This one took me much longer time not gonna lie.😮‍💨

✅ Solution : this is as efficient as it can get

python 
def countSpaceialSubString(s):
    seen = set()
    l = 0
    count = 0
    
    for r in range(len(s)):
        if s[r] in seen:
            while s[l] != s[r]:
                seen.remove(s[l])
                l += 1
            l += 1
        
        seen.add(s[r])
        count += r - l + 1
    
    return count 
        
        
print(countSpaceialSubString("abcd"))
print(countSpaceialSubString("ooo"))
print(countSpaceialSubString("abab"))
print(countSpaceialSubString("abcabc"))

Question: Can You Make This String a Palindrome?

A palindrome is a string that reads the same forwards and backwards. Given a string, determine if it's possible to rearrange the characters to form a palindrome.

Examples:

1. Input: "civic"

   • Output: True

   • Explanation: The string is already a palindrome.

2. Input: "ivicc"

   • Output: True

   • Explanation: Rearranging the characters can form the palindrome "civic".

3. Input: "hello"

   • Output: False

   • Explanation: No rearrangement can form a palindrome.

4. Input: "aabbcc"

   • Output: True

   • Explanation: Rearranging the characters can form the palindrome "abcba".

5. Input: "racecar"

   • Output: True

   • Explanation: The string is already a palindrome.

Challenge:
Write a function that takes a string as input and returns True if the string can be rearranged to form a palindrome, and False otherwise.

Question Description
Given a string s and an integer k, return the number of substrings in s of length k with no repeated characters.
Example 1:
Input: s = "unonleetcode", k = 5
Output: 2
Explanation: There are 6 substrings they are: 'havef','avefu','vefun','efuno','etcod','tcode'.

Example 2:
Input: s = "home", k = 5
Output: 0
Explanation: Notice k can be larger than the length of s. In this case, it is not possible to find any substring.

Example 3:
Input: s = "havefunonleetcode", k = 5
Output: 6
Explanation: There are 6 substrings they are: 'havef','avefu','vefun','efuno','etcod','tcode'.

✅ Solution: as I have told you this pattern repeats a lot so you got this

def subStringK(s, k):
    seen = set()
    count = 0
    l = 0
    
    for r in range(len(s)):
        if s[r] in seen:
            while s[l] != s[r]:
                seen.remove(s[l])
                l += 1
            l += 1
        
        if r - l + 1 == k:
            count += 1
            seen.remove(s[l])
            l += 1
        
        seen.add(s[r])
    return count


print(subStringK("unonleetcode", 5))
print(subStringK("havefunonleetcode", 5))
print(subStringK("aaabbaaa", 2))
print(subStringK("aaabbaaa", 100))

✅ Hello everyone, today was my interview date, and I was asked the following question: At first, I thought I could use the two pointers technique to solve it, but then I realized that that would make the algorithm inefficient. Then I noticed that the number of 1s will be the length of the subarray with grouped 1s. This changed my approach to a fixed sliding window, and then the rest was easy. My interviewer was very nice and guided me the whole way.

'''
Given a binary array data, return the minimum number of swaps required to group all 1’s
present in the array together in any place in the array.


Example 1:

Input: data = [1,0,1,0,1]
Output: 1
Explanation: There are 3 ways to group all 1's together:
[1,1,1,0,0] using 1 swap.
[0,1,1,1,0] using 2 swaps.
[0,0,1,1,1] using 1 swap.
The minimum is 1.

Example 2:
Input: data = [0,0,0,1,0]
Output: 0
Explanation: Since there is only one 1 in the array, no swaps are needed.

Example 3:
Input: data = [1,0,1,0,1,0,0,1,1,0,1] count_ones = 6 count_zeros = 3 curr_zeros = 3 min of count_zeros and curr_zeros
l
r
time comp = O(n)
space comp = O(1)

Output: 3
Explanation: One possible solution that uses 3 swaps is [0,0,0,0,0,1,1,1,1,1,1].


Constraints:

1 <= data.length <= 10**5
data[i] is either 0 or 1.
'''

"""
1. count 1's store one count_ones
2. assign count_zeros = inf curr_zeros = 0
3. l, r = 0
4. check for a valid window
5. update curr_zeros
6. take the min of the count_zeros and curr_zeros
7. check if the values at the indexes are zeros if so decrement curr_zeros
8. update pointers
9. return count_zeros
"""
# my code

def minNumberOfSwaps(arr):
    count_ones = arr.count(1)
    count_zeros, curr_zeros = float('inf'), 0
    l = 0
    
    for r in range(len(arr)):
        if arr[r] == 0:
            curr_zeros += 1
        # check for a valid window
        if r - l + 1 == count_ones:
            count_zeros = min(count_zeros, curr_zeros)
            if arr[l] == 0:
                curr_zeros -= 1
            l += 1

    return count_zeros if count_zeros != float('inf') else 0
       

"""
1= 6
curr_zeros = 3
count_zeros = 3
1,0,1,0,1,0,0,1,1,0,1
l
r
"""


#A2SV #a2sv #a2sv2024
A2SV a2sv 2024 In person

🚀 @AceCoding Presents! 🚀

"""
You are given a string s consisting only of the letters 'a' and 'b', and an integer k.
What is the minimum number of characters you need to change to obtain a substring of length ≥ k where all characters are the same?

