🇺🇿 Amaliyot.

Hammaga salom! Bugun masala yechamiz. Shu masalaga men yozgan yechimni beraman va keyingi postda izohini berishga harakat qilaman.

🟡 O'rta daraja: 75. Sort Colors
👨‍💻 Mening yechimim:
Java 100% Faster | O(N + N) Solution

O'zingiz yechishga harakat qiling.
Omad!✌️

🇬🇧 Problem 75. Sort Colors Explained

🟡 Medium: 75. Sort Colors
👨‍💻 My solution:
Java 100% Faster | O(N + N) Solution

Hi, guys! It's time to show solution for yesterday's task and explain it. Let's gooo!

👨‍💻 Task.

Basically, you are given array which has only values 0, 1, 2 and not ordered. Your task is to sort the array without using any library sorting algo.
Pay attention to that we don't return any array and we need to provide result inside given array.

👨‍💻 My solution:

class Solution {
    public void sortColors(int[] nums) {
        int n0 = 0, n1 = 0;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] == 0) n0++;
            else if (nums[i] == 1) n1++;
        }
        for (int i = 0; i < nums.length; i++) {
            if (n0-- > 0) nums[i] = 0;
            else if (n1-- > 0) nums[i] = 1;
            else nums[i] = 2;
        }
    }
}

👨‍💻 Explanation.

- I run thru array two times.
- In the first run, I count the number of 0's and 1's and store them in n0 and n1 respectively.
- In the second run, I insert 0's from the beginning until we run out of 0's. Then I insert 1's into array until we run out of 1's. Then I just fill the array with 2's till the end.

That's it.

Hope you enjoyed it! 😁✌️

🇺🇿 75. Sort Colors Masalasi

🟡 O'rta daraja: 75. Sort Colors
👨‍💻 Mening yechimim:
Java 100% Faster | O(N + N) Solution

Hammaga salom! Kechagi masalaning yechimini ko'rsatib, uni tushuntirish vaqti keldi. Boshladik!🚀🚀🚀

👨‍💻 Vazifa.

Oddiy qilib aytganda, bizga 0,1 va 2 sonlari bilan tartibsiz holda to'ldirilgan massiv berilgan. Bizning vazifamiz bu massivni tartibga solish (sort).
Ahamiyat bering: natijani berilgan massivning o'ziga saqlashimiz lozim.

👨‍💻 Mening yechimim:

class Solution {
public void sortColors(int[] nums) {
int n0 = 0, n1 = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] == 0) n0++;
else if (nums[i] == 1) n1++;
}
for (int i = 0; i < nums.length; i++) {
if (n0-- > 0) nums[i] = 0;
else if (n1-- > 0) nums[i] = 1;
else nums[i] = 2;
}
}
}


👨‍💻Izoh
- Massivni ikki marta aylanib chiqaman.
- Birinchi marta aylanganda 0lar va 1lar sonini hisoblab, n0 va n1 o'zgaruvchilari orqali saqlayman.
- Ikkinchi marta aylanganda massivni boshida 0lar bilan to'ldiraman. 0lar tugaganda, 1lar bilan to'ldirib boshlayman. 1lar ham tugaganda - massivning qolgan qismini 2'lar bilan to'ldirib chiqaman.

Bo'ldi. Tugadi.

Sizlarga yoqdi degan umiddaman! 😁✌️

🇬🇧 Practice.

Hi, guys! Today's task is Tribonacci task. It's like Fibonacci, but Tribonacci.😁

🟢 Easy: 1137. N-th Tribonacci Number
👨‍💻 My solution: Java 100% Faster | Array As A Cache

Try to write the fastest solution. And we'll analyze the solution in the next post.
Peace!✌️

🇺🇿 Amaliyot.

Hammaga salom! Bugun oson darajali masala yechamiz - Tribonachi ketma-ketligi. U xuddi Fibonachi kabi, lekin Tribonachidir.😁

🟢 Oson daraja: 1137. N-th Tribonacci Number
👨‍💻 Mening yechimim: Java 100% Faster | Array As A Cache

Eng tez yechim yozishga harakat qiling.
Keyingi postda sizlar bilan yechimni tahlil qilib chiqamiz.
Peace!✌️

🇬🇧 Practice.

Hi, guys! Let me explain my solution for yesterday's problem.

🟢 Easy: 1137. N-th Tribonacci Number
👨‍💻 My solution: Java 100% Faster | Array As A Cache

👨‍💻 Task.
The task is to find the n-th element in tribonacci sequence. It's the same as Fibonacci, but here the element is the sum of previous three element. More on Fibonacci sequence you can read here.

👨‍💻 My solution:
Solution is below in the picture.

