#leet_code_Q1 #Array Given a non-empty array of integers nums, every element appears twice except for one. Find that single one.

You must implement a solution with a linear runtime complexity and use only constant extra space.



Example 1:

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

Input: nums = [4,1,2,1,2]
Output: 4
Example 3:

Input: nums = [1]
Output: 1 Each element in the array appears twice except for one element which appears only once.

give your answer in the bot below or in the comment section

no_136 on leetcode.com

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

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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