Completed Binary-Search-2

Find_Minimum.py

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+# Time Complexity : O(logn)
+# Space Complexity : O(1)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No 
+
+# Your code here along with comments explaining your approach
+# Do binary serach and check if mid is greater than right , then minimum is in right elese left.
+# One side is always sorted.
+import sys
+class Solution(object):
+    def findMin(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: int
+        """
+        l, r = 0, len(nums) - 1
+        while l < r:
+            mid = (l + r) // 2
+            #right half
+            if nums[mid] > nums[r]:
+                l = mid + 1
+            #left half
+            else:
+                r = mid
+        return nums[r]
+

Find_peak.py

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+# Time Complexity : O(log n)
+# Space Complexity : O(1)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : missed first and last edge case.
+
+# Your code here along with comments explaining your approach
+# 2 stacks - 1 for storing actual pushed values,
+# 2 - for storing minimum values at each level,
+# These 2 stacks must have 1:1 mapping
+
+class Solution(object):
+    def findPeakElement(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: int
+        """
+        l = 0
+        n = len(nums)
+        r =  n - 1
+
+        while l <= r:
+            mid = l + ( r - l ) // 2
+            if (mid == 0 or nums[mid] > nums[mid-1]) and (mid == n-1 or nums[mid] > nums[mid+1]):
+                return mid
+            elif nums[mid + 1] > nums[mid] and mid < n-1:
+                l = mid + 1
+            else:
+                r = mid - 1
+        return - 1

First_Last.py

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+# Time Complexity : O(logn) -> 2 binary search = O(2logn) = O(logn)
+# Space Complexity : O(1)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No 
+
+# Your code here along with comments explaining your approach
+# Find first occurence of target using firstposition function- returns the first index
+# Check if firstposition function returns -1 then no target exists.
+# Else find the last position of the target using other lastposition function similar to first.
+class Solution(object):
+    def searchRange(self, nums, target):
+        """
+        :type nums: List[int]
+        :type target: int
+        :rtype: List[int]
+        """
+        def firstposition(nums,target,l,r):
+            while l <= r:
+                mid = l + (r - l)// 2
+                if nums[mid] == target:
+                    if mid == 0 or nums[mid -1] != target:
+                        return mid
+                    else:
+                        r = mid - 1
+                elif nums[mid] < target:
+                    l = mid + 1
+                else:
+                    r = mid - 1
+            return -1
+
+        def lastposition(nums,target,l ,r):
+            while l <= r:
+                mid = l + (r - l) // 2
+                if nums[mid] == target:
+                    if mid == n or nums[mid+1] != target:
+                        return mid
+                    else:
+                        l = mid + 1
+                elif nums[mid] < target:
+                    l = mid + 1
+                else:
+                    r = mid - 1
+            return -1
+
+        n = len(nums) -1
+        first = firstposition(nums,target,0,n)
+        if first == -1:
+            return [-1,-1]
+        last = lastposition(nums,target,0 ,n)
+        return [first,last]
+