Coding Interview Questions

Rama A. - "1. What is an LFU Cache? An LFU (Least Frequently Used) cache is a memory management system with a fixed capacity. It tracks how often items are accessed and evicts the least popular items when it runs out of space. It has one critical rule and one tie-breaker rule: The Core Rule: When the cache is full and you need to add a new item, you must kick out the item with the lowest access frequency. The Tie-Breaker (LRU): If multiple items share that same lowest frequency"See full answer

Alok S. - "I’d clarify the scope first. I’ll assume they want: Given a root folder and a search text, recursively find all files whose filename contains that text. Code: #include #include #include #include using namespace std; namespace fs = std::filesystem; vector searchFiles(const string& rootPath, const string& target) { vector ans; if(!fs::exists(rootPath)) { return ans; } // recursively go through all folder"See full answer

叶 路. - "from collections import deque def updateword(words, startword, end_word): if end_word not in words: return None # Early exit if end_word is not in the dictionary queue = deque([(start_word, 0)]) # (word, steps) visited = set([start_word]) # Keep track of visited words while queue: word, steps = queue.popleft() if word == end_word: return steps # Found the target word, return steps for i in range(len(word)): "See full answer

Ravi K. - "select e.firstname as firstname, m.salary as manager_salary from employees e join employees m on e.manager_id = m.id where e.salary > m.salary; `"See full answer

Dev S. - "#include #include bool palindrome(std::string &str, int left, int right, int error) { if (left >= right) { return true; } if (str[left] == str[right]) { return palindrome(str, left + 1, right - 1, error); } else if (error == 0) { return (palindrome(str, left + 1, right, 1) || palindrome(str,left, right -1,1)); } else { return false; } } int main() { std::string str = "abcbca"; int size = str.size() - 1; if"See full answer

Yeshwanth D. - "Was this for an entry level engineer role?"See full answer

Alok B. - "seeing in a small demo any big events that make easy to understand situation in that time."See full answer

E B. - "Clarifying Points Number of users: unique user id “More than three people” we have user id for the call recipient, and we want number of unique recipients called. We would not count each instance of calling someone. Last week: last 7 days from today SELECT User_id, COUNT(distinct calleduserid) as userscalledcount FROM calls WHERE calldate between currentdate - interval ‘7 day’ and current_date Group by user_id HAVING COUNT(distinct calleduserid) >= 3 This return"See full answer

Salome L. - "SELECT MIN(id) AS id, TRIM(LOWER(email)) AS cleaned_email FROM users GROUP BY cleaned_email ORDER BY id `"See full answer

Sravanthi M. - "public static boolean isPalindrome(String str){ boolean flag = true; int len = str.length()-1; int j = len; for(int i=0;i<=len/2;i++){ if(str.charAt(i)!=str.charAt(j--)){ flag = false; break; } } return flag; }"See full answer

Alfred O. - "We can use dictionary to store cache items so that our read / write operations will be O(1). Each time we read or update an existing record, we have to ensure the item is moved to the back of the cache. This will allow us to evict the first item in the cache whenever the cache is full and we need to add new records also making our eviction O(1) Instead of normal dictionary, we will use ordered dictionary to store cache items. This will allow us to efficiently move items to back of the cache a"See full answer

Rick E. - " O(n) - characters in the string O(n) - stack def identify_adjacent(s: str, k: int) -> str: stack = [] n = len(s) for ch in s: if stack: topch, topcnt = stack[-1] if top_ch == ch: top_cnt += 1 stack[-1] = (ch, top_cnt) else: top_cnt = 1 stack.append((ch,1)) if top_cnt == k: stack.pop() else: stack.append"See full answer

Alvis F. - "Requirements and Goals Primary Goal:Store key-value pairs in a cache with efficient access (reads/writes). Evict items based on a certain “rank,” which might reflect popularity, frequency, or custom ranking logic. Functional Requirements:Put(key, value, rank): Insert or update a key with the given value and rank. Get(key): Retrieve the value associated with the key if it exists. Evict(): If the cache is at capacity, evict the item with the lowest rank (or according"See full answer

Ankush G. - "=>user => window {addfile(void * file, userID, chatID) => add to fileList=> Display(file pointer)} => file=> pointer to DB, thumbnail => file will stores in database, pointer to the file+ thumbnail to be visible to chat => file is associated with chat => chat : users list, notepad: post { , user posted} class file{ void * buffer = nullptr; string fileName = nullptr; public: file( void content, string filename ):fileName ("See full answer

Bamboo Y. - "Since the problem asks for a O(logN) solution, I have to assume that the numbers are already sorted, meaning the same number are adjacent to each other, the value of the numbers shouldn't matter, and they expect us to use Binary Search. First, we should analyze the pattern of a regular number array without a single disrupter. Index: 0 1 2 3 4. 5 6. 7. 8. 9 Array:[1, 1, 2, 2, 4, 4, 5, 5, 6, 6] notice the odd indexes are always referencing the second of the reoccurring numbers and t"See full answer

Richard W. - "int getMaxWater(vector& nums) { int n = nums.size(); int mx = INT_MIN; int i=0, j=n-1; while(i<j) { int water = (j - i) * min(nums[i], nums[j]); mx = max(mx, water); if(nums[i] < nums[j]){ i++; } else { j--; } } return mx; } `"See full answer