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javadev merged 3 commits intoLeetCode-in-Python:mainfromThanhNIT:tasks-199-209
Sep 15, 2025
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30 changes: 30 additions & 0 deletionsREADME.md
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# #Medium #Top_100_Liked_Questions #Depth_First_Search #Breadth_First_Search #Tree #Binary_Tree
# #LeetCode_75_Binary_Tree/BFS #Data_Structure_II_Day_16_Tree #Level_2_Day_15_Tree
# #Top_Interview_150_Binary_Tree_BFS #2025_09_14_Time_0_ms_(100.00%)_Space_17.76_MB_(60.72%)

from typing import List, Optional

class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right


class Solution:
def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
result = []
self._recurse(root, 0, result)
return result

def _recurse(self, node: Optional[TreeNode], level: int, result: List[int]):
if node is not None:
if len(result) < level + 1:
result.append(node.val)
self._recurse(node.right, level + 1, result)
self._recurse(node.left, level + 1, result)

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import unittest
from Solution0199 import Solution, TreeNode

class SolutionTest(unittest.TestCase):
def test_rightSideView(self):
left = TreeNode(2)
left.right = TreeNode(5)
right = TreeNode(3)
right.right = TreeNode(4)
root = TreeNode(1, left, right)
self.assertEqual(Solution().rightSideView(root), [1, 3, 4])

def test_rightSideView2(self):
root = TreeNode(1)
root.right = TreeNode(3)
self.assertEqual(Solution().rightSideView(root), [1, 3])

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199\. Binary Tree Right Side View

Medium

Given the `root` of a binary tree, imagine yourself standing on the **right side** of it, return _the values of the nodes you can see ordered from top to bottom_.

**Example 1:**

![](https://assets.leetcode.com/uploads/2021/02/14/tree.jpg)

**Input:** root = [1,2,3,null,5,null,4]

**Output:** [1,3,4]

**Example 2:**

**Input:** root = [1,null,3]

**Output:** [1,3]

**Example 3:**

**Input:** root = []

**Output:** []

**Constraints:**

* The number of nodes in the tree is in the range `[0, 100]`.
* `-100 <= Node.val <= 100`
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# #Medium #Bit_Manipulation #Algorithm_II_Day_19_Bit_Manipulation
# #Top_Interview_150_Bit_Manipulation #2025_09_14_Time_1_ms_(82.83%)_Space_18.01_MB_(19.55%)

class Solution:
def __init__(self):
# Precomputed masks for different bit positions
self.masks = [
0,
0x80000000,
0xC0000000,
0xE0000000,
0xF0000000,
0xF8000000,
0xFC000000,
0xFE000000,
0xFF000000,
0xFF800000,
0xFFC00000,
0xFFE00000,
0xFFF00000,
0xFFF80000,
0xFFFC0000,
0xFFFE0000,
0xFFFF0000,
0xFFFF8000,
0xFFFFC000,
0xFFFFE000,
0xFFFFF000,
0xFFFFF800,
0xFFFFFC00,
0xFFFFFE00,
0xFFFFFF00,
0xFFFFFF80,
0xFFFFFFC0,
0xFFFFFFE0,
0xFFFFFFF0,
0xFFFFFFF8,
0xFFFFFFFC,
0xFFFFFFFE
]

def rangeBitwiseAnd(self, left: int, right: int) -> int:
if left == right:
return left
return right & self.masks[self._numberOfLeadingZeros(left ^ right)]

def _numberOfLeadingZeros(self, n: int) -> int:
"""Count leading zeros in 32-bit integer"""
if n == 0:
return 32
count = 0
if n < 0:
return 0
while n > 0:
n = n >> 1
count += 1
return 32 - count

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import unittest
from Solution0201 import Solution

class SolutionTest(unittest.TestCase):
def test_rangeBitwiseAnd(self):
self.assertEqual(Solution().rangeBitwiseAnd(5, 7), 4)

def test_rangeBitwiseAnd2(self):
self.assertEqual(Solution().rangeBitwiseAnd(0, 0), 0)

def test_rangeBitwiseAnd3(self):
self.assertEqual(Solution().rangeBitwiseAnd(1, 2147483647), 0)

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201\. Bitwise AND of Numbers Range

Medium

Given two integers `left` and `right` that represent the range `[left, right]`, return _the bitwise AND of all numbers in this range, inclusive_.

**Example 1:**

**Input:** left = 5, right = 7

**Output:** 4

**Example 2:**

**Input:** left = 0, right = 0

**Output:** 0

**Example 3:**

**Input:** left = 1, right = 2147483647

**Output:** 0

**Constraints:**

* <code>0 <= left <= right <= 2<sup>31</sup> - 1</code>
22 changes: 22 additions & 0 deletionssrc/main/python/g0201_0300/s0202_happy_number/Solution0202.py
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# #Easy #Top_Interview_Questions #Hash_Table #Math #Two_Pointers #Algorithm_II_Day_21_Others
# #Programming_Skills_I_Day_4_Loop #Level_2_Day_1_Implementation/Simulation
# #Top_Interview_150_Hashmap #2025_09_15_Time_0_ms_(100.00%)_Space_17.57_MB_(97.85%)

class Solution:
def isHappy(self, n: int) -> bool:
a = n
rem = 0
sum_val = 0
if a == 1 or a == 7:
return True
elif 1 < a < 10:
return False
else:
while a != 0:
rem = a % 10
sum_val += rem * rem
a //= 10
if sum_val != 1:
return self.isHappy(sum_val)
else:
return True
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import unittest
from Solution0202 import Solution

class SolutionTest(unittest.TestCase):
def test_isHappy(self):
self.assertTrue(Solution().isHappy(19))

def test_isHappy2(self):
self.assertFalse(Solution().isHappy(2))

