From 80eb1aedbab8555aaee04e37f203d9857e0b9a73 Mon Sep 17 00:00:00 2001
From: Shubham Jain <shubham@Shubhams-MacBook-Air.local>
Date: Sat, 27 Dec 2025 17:37:59 -0800
Subject: [PATCH] Completed BinarySearch2

---
 Solutions.py | 154 +++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 154 insertions(+)
 create mode 100644 Solutions.py

diff --git a/Solutions.py b/Solutions.py
new file mode 100644
index 00000000..6fc8d5a2
--- /dev/null
+++ b/Solutions.py
@@ -0,0 +1,154 @@
+# Problem 1: Problem 1: (https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/)
+# Time Complexity : O(Log n)
+# Space Complexity : O(1)
+# Did this code successfully run on Leetcode :yes
+# Any problem you faced while coding this :no
+
+
+class Solution(object):
+    def searchRange(self, nums, target):
+        """
+        :type nums: List[int]
+        :type target: int
+        :rtype: List[int]
+        """
+        ## Start with trying to find the first occourance of the target
+        ## since the search space can be reduced to half, intution is to use binary search 
+        ## how to search second occurance ? again search by using binary search
+
+        def first_binary_search(self,nums,target):
+            l=0
+            r= len(nums)-1
+            while (l<=r):
+    
+                mid = l + (r-l)/2
+                ## check if mid is the target, now that we have the target we need to figure if this is first occurance
+                if nums[mid] == target:
+                    ## checking first occurance by checking with neighbors since array is non decreasing
+                    if (mid ==0 or nums[mid] > nums[mid-1]):
+                        return mid
+                    ## if not first occurance keep moving 
+                    else:
+                        r = mid -1
+                ## moving left since mid is bigger never going to find in right side        
+                elif nums[mid] > target:
+                    r=mid-1
+                ## move right    
+                else:
+                    l= mid+1
+            return -1
+        
+        def second_binary_search(self,nums,target,first):
+            ## once we are done with first occurance its time to find the last ocurance
+
+            l=first
+            r= len(nums)-1
+            while (l<=r):
+                mid = l + (r-l)/2
+                ## if mid is target check if last occurance by comparing to right neighbour 
+                if nums[mid] == target and (mid== len(nums)-1 or nums[mid] < nums[mid+1]):
+                    return mid
+                ## move left if greater than target    
+                elif nums[mid] > target:
+                    r = mid-1
+                ## move right    
+                else:
+                    l = mid+1
+            return -1
+        
+        ## start with the base condition, if size 0 return
+        if len(nums) ==0 :
+            return [-1,-1]
+
+        first = first_binary_search(self,nums,target)
+        ## if no first index found, no need to do second binary search
+        if first == -1:
+            return [-1,-1]
+
+        second= second_binary_search(self,nums,target,first)
+        return [first,second]
+
+
+
+
+#####################################
+# Problem 2: (https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/)
+# Time Complexity : O(Logn)
+# Space Complexity : O(1)
+# Did this code successfully run on Leetcode :yes
+# Any problem you faced while coding this :no
+
+
+class Solution(object):
+    def findMin(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: int
+        """
+        ## We can reduce the search space by half either looking into left or right side since we have 
+        ## rotated sorted, Intution is to use binary search
+        left, right = 0, len(nums)-1
+    
+        while left <= right:
+            
+            if nums[left]<= nums[right]:
+                return nums[left]
+
+            mid = left + (right-left)/2
+            ## checking if mid is smaller than both the neighbours, if yes return index
+            if (nums[mid] < nums[mid+1]) and nums[mid] < nums[mid-1]:
+                return nums[mid]
+            
+            ## check if left side is sorted and check there
+            if nums[left] <= nums[mid]:
+                left = mid+1
+            ## else check the right side    
+            else:
+                right= mid -1
+        return -1
+
+
+            
+
+
+#####################################
+
+# Problem 3: (https://leetcode.com/problems/find-peak-element/)
+# Time Complexity : O(logn)
+# Space Complexity : O(1)
+# Did this code successfully run on Leetcode :yes
+# Any problem you faced while coding this :no
+
+
+class Solution(object):
+    def findPeakElement(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: int
+        """
+
+        ## we can start with binary search why ?
+        ## because we want to reduce the search space everytime by half
+        ## how do I know if I have reached the peak? - check the neighbors
+        ## if highest from both the neighbours then we can say we have reached a peak
+        ## where to go in case we are not at peak? go to the higher side and we will surely reach the peak
+        ## we can go either side, to find a peak, but for this we will traverse to higher side
+        l=0
+        r=len(nums)-1
+
+        while (l<=r):
+            if len(nums) ==1:
+                return 0
+
+            mid = l + (r-l)/2
+            ## if mid is greater than both left and right neighbour
+            if (mid ==0 or nums[mid] > nums[mid-1]) and (mid ==len(nums)-1 or nums[mid] > nums[mid+1]):
+                return mid
+            ## if left is higher move right
+            if mid!=0 and nums[mid-1] > nums[mid]:
+                r = mid-1
+            ## if right is higher move left
+            else:
+                l = mid+1
+        return -1
+        
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