Software Engineer Coding Interview Questions

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

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

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

Anonymous Roadrunner - "def is_palindrome(s: str) -> bool: new = '' for a in s: if a.isalpha() or a.isdigit(): new += a.lower() return (new == new[::-1]) debug your code below print(is_palindrome('abcba')) `"See full answer

Jatin S. - "public int minInsertions(String s) { int n = s.length(); int dp = new intn; for (int len = 2; len <= n; n++) { for (int i = 0; i + len - 1 < n: i++) { int j = i + len - 1; if (s.charAt(i) == s.charAt(j)) { dpi = dpi + 1; } else { dpi = 1 + Math.min(dpi + 1, dpi); } } } return dp0; } `"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

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

Ashish D. - "This is a geometric variation of the "Sliding Window" technique. The core challenge is converting a 2D coordinate problem into a 1D linear problem using angles. The standard approach is the Angular Sweep. By converting the Cartesian coordinates (x,y) of every tree into Polar coordinates (specifically the angle θ), we can process the trees based on the direction they lie in relative to the observer. The Strategy Coordinate Translation: Calculate the relative coordinates (Δx,Δy) for"See full answer

devopsjesus - "This was a 60 minute assessment. The clock is ticking and you're being observed by a senior+ level engineer. Be ready to perform for an audience. The implementation for the system gets broken up into three parts: Implement creating accounts and depositing money into an account by ID Implement transferring money with validation to ensure the accounts for the transfer both exist and that the account money is being removed from has enough money in it to perform the transfer Implement find"See full answer

Ramachandra N. - "def sortedSquares(nums): n = len(nums) result = [0] * n left, right = 0, n - 1 pos = n - 1 while left abs(nums[right]): result[pos] = nums[left] ** 2 left += 1 else: result[pos] = nums[right] ** 2 right -= 1 pos -= 1 return result `"See full answer

Maykon henrique D. - "function longestCommonPrefix(arr1, arr2) { const prefixSet = new Set(); for (let num of arr1) { let str = num.toString(); for (let i = 1; i <= str.length; i++) { prefixSet.add(str.substring(0, i)); } } let longestPrefix = ""; for (let num of arr2) { let str = num.toString(); for (let i = 1; i <= str.length; i++) { let prefix = str.substring(0, i); if (prefixSet.has(prefix)) { "See full answer

Pushkar G. - "Assumptions - Multiple reader/writer threads issue set() [with transaction] and get() operations concurrently All data can fit into memory. Requirements - Good performance (ex. less lock contention) Optimize memory footprint (ex. don't store stale entries in memory) Naive Approach - Wrap a standard hashmap/dictionary with reader-writer lock. We can also shard the map to reduce lock contention (ex. 32 segments for 32 core cpu) Allocate local buffer to a writer when a transacti"See full answer

Ankush G. - " max Min 4, 3, 1 , 6, 7, 8 1 3 4 6 7 8 9 0 1 2 3 0 1 1 2 3 6 7 8 9 class MedianFinder{ std::priority_queue minHeap; std::priority_queue, greater> maxHeap; int numEleMaxheap = 0, numEleMinHeap = 0; public: void addNum( int n) { if(numEleMaxheap == numEleMinHeap ) { maxHeap.push(n); int maxofmaxheap = maxHeap.top(); maxHeap.pop(); "See full answer