Try this question. i am not getting enough answers from you

Hint: try to do it with one of the bitwise operationd

Hint2: try Xor operation

lets submit our code

it's good. we did better from 78 % of submitted code in python by running time

Level Up Problem-Solving: LeetCode trains your brain to break down problems, a skill used by 80% of software engineers according to a [Payscale survey]( Indeed salary survey for Software Engineer).

lets do another Question on Leet code

#leet_code_Q2 #Array #leet_code_169 #Majority_Element Given an array nums of size n, return the majority element.

The majority element is the element that appears more than (n / 2)times. You may assume that the majority element always exists in the array.

Example 1:

Input: nums = [3,2,3]
Output: 3
Example 2:

Input: nums = [2,2,1,1,1,2,2]
Output: 2

Let's tackle the question of finding the number that appears only once in a list.

if you want to figure it out your self, dont see the answer!!!!!!!!!!!!!!!!!!

We could iterate through the list, comparing each element to all others to count occurrences. This approach would work, but it has a time complexity of n^2, exceeding the given constraint of linear time (O(n)).



Since the question guarantees that all other elements appear twice, we can exploit this property. Here's how:



Convert the list to a set. Sets eliminate duplicates, leaving only the unique element.

Calculate the sum of the elements in the original list and the set.

Unique number = Sum of list elements - (2 * Sum of set elements)

This method efficiently finds the unique element in linear time.



There's an even better approach called the XOR method. If you'd like an explanation of how it works, I can provide the code and break it down for you class Solution:

def singleNumber(self, nums: List[int]) -> int:

result = 0

for i in nums:

result^=i

return result

it's good. we did better from 78 % of submitted code in python by running time

give me your ideas about this question. how can we solve it???

so let's tackle this question. so how can we know the element is maority element ,if it is the most frequent element in the array yea . so we can calculate each element freaquency by creating a dictionary like this my_dicti = {} for i in nums: if nums[i] in my_dict: my_dict[i] +=1 else: my_dict[i] = 1 and we can print the find the maximum frequent number in the dictionary and return it but there is also another efficient method to do it by declearing two variabes one to contain the number and the second to cotain it's occurence like this

nums = [1,2,3,4,5,4,4,3]
contain = nums[0]
occurence = 0
for i in nums:
    if i == contain:
        occurence += 1
    else:
        if occurence == 0:
            contain = i
            occurence =1
        else:
            occurence -= 1

Try this question on leet code Majority Element

557. Reverse Words in a String III Given a string s, reverse the order of characters in each word within a sentence while still preserving whitespace and initial word order.
Example 1:

Input: s = "Let's take LeetCode contest"
Output: "s'teL ekat edoCteeL tsetnoc"
Example 2:

Input: s = "Mr Ding"
Output: "rM gniD" #leet_code_Q3 #String

Answer:

 words = s.split()
  reversed_words = [word[::-1] for word in words] 
  return ' '.join(reversed_words)

9. Palindrom number #leet_code_Q4 #math #my_twopointer Given an integer x, return true if x is a
palindrome, and false otherwise.
Example 1:
Input: x = 121
Output: true
Explanation: 121 reads as 121 from left to right and from right to left.
Example 2:
Input: x = -121
Output: false
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.
Example 3:
Input: x = 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.