39 changes: 39 additions & 0 deletionssrc/main/python/g0201_0300/s0202_happy_number/readme.md
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202\. Happy Number

Easy

Write an algorithm to determine if a number `n` is happy.

A **happy number** is a number defined by the following process:

* Starting with any positive integer, replace the number by the sum of the squares of its digits.
* Repeat the process until the number equals 1 (where it will stay), or it **loops endlessly in a cycle** which does not include 1.
* Those numbers for which this process **ends in 1** are happy.

Return `true` _if_ `n` _is a happy number, and_ `false` _if not_.

**Example 1:**

**Input:** n = 19

**Output:** true

**Explanation:**

1<sup>2</sup> + 9<sup>2</sup> = 82

8<sup>2</sup> + 2<sup>2</sup> = 68

6<sup>2</sup> + 8<sup>2</sup> = 100

1<sup>2</sup> + 0<sup>2</sup> + 0<sup>2</sup> = 1

**Example 2:**

**Input:** n = 2

**Output:** false

**Constraints:**

* <code>1 <= n <= 2<sup>31</sup> - 1</code>
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# #Easy #String #Hash_Table #Level_1_Day_2_String #Top_Interview_150_Hashmap
# #2025_09_14_Time_3_ms_(95.32%)_Space_18.38_MB_(7.18%)

from typing import List

class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
map_array = [0] * 128
str_chars = list(s)
tar_chars = list(t)
n = len(str_chars)
for i in range(n):
if map_array[ord(tar_chars[i])] == 0:
if self._search(map_array, ord(str_chars[i]), ord(tar_chars[i])) != -1:
return False
map_array[ord(tar_chars[i])] = ord(str_chars[i])
else:
if map_array[ord(tar_chars[i])] != ord(str_chars[i]):
return False
return True

def _search(self, map_array: List[int], target: int, skip: int) -> int:
for i in range(128):
if i == skip:
continue
if map_array[i] != 0 and map_array[i] == target:
return i
return -1
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import unittest
from Solution0205 import Solution

class SolutionTest(unittest.TestCase):
def test_isIsomorphic(self):
self.assertTrue(Solution().isIsomorphic("egg", "add"))

def test_isIsomorphic2(self):
self.assertFalse(Solution().isIsomorphic("foo", "bar"))

def test_isIsomorphic3(self):
self.assertTrue(Solution().isIsomorphic("paper", "title"))

33 changes: 33 additions & 0 deletionssrc/main/python/g0201_0300/s0205_isomorphic_strings/readme.md
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205\. Isomorphic Strings

Easy

Given two strings `s` and `t`, _determine if they are isomorphic_.

Two strings `s` and `t` are isomorphic if the characters in `s` can be replaced to get `t`.

All occurrences of a character must be replaced with another character while preserving the order of characters. No two characters may map to the same character, but a character may map to itself.

**Example 1:**

**Input:** s = "egg", t = "add"

**Output:** true

**Example 2:**

**Input:** s = "foo", t = "bar"

**Output:** false

**Example 3:**

**Input:** s = "paper", t = "title"

**Output:** true

**Constraints:**

* <code>1 <= s.length <= 5 * 10<sup>4</sup></code>
* `t.length == s.length`
* `s` and `t` consist of any valid ascii character.
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# #Medium #Array #Binary_Search #Prefix_Sum #Sliding_Window #Algorithm_II_Day_5_Sliding_Window
# #Binary_Search_II_Day_1 #Top_Interview_150_Sliding_Window
# #2025_09_14_Time_7_ms_(91.69%)_Space_28.40_MB_(10.62%)

from typing import List

class Solution:
def minSubArrayLen(self, target: int, nums: List[int]) -> int:
min_res=float('inf')
n=len(nums)
l=0
r=0
curr=0
while r<n:

curr+=nums[r]
while curr>=target:
curr-=nums[l]
min_res=min(min_res,r-l+1)
l+=1
r+=1


return min_res if min_res!=float('inf') else 0
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import unittest
from Solution0209 import Solution

class SolutionTest(unittest.TestCase):
def test_minSubArrayLen(self):
self.assertEqual(Solution().minSubArrayLen(7, [2, 3, 1, 2, 4, 3]), 2)

def test_minSubArrayLen2(self):
self.assertEqual(Solution().minSubArrayLen(4, [1, 4, 4]), 1)

def test_minSubArrayLen3(self):
self.assertEqual(Solution().minSubArrayLen(11, [1, 1, 1, 1, 1, 1, 1, 1]), 0)

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