👨‍💻 Explanation.
1. My solution is based on recursive approach with memoization. I wrote a post on this here. So, what I do is first cover 3 base cases: for n = 0, n = 1, n = 2.
2. As you can see I used an array of size 38 as a cache. We need a cache to store the result for cases which we have already gone through. This is done for optimization purposes.
Q: Why the size of the cache array is 38?
A: Because in the task description n is between 1 and 37. The size is 38 because I want to store value by its index, not index - 1.
Q: Why you use array instead of Hashtable data structure implementation?
A: Simply saying, I don't need it 😂😂😂. HashMap, Hashtable implementation in Java, needs some more memory and, even if its access speed to element is O(1) [if no collision], has some extra operations for hashing, inserting, and retrieving elements. It takes time. Array requires less resources. Choose Arrays!
Just choose it!😁
3. if (cache[n] > 0) return cache[n]; - let's see what is going on here. Note that the array is an array with primitive types. When such array is created, all values will be equal to the default value of that type. In our case, the primitive type is integer and the default for it is 0. So, this if expression means that if value for index n is added, then return its value (we have already gone through this case).
4. else case is the recursive expression. Mathematically we can write: f(x) = f(x-1) + f(x - 2) + f(x - 3).

This is it. I hope you liked the solution.
Peace!✌️

🇺🇿 Amaliyot.

Hammaga salom! Keling bugun kechagi masalaning yechimini tahlil qilaylik.

🟢 Oson daraja: 1137. N-th Tribonacci Number
👨‍💻 Mening yechimim: Java 100% Faster | Array As A Cache

👨‍💻 Vazifa.
Bizning vazifamiz bu Tribonacci ketma-ketligidagi n-elementini topish. U xuddi Fibonachi sonlar ketma-ketligiga o'xshaydi, lekin farqi bu har bir element undan oldin turgan uchta elementning yig'indisiga tengdir. Fibonachi sonlare haqida bu yerda o'qisangiz bo'ladi.

👨‍💻 Mening yechimim:
Yechimni rasm orqali ko'rsatganman.

👨‍💻Izoh
1. Mening yechimim memoizatsiyani ishlatuvchi rekursiv usulga asoslangan. Bu usul haqida yozganman: havola. Shunday qilib, asosiy holatlardan boshlaymiz. Bular: n = 0, n = 1, n = 2 holatlari.
2. Men hajmi 38 elementlik massivni kesh sifatida ishlatdim. Keshni ishlatishimiz sababi - bu bir marta hisoblangan qiymatlarni saqlab qolib, qayta ularni hisoblashga vaqt ketkazmaslik. Bu orqali optimizatsiyaga erishamiz.
S: Nimaga massivning hajmi 38ga teng?
J: Vazifani o'qisak, berilayotgan n soni 1 va 37 orasida bo'ladi. Hajmini esa 38ga teng qildim, chunki qiymatlarni n ga teng bo'lgan indekslar orqali saqlamoqchiman, indeks - 1 bo'yicha emas.
S: Nimaga massiv ishlatilgan? Hashtable ma'lumotlar tuzilmasini ishlatsa bo'ladiku.
J: Oddiy qilib aytadigan bo'lsam, menga u kerak emas edi ham (ishlatgim ham kelmadi, ochig'i).😂😂😂 Java-da Hashtable-ning implementatsiyalaridan biri HashMap ko'proq xotira va, xeshlash, kiritish va olish operatsiyalariga, vaqt talab qiladi. HashMap-ning elementni olish tezligi O(1) deyilsa ham [kolliziya bo'lmagan holda], bu O(1) konstanta massivning o'xshash operatsiyalaridagi O(1) dan kattaroq.
Massivni tanlang! Massivga ovoz bering!😁
3. if (cache[n] > 0) return cache[n]; - bu yerda nima bo'layotganini ko'raylik. Ahamiyat bering, men ishlatgan massiv - bu primitiv ma'lumot turidagi elementlardan iborat massiv. Qachonki shunday massiv yaratsak, u ishlatilgan primitiv turining default [tarjimasini topsangiz - yozib yuboring] qiymati bilan to'ldiriladi. Bizning holda bu 0. Shunday qilib, if -ning ma'nosi: agar n uchun qiymat topgan bo'lsak, shu qiymatni ishlataman, agar topmagan bo'lsak - qidiramiz.
4. else-da esa rekursiyaviy ifodamiz. Uni matematik holda yozadigan bo'lsak, f(x) = f(x-1) + f(x - 2) + f(x - 3) - ko'rinishda bo'ladi. Bu ko'rinishda tushunish osonroq :).

Sizga yoqdi degan umiddaman.
Omad!✌️

🇬🇧 Practice [makes it better]

Hi, guys! Let's solve one interesting medium problem today. Shortly, you will work with interesting palindromes.

🟡 Medium: 2396. Strictly Palindromic Number
👨‍💻 My solution: Java 0ms 100% faster solution

As usual, solution will be explained in the next post. 😄
Peace! ✌️

🇺🇿 Amaliyot [makes is better]

Hammaga salom! Keling bitta juda ham qiziqarli bo'lgan masalani yechamiz. Qisqa qilib aytganda, sizlar bir sonning har xil palindromlari bilan ishlaysiz.