Example 1:
s = “aabaabaa”, k = 3
Output: 1
Explanation: s can be transformed to “aaaaabaa”

Example 2:
s = “bbabbaba”, k = 8
Output: 3
Explanation: s can be transformed to “bbbbbbbb”

Constraints:
1 <= s.length <= 10^5
1 <= k <= s.length

aaab, k = 4
aaaa
abs(3 - 1) = 2
"""

Given an array of integer arrays arrays where each arrays[i] is sorted in strictly increasing order,
return an integer array representing the longest common subsequence among all the arrays.
A subsequence is a sequence that can be derived from another sequence by deleting some elements (possibly none)
without changing the order of the remaining elements.

Example 1:
Input: arrays = [[1,3,4],
                 [1,4,7,9]]
Output: [1,4]
Explanation: The longest common subsequence in the two arrays is [1,4].
Example 2:
Input: arrays = [[2,3,6,8],
                 [1,2,3,5,6,7,10],
                 [2,3,4,6,9]]
Output: [2,3,6]
Explanation: The longest common subsequence in all three arrays is [2,3,6].

Example 3:
Input: arrays = [[1,2,3,4,5],
                 [6,7,8]]
Output: []
Explanation: There is no common subsequence between the two arrays.


Constraints:
2 <= arrays.length <= 100
1 <= arrays[i].length <= 100
1 <= arrays[i][j] <= 100
arrays[i] is sorted in strictly increasing order.

'''

#Registration


Register Here


Requirement:

A student of AASTU
(Any department)

Interest for puzzles, games, problem solving


Join us @aastucpc

ይህ በምስሉ የምታዩት ልጅ: ያቤፅ ሰለሞን ይባላል። የ8ኛ ክፍል ተማሪ ነው። የሚኖረው የካ አባዶ ሲሆን ካጋጠመው የኩላሊት በሽታ የመጨረሻ ደረጃ ላይ ስለደረሰ ወደ ውጪ ሀገር ሄዶ transplant ማድረግ እንዳለበት ለቤተሰቦቹ ተነግሯል። ለህክምናውም ወደ 5 ሚሊየን ብር ያስፈልገዋል። አባቱ የአባዶ መሰረተ ክርስቶስ ቤ/ክ አገልጋይ ሲሆን ሌላ የገቢ ምንጭ ስለሌለው ልጁ እንዲታከም እርዳታ ጠይቋል።
ስለዚህ የዚህ ብላቴና ህይወት ባለን አቅም በመደገፍ እንታደግ፤ እንፀልይለት!

ለመደገፍ የምትፈልጉ: Account holder: Solomon Alemayehu (Yabets’ Father)
CBE: 1000057164143
Awash: 01320037587500
Abyssinia: 102845781
Cooperative : 1000072492677
Go fund me: https://gofund.me/698c8f50
Phone: 0911670476/ 0962157832

🎉 G6 A2SV In-Person Education Program results 🚀🎓

The G6 A2SV In-Person Education Program results will be revealed by the end of next week! 🗓✨

Source: A2SV - Weekly Wins and Demos - December 27
👉 See the updates on the Remote Education too.

Stay tuned, and get ready to celebrate your hard work and achievements! 🧑‍💻🌟

#A2SVResults 🎊

@AceCoding presents

✅ Mastering Browsers Dev tools with a single video:

📗Recommendation source 👉 Evangadi Fullstack web dev course.

https://youtu.be/x4q86IjJFag?si=UIQxXYxzXHYcr96Q

💻 Call for Registration for Winter Bootcamp

[UNESCO UNITWIN] 2025 Winter Digital Innovation BootCamp
Theme: Artificial Intelligence and K-eGov Open Source Framework in Ethiopia

The Bootcamp registration is now open!

Due Date: January 25 (Saturday)

👉🏾 Registration Link

You can also access the registration link via the QR Code provided on the poster above.

For inquiries, feel free to contact us via email at kimyena@handong.edu

ጥያቄ ካላችሁ እዚህ ላይ ማቅረብ ትችላላችሁ  🏃‍♂️👇👇👇👇👇👇👇👇👇👇👇👇👇

⚗️⚗️@Aastu_poll_and_quiz_bot⚗️⚗️

☝️☝️☝️☝️☝️☝️☝️☝️☝️☝️☝️☝️

   📤📤📕📕📕📕📕📕📕📤📤
   📊📊📚  @Aastu_poll   📊📊
   📊📊📚  @Aastu_poll   📊📊
   📊📊📚  @Aastu_poll  📊📊
   📊📊 📒📒📒📒📒📒📊📊

Calling All AASTU Students: Join UniHack 2025!

Are you ready to innovate, solve challenges, and showcase your skills? UniHack 2025 is exclusively for AASTU students, offering you the platform to turn your ideas into impactful projects.

🗓 Event Date: February 19, 2025
📍 Venue: Old graduation hall(AASTU)
🌐 Apply Now: LINK

Who Can Apply?

This event is exclusively for AASTU students from any department. Whether you’re a coder, designer, or idea generator, there’s a place for you at UniHack!

Don’t miss this incredible opportunity to represent AASTU’s innovation and talent.

📌 Visit our Website LINK for more details and to apply.

Spaces are limited, so apply today and get ready to innovate!

Check your emails🚨

A2SV is sending acceptance emails for G6 In person Education🎉🎉

#A2SVInPersonEducation #G6
@AceCoding

How to deploy static websites using GitHub : 👉 Watch This

For Dynamic websites, use other deploying platforms like Netlify, Vercel, or Hostinger alongside bundling tools like vite, parcel, or Webpack. You can use this deploying sites or platforms for statics websites too without bundling tools.

@AceCoding