Completed Binary Search 2

problem1.py

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+# LEETCODE PROBLEM 34. FIND FIRST AND LAST POSITION OF ELEMENT IN SORTED ARRAY
+# TIME COMPLEXITY: O(log N) where N is the number of elements in the array
+# SPACE COMPLEXITY: O(1)
+# Any problem you faced while coding this: Tried a new methdology on how can I avoid using
+# two different function, I had some hiccups with the logic but later was successful
+
+class Solution(object):              
+    def searchRange(self, nums, target):
+        """
+        :type nums: List[int]
+        :type target: int
+        :rtype: List[int]
+        """
+
+        def findTarget(nums,start,end,target,isFirst):
+            bestFound=-1
+            while start<=end:
+                mid=(start+end)//2
+                if nums[mid]<target:
+                    start=mid+1
+                elif nums[mid]>target:
+                    end=mid-1
+                else:
+                    bestFound=mid
+                    if isFirst:
+                        end=mid-1
+                    else:
+                        start=mid+1
+            return bestFound
+
+        high=len(nums)-1
+        firstValue=findTarget(nums,0,high,target,True)
+        if firstValue==-1:
+            return[-1,-1]
+        secondValue=findTarget(nums,firstValue,high,target,False)
+        return [firstValue,secondValue]
+

problem2.py

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+# LEETCODE PROBLEM 153. FIND MINIMUM IN ROTATED SORTED ARRAY
+# TIME COMPLEXITY: O(log N) where N is the number of elements in the array
+# SPACE COMPLEXITY: O(1) 
+# Any problem you faced while coding this: None
+
+class Solution(object):
+    def findMin(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: int
+        """
+        low=0
+        high=len(nums)-1
+        while low<high:
+            mid=(low+high)//2
+            if nums[mid]<=nums[high]:
+                high=mid
+            else:
+                low=mid+1
+        return nums[low]

problem3.py

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+# LEETCODE PROBLEM 162. FIND PEAK ELEMENT
+# TIME COMPLEXITY: O(log N) where N is the number of elements in the array
+# SPACE COMPLEXITY: O(1)
+# Any problem you faced while coding this: Had a bit issue when I was making the edge cases 
+# ie the index 0 and index n-1 logic
+
+class Solution(object):
+    def findPeakElement(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: int
+        """
+        low=0
+        high=len(nums)-1
+        while (low<=high):
+            mid=(low+high)//2
+            if (mid==0 or nums[mid-1]<nums[mid]) and (mid==len(nums)-1 or nums[mid]>nums[mid+1]):
+                return mid
+            elif (mid==len(nums)-1 or nums[mid+1]>nums[mid]):
+                low=mid+1
+            else:
+                high=mid-1
+        return -1