🟡 O'rta daraja: 2396. Strictly Palindromic Number
👨‍💻 Mening yechimim: Java 0ms 100% faster solution

Har doimgidek yechimini ertaga izohlab berishga harakat qilaman. 😄😁👀
Omad! ✌️

Oldik session ☕️

Sardor aka nega PHP eng zo’r tilligini tushuntirib berdilar.

To'dalashib kelinglar 😁

#oldik

Bugun zo’r dasturchilar bilan tanishdim.

Today I was happy to get to know these great programmers.

Oldik!🥂

🇬🇧 2396. Strictly Palindromic Number

Hi, guys! Today I wanted to explain my solution, but realized that there is 1 line solution for the problem. So, let's analyze it.

🟡 Medium: 2396. Strictly Palindromic Number

👨‍💻 Task:
We need to find all base representations from 2 to n - 2 for given n. And we need to return true if all of these representations are palindromic, else return false.

👨‍💻 Best solution:

public boolean isStrictlyPalindromic(int n) {
return false;
}

👨‍💻 Explanation:

Why does it always return false? Because base(n - 2) representation of any number (given by task) will be "12", which is not palindromic.
For example, let's take n = 21. Then base(n-2) will be base(19). Let's find base(19) representation for 21.
1. 21 / 19 is 1 with remainder 2. We write "2" and then "1". We get "21" and then revert it according to rules. Finally, we get "12".

I hope this made it clear!
Peace!✌️

🇺🇿 2396. Strictly Palindromic Number

Hammaga salom! Men o'zimning yechimimning izohini bermoqchi edim, ammo menikidan ancha yaxshiroq yechim borligini tushundim. Shu yaxshiroq yechimning izohini beraman.

🟡 Medium: 2396. Strictly Palindromic Number

👨‍💻 Vazifa
Biz berilgan n sonining har bir base(2) dan base(n-2) gacha bo'lgan bo'lgan ko'rinishi palindrom bo'lishligini aniqlashimiz kerak.
Bu degani agar base(10) dagi 4 sonini olsak uning base(2) ko'rinishi "100" bo'ladi va, ko'rib turganingizdek, bu palindrom emas.

👨‍💻 Eng zo'r yechim:

public boolean isStrictlyPalindromic(int n) {
return false;
}

👨‍💻 Izoh.
Nimaga har doim false qaytaradi? Chunki berilayotgan har qanday sonning base(n-2) ko'rinishi "12" teng bo'lib qoladi va bu palindrom emas.
Keling base(10) dagi bo'lgan 21 sonining base(n-2) yoki base(19) ko'rinishini topamiz.
Boshida 21ni 19ga bo'lamiz: 21 / 19 = 1 va qoldiqda 2. Biz boshida "2" ni yozamiz, keyin esa "1" ni. Shunda "21" bo'ladi. Endi bu ko'rinishni orqadan oldinga yozamiz: "12". Ko'rganingizdek, biz 21 sonining base(19) da ko'rinishi "12" bo'lishini topdik. Bitta base-dan ikkinchi base-ga aylantirish qoidalari bu yerda: havola.

Base o'zbekchada nima bo'lishini yozvoringizlar🙌
Yaxshi tushuntira oldim degan umiddaman!
Omad!✌️

🇬🇧 706. Design HashMap

Hi, guys! I wrote several posts where I explained the way Hashtables work on a basic level. You can check those posts here: Hashtable.
Let's check how we mastered the basics of Hashtables.

🟢 Easy: 706. Design HashMap
👨‍💻 My solution: Java | Built HashMap Based On Array

Please, try to build a HashMap on your own before checking solutions of other people.
As usual, explanation to be provided 😀.
Good luck!🤓

🇺🇿 706. Design HashMap

Hammaga salom! Men Hashtable ma'lumotlar tuzilmasini qanday ishlashi haqida postlar seriyasini yozgan edim. Postlarni bu yerda qarab chiqishingiz mumkin: Hashtable.
Keling Hashtable mavzusini qanchalik yaxshi o'zlashtirganligimizni tekshirib ko'raylik.

🟢 Oson daraja: 706. Design HashMap
👨‍💻 Mening yechimim: Java | Built HashMap Based On Array

Iltimos, tayyor yechimga qarashdan avval, o'zingiz yaxshilab urinib ko'ring.
Har doimgidek, izohni keyingi postlarda kutamiz😀.
Omad! 🤓

🇬🇧 500 solved problems

Guys, I just wanted to share that I was able to achieve a new milestone today - 500 solved problem on LeetCode.
Yaaaay!🤓

🇺🇿 500 yechilgan masala

Sizlar bilan yana bitta muhim bosqichga erishganim bilan bo'lishmoqchiman - bu LeetCode-da yechilgan masalalarning soni 500ga yetdi!
Oldik!🥂