Leetcode 226. Invert Binary Tree. Python (Very Easy)

226. Invert Binary Tree

class Solution:

def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:

if not root:

return None

temp = root.left

root.left = root.right

root.right = temp

self.invertTree(root.left)

self.invertTree(root.right)

return root

Explaination :

Comments

Leetcode 371. Sum of Two Integers. C++ / Java

371 . Sum of Two Integers Given two integers a and b , return the sum of the two integers without using the operators + and - . Example 1: Input: a = 1, b = 2 Output: 3 Example 2: Input: a = 2, b = 3 Output: 5 Constraints: -1000 <= a, b <= 1000 Solution : C++ : class Solution { public: int getSum(int a, int b) { if (b==0) return a; int sum = a ^ b; int cr = (unsigned int) (a & b) << 1; return getSum(sum, cr); } }; Java : class Solution { public int getSum(int a, int b) { while(b != 0){ int tmp = (a & b) << 1; a = a ^ b; b = tmp; } return a; } } Explaination :

Leetcode 347. Top K Frequent Elements. Python (Bubble Sort)

Top K Frequent Elements Given an integer array nums and an integer k , return the k most frequent elements . You may return the answer in any order . Example 1: Input: nums = [1,1,1,2,2,3], k = 2 Output: [1,2] Example 2: Input: nums = [1], k = 1 Output: [1] Constraints: 1 <= nums.length <= 10 5 -10 4 <= nums[i] <= 10 4 k is in the range [1, the number of unique elements in the array] . It is guaranteed that the answer is unique . Follow up: Your algorithm's time complexity must be better than O(n log n) , where n is the array's size. Solution : class Solution: def topKFrequent(self, n: List[int], k: int) -> List[int]: # [1,1,1,2,2,3] & k = 2 f = [[] for i in range(len(n) + 1)] # f = [...

Leetcode 338. Counting Bits. Python (Bubble Sort)

Given an integer n , return an array ans of length n + 1 such that for each i ( 0 <= i <= n ) , ans[i] is the number of 1 's in the binary representation of i . Example 1: Input: n = 2 Output: [0,1,1] Explanation: 0 --> 0 1 --> 1 2 --> 10 Example 2: Input: n = 5 Output: [0,1,1,2,1,2] Explanation: 0 --> 0 1 --> 1 2 --> 10 3 --> 11 4 --> 100 5 --> 101 Constraints: 0 <= n <= 10 5 Follow up: It is very easy to come up with a solution with a runtime of O(n log n) . Can you do it in linear time O(n) and possibly in a single pass? Can you do it without using any built-in function (i.e., like __builtin_popcount in C++)? Solution : class Solution: def countBits(self, n: int) -> List[int]: dp = [0] * (n+1) c = 1 ...

LEETCODE SOLUTIONS

Search This Blog