feat(work): add new LeetCode solutions for median, largest rectangle, and simplify path

This commit is contained in:
Prad Nukala committed 2026-10-05 13:51:47 -04:00
1 parent f14dcf2723
commit d388b3000e
3 files changed
+203

No files matched your search

@@ -0,0 +1,46 @@
"""
4. Median of Two Sorted Arrays
Difficulty: Hard
https://leetcode.com/problems/median-of-two-sorted-arrays/
──────────────────────────────────────────────────
Given two sorted arrays nums1 and nums2 of size m and n respectively,
return the median of the two sorted arrays.
The overall run time complexity should be O(log (m+n)).
Example 1:
Input: nums1 = [1,3], nums2 = [2]
Output: 2.00000
Explanation: merged array = [1,2,3] and median is 2.
Example 2:
Input: nums1 = [1,2], nums2 = [3,4]
Output: 2.50000
Explanation: merged array = [1,2,3,4] and median is (2 + 3) / 2 = 2.5.
Constraints:
• nums1.length == m
• nums2.length == n
• 0 <= m <= 1000
• 0 <= n <= 1000
• 1 <= m + n <= 2000
• -10^6 <= nums1[i], nums2[i] <= 10^6
"""
class Solution:
def findMedianSortedArrays(self, nums1: list[int], nums2: list[int]) -> float:
@@ -0,0 +1,38 @@
"""
84. Largest Rectangle in Histogram
Difficulty: Hard
https://leetcode.com/problems/largest-rectangle-in-histogram/
──────────────────────────────────────────────────
Given an array of integers heights representing the histogram's bar
height where the width of each bar is 1, return the area of the
largest rectangle in the histogram.
Example 1:
Input: heights = [2,1,5,6,2,3]
Output: 10
Explanation: The above is a histogram where width of each bar is 1.
The largest rectangle is shown in the red area, which has an area =
10 units.
Example 2:
Input: heights = [2,4]
Output: 4
Constraints:
• 1 <= heights.length <= 10^5
• 0 <= heights[i] <= 10^4
"""
class Solution:
def largestRectangleArea(self, heights: list[int]) -> int:
+119
View File
@@ -0,0 +1,119 @@
"""
71. Simplify Path
Difficulty: Medium
https://leetcode.com/problems/simplify-path/
──────────────────────────────────────────────────
You are given an absolute path for a Unix-style file system, which
always begins with a slash '/'. Your task is to transform this
absolute path into its simplified canonical path.
The rules of a Unix-style file system are as follows:
• A single period '.' represents the current directory.
• A double period '..' represents the previous/parent directory.
• Multiple consecutive slashes such as '//' and '///' are treated as
a single slash '/'.
• Any sequence of periods that does not match the rules above should
be treated as a valid directory or file name. For example, '...' and
'....' are valid directory or file names.
The simplified canonical path should follow these rules:
• The path must start with a single slash '/'.
• Directories within the path must be separated by exactly one slash
'/'.
• The path must not end with a slash '/', unless it is the root
directory.
• The path must not have any single or double periods ('.' and '..')
used to denote current or parent directories.
Return the simplified canonical path.
Example 1:
Input: path = "/home/"
Output: "/home"
Explanation:
The trailing slash should be removed.
Example 2:
Input: path = "/home//foo/"
Output: "/home/foo"
Explanation:
Multiple consecutive slashes are replaced by a single one.
Example 3:
Input: path = "/home/user/Documents/../Pictures"
Output: "/home/user/Pictures"
Explanation:
A double period ".." refers to the directory up a level (the parent
directory).
Example 4:
Input: path = "/../"
Output: "/"
Explanation:
Going one level up from the root directory is not possible.
Example 5:
Input: path = "/.../a/../b/c/../d/./"
Output: "/.../b/d"
Explanation:
"..." is a valid name for a directory in this problem.
Constraints:
• 1 <= path.length <= 3000
• path consists of English letters, digits, period '.', slash '/' or
'_'.
• path is a valid absolute Unix path.
"""
class Solution:
def simplifyPath(self, path: str) -> str:
stack = []
segments = [item for item in path.split("/") if item]
for p in segments:
if stack and p == "..":
stack.pop()
elif not stack and p == ".." or p == ".":
continue
else:
stack.append(p)
return "/" + "/".join(stack)