
+        while low<=high:
+            mid = (low+high)//2
+            if nums[mid] == target:
+                if mid == len(nums)-1 or nums[mid]<nums[mid+1]:
+                    last_ind=mid
+                    break
+                else:
+                    low=mid+1
+            else:
+                if nums[mid] < target:
+                    low = mid+1
+                else:
+                    high = mid-1
+
+        return[first_ind,last_ind]
+
+
+
+