binary 2

find-min-in-rotated.py

@@ -0,0 +1,13 @@
+class Solution:
+    def findMin(self, nums: List[int]) -> int:
+        left = 0
+        right = len(nums)-1
+
+        while left < right:
+            mid = left + (right - left)//2
+            if nums[mid] > nums[right]:
+                left = mid + 1
+            else:
+                right = mid
+
+        return nums[left]

find-peak-element.py

@@ -0,0 +1,17 @@
+class Solution:
+    def findPeakElement(self, nums: List[int]) -> int:
+        # -inf [1,2,1,3,5,6,4] -inf
+        #       l     m l lr
+        # -inf [1,2,1,3,5,6,7] -inf, 7 is the peak
+        #       l       l m l
+        left = 0
+        right = len(nums)-1
+        while left < right:
+            mid = left + (right - left)//2
+
+            if nums[mid] < nums[mid+1]:
+                left = mid + 1
+            else:
+                right = mid
+
+        return left

first-and-last.py

@@ -0,0 +1,50 @@
+class Solution:
+    # wrong - if you do binary search, then find 1, and expand window -> that is o(n) bc array could be all same number like [8,8,8,8,..]
+        def searchRange(self, nums: List[int], target: int) -> List[int]:
+            lower_bound = self.binarySearchLower(nums, target)
+            if lower_bound == -1:
+                return [-1, -1]
+
+            upper_bound = self.binarySearchUpper(nums, target)
+
+            return [lower_bound, upper_bound]
+
+        def binarySearchLower(self,nums, target):
+            left = 0
+            right = len(nums)-1
+
+            while left <= right:
+                mid = left + (right-left)//2
+                if nums[mid]==target:
+                    # we're at lower
+                    if left == mid or nums[mid-1] < target:
+                        return mid
+                    else:
+                        right = mid - 1 
+                elif nums[mid] < target:
+                    left = mid + 1
+                else:
+                    right = mid-1
+
+            return -1
+
+        def binarySearchUpper(self,nums, target):
+            left = 0
+            right = len(nums)-1
+
+# [5,7,7,8,8,8,10]
+ #.l     l m    r
+            while left <= right:
+                mid = left + (right-left)//2
+                if nums[mid]==target:
+                    # we're at upper bound 
+                    if right == mid or nums[mid+1] > target:
+                        return mid
+                    else:
+                        left = mid + 1 
+                elif nums[mid] < target:
+                    left = mid + 1
+                else:
+                    right = mid-1
+
+            return -1