14-kun
Masala: 235. Lowest Common Ancestor of a Binary Search Tree
Qiyinlik darajasi: O'rta
Bizni kuzatishda davom eting
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Masala: 235. Lowest Common Ancestor of a Binary Search Tree
Qiyinlik darajasi: O'rta
Bizni kuzatishda davom eting
Telegram | Youtube
Leetcode
Lowest Common Ancestor of a Binary Search Tree - LeetCode
Level up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview.
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Kun masalasi: 150. Evaluate Reverse Polish Notation
Qiyinlik darajasi: O'rta
Bizni kuzatishda davom eting
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LeetCode
Evaluate Reverse Polish Notation - LeetCode
Can you solve this real interview question? Evaluate Reverse Polish Notation - You are given an array of strings tokens that represents an arithmetic expression in a Reverse Polish Notation [http://en.wikipedia.org/wiki/Reverse_Polish_notation].
Evaluate…
Evaluate…
JavaHere
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150. Evaluate Reverse Polish Notation | Leetcode
#leetcode #algoritm_uz #dasturlash #javokhir_akramjonov
Shiorimiz: Ertaga, Dushanbadan, Yanagi yildan, 18 kirsangiz emas, Bugundan boshlang.
O'zbekcha kontentlar ko'payishiga va tarqalishiga hissa qo'shing.
Obuna bo'lishni unutmang.
Telegram: https://…
Shiorimiz: Ertaga, Dushanbadan, Yanagi yildan, 18 kirsangiz emas, Bugundan boshlang.
O'zbekcha kontentlar ko'payishiga va tarqalishiga hissa qo'shing.
Obuna bo'lishni unutmang.
Telegram: https://…
Forwarded from Sardor Salimov | IT teacher
Dasturchilar Klubi asoschisi Sardor Salimov va uning jamoasidan "Online C++" loyihasi.
Endi siz C++ ni 6 oyda emas, balki 2 oyda to'liq o'rganishingiz mumkin.
Buning uchun bizning yopiq kurslarimizga a'zo bo'ling.
JOYLAR CHEGARALANGAN
🌐 Kurs formati : Online (har kuni)
📅 Davomiyligi: 2 oy (45 ta dars)
💲 Kurs narxi: 499 000 so'm (plastik karta orqali)
👨💼 O'qituvchi: Sardor Salimov
🧑💻 7 yillik C++ tili tajribasi va 5 yillik pedagogik tajriba
🎓 500 dan ortiq o'quvchi
💻 ICPC va IOI bo'yicha murabbiy
Tezda ro'yhatdan o'ting:
👉Murojaat uchun: @dsforrest, @dasturchilarklubi_HR
📞Telefon: +998933514665, +998934940765
Endi siz C++ ni 6 oyda emas, balki 2 oyda to'liq o'rganishingiz mumkin.
Buning uchun bizning yopiq kurslarimizga a'zo bo'ling.
JOYLAR CHEGARALANGAN
🌐 Kurs formati : Online (har kuni)
📅 Davomiyligi: 2 oy (45 ta dars)
💲 Kurs narxi: 499 000 so'm (plastik karta orqali)
👨💼 O'qituvchi: Sardor Salimov
🧑💻 7 yillik C++ tili tajribasi va 5 yillik pedagogik tajriba
🎓 500 dan ortiq o'quvchi
💻 ICPC va IOI bo'yicha murabbiy
Tezda ro'yhatdan o'ting:
👉Murojaat uchun: @dsforrest, @dasturchilarklubi_HR
📞Telefon: +998933514665, +998934940765
👍2
Kun masalasi: 739. Daily Temperatures
Qiyinlik darajasi: O'rta
Bizni kuzatishda davom eting
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LeetCode
Daily Temperatures - LeetCode
Can you solve this real interview question? Daily Temperatures - Given an array of integers temperatures represents the daily temperatures, return an array answer such that answer[i] is the number of days you have to wait after the ith day to get a warmer…
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Kun masalasi: 1971. Find if Path Exists in Graph
Qiyinlik darajasi: Oson
Bizni kuzatishda davom eting
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LeetCode
Find if Path Exists in Graph - LeetCode
Can you solve this real interview question? Find if Path Exists in Graph - There is a bi-directional graph with n vertices, where each vertex is labeled from 0 to n - 1 (inclusive). The edges in the graph are represented as a 2D integer array edges, where…
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Kun masalasi: 841. Keys and Rooms
Qiyinlik darajasi: O'rta
Bizni kuzatishda davom eting
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LeetCode
Keys and Rooms - LeetCode
Can you solve this real interview question? Keys and Rooms - There are n rooms labeled from 0 to n - 1 and all the rooms are locked except for room 0. Your goal is to visit all the rooms. However, you cannot enter a locked room without having its key.
When…
When…
👌1
Keyingi video darslik nima haqida bo'lishini xohlaysiz?
Anonymous Poll
65%
DFS va BFS ning 5 6 ta tadbiqlari
35%
Floyd Warshall algoritmi
Kun masalasi: 886. Possible Bipartition
Qiyinlik darajasi: O'rta
Bizni kuzatishda davom eting
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LeetCode
Possible Bipartition - LeetCode
Can you solve this real interview question? Possible Bipartition - We want to split a group of n people (labeled from 1 to n) into two groups of any size. Each person may dislike some other people, and they should not go into the same group.
Given the integer…
Given the integer…
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