diff --git a/leetcode153.py b/leetcode153.py
new file mode 100644
index 00000000..5eb01b44
--- /dev/null
+++ b/leetcode153.py
@@ -0,0 +1,73 @@
+#!/usr/bin/env python3
+# -*- coding: utf-8 -*-
+"""
+Created on Sun Feb  8 15:37:19 2026
+
+@author: rishigoswamy
+
+    Problem:
+    ----------
+    https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/
+
+    Approach:
+    ----------
+    The array is originally sorted in ascending order and then rotated at
+    an unknown pivot, which causes the minimum element to be the rotation
+    point.
+    
+    If the array is already sorted (nums[low] < nums[high]), the first
+    element is the minimum and can be returned immediately.
+    
+    Otherwise, we use binary search to locate the pivot by checking the
+    relationship between nums[mid] and its neighboring elements.
+    If nums[mid] is smaller than nums[mid - 1], it is the minimum;
+    if nums[mid + 1] is smaller than nums[mid], then nums[mid + 1] is the
+    minimum.
+    
+    Based on whether nums[mid] lies in the left sorted portion or the right
+    rotated portion, we discard one half of the search space and continue.
+    
+    Time Complexity:
+    ----------------
+    O(log n), where n is the number of elements in the array.
+    
+    Space Complexity:
+    -----------------
+    O(1), since the search is performed in-place using constant extra space.
+    
+    Did this code successfully run on Leetcode:
+    ------------------------------------------
+    Yes
+    
+    Any problem you faced while coding this:
+    ---------------------------------------
+    Carefully handling boundary conditions when accessing nums[mid - 1]
+    and nums[mid + 1] to avoid index out-of-range errors.
+
+"""
+
+
+class Solution:
+    def findMin(self, nums) -> int:
+        low = 0
+        high = len(nums) - 1
+
+        if low == high:
+            return nums[0]
+        
+        if nums[low] < nums[high]:
+            return nums[low]
+
+        while low <= high:
+            mid = (low + high)//2
+
+            if nums[mid] < nums[mid-1]:
+                return nums[mid]
+            if nums[mid+1] < nums[mid]:
+                return nums[mid+1]
+            
+            if nums[mid] > nums[low]:
+                low = mid + 1
+            else:
+                high = mid - 1
+        
diff --git a/leetcode162.py b/leetcode162.py
new file mode 100644
index 00000000..b67d64ff
--- /dev/null
+++ b/leetcode162.py
@@ -0,0 +1,81 @@
+#!/usr/bin/env python3
+# -*- coding: utf-8 -*-
+"""
+Created on Sun Feb  8 15:43:49 2026
+
+@author: rishigoswamy
+
+    Problem:
+    ----------
+    https://leetcode.com/problems/find-peak-element/
+    
+    Approach:
+    ----------
+    A peak element is defined as an element that is strictly greater than
+    its neighboring elements, and the problem guarantees that a peak
+    always exists.
+    
+    We use binary search to efficiently find a peak by comparing the
+    middle element with its neighbors.
+    If the middle element is greater than both neighbors, it is a peak
+    and can be returned immediately.
+    
+    If the left neighbor is greater than the middle element, a peak must
+    exist in the left half of the array; otherwise, a peak must exist in
+    the right half.
+    By discarding half of the search space at each step, we maintain
+    logarithmic time complexity.
+    
+    Special boundary checks are performed for the first and last elements
+    to safely handle edge cases.
+    
+    Time Complexity:
+    ----------------
+    O(log n), where n is the number of elements in the array.
+    
+    Space Complexity:
+    -----------------
+    O(1), since the search is performed in-place using constant extra space.
+    
+    Did this code successfully run on Leetcode:
+    ------------------------------------------
+    Yes
+    
+    Any problem you faced while coding this:
+    ---------------------------------------
+    Handling boundary conditions for the first and last indices and
+    correctly deciding which half of the array to discard during the
+    binary search.
+
+
+"""
+import List
+
+class Solution:
+    def findPeakElement(self, nums: List[int]) -> int:
+        low = 0
+        high = len(nums) - 1
+
+        if high == low:
+            return 0
+
+        while low <= high:
+            
+            mid = (low+high)//2
+
+            if mid == 0 and nums[mid+1] < nums[mid]:
+                return mid
+            
+            elif mid == len(nums) - 1 and nums[mid-1] < nums[mid]:
+                return mid
+            
+            elif nums[mid-1] < nums[mid] and nums[mid+1] < nums[mid]:
+                return mid
+
+            if mid > 0 and nums[mid] < nums[mid-1]:
+                high = mid - 1
+            
+            elif mid < len(nums):
+                low = mid + 1
+        
+
diff --git a/leetcode34.py b/leetcode34.py
new file mode 100644
index 00000000..b3f6052e
--- /dev/null
+++ b/leetcode34.py
@@ -0,0 +1,84 @@
+#!/usr/bin/env python3
+# -*- coding: utf-8 -*-
+"""
+Created on Sun Feb  8 15:39:30 2026
+
+@author: rishigoswamy
+    
+    Problem:
+    ----------
+    https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/
+
+    Approach:
+    ----------
+    Since the array is sorted in ascending order, binary search can be used
+    to locate the first and last occurrences of the target efficiently.
+    
+    We perform two separate binary searches:
+    - The first search finds the leftmost (first) position of the target by
+      continuing the search on the left half even after the target is found.
+    - The second search finds the rightmost (last) position of the target by
+      continuing the search on the right half after the target is found.
+    
+    Each search carefully checks boundary conditions to ensure the correct
+    extreme index is identified.
+    
+    Time Complexity:
+    ----------------
+    First binary search  : O(log n)
+    Second binary search : O(log n)
+    Overall              : O(log n)
+    
+    Space Complexity:
+    -----------------
+    O(1), since only constant extra space is used.
+    
+    Did this code successfully run on Leetcode:
+    ------------------------------------------
+    Yes
+    
+    Any problem you faced while coding this:
+    ---------------------------------------
+    Handling edge cases where the target appears at the beginning or end
+    of the array, and ensuring both searches terminate correctly when the
+    target does not exist.
+
+"""
+
+class Solution:
+    def searchRange(self, nums, target):
+        low = 0
+        high = len(nums) - 1
+        mid = (low + high)//2
+        first, second = -1, -1
+
+        while low <= high:
+            mid = (low + high)//2
+            if nums[mid] == target:
+                if mid == 0 or nums[mid-1] != nums[mid]:
+                    first = mid
+                    break
+                else:
+                    high = mid - 1
+            elif nums[mid] > target:
+                high = mid - 1
+            else:
+                low = mid + 1
+        
+
+        low = 0
+        high = len(nums) - 1
+        while low <= high:
+            mid = (low + high)//2
+            if nums[mid] == target:
+                if mid == len(nums)-1 or nums[mid+1] != target:
+                    second = mid
+                    break
+                else:
+                    low = mid + 1
+            elif nums[mid] > target:
+                high = mid - 1
+            else:
+                low = mid + 1
+        
+        return [first, second]
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