Microsoft 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

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

Kishor J. - "we can use two pointer + set like maintain i,j and also insert jth character to set like while set size is equal to our window j-i+1 then maximize our answer and increase jth pointer till last index"See full answer

Rahul M. - "Idea for solution: Reverse the complete char array Reverse the words separated by space. i.e. Find the space characters and the reverse the subarray between two space characters. vector reverseSubarray(vector& arr, int s, int e) { while (s reverseWords(vector& arr ) { int n = arr.size(); reverse(arr, 0, n - 1"See full answer

Nicholas S. - "First thing the interviewee did wrong is not asking clarifying questions. This is the most vague problem I have every heard, and the interviewee just made assumptions and started programming."See full answer

Aashka C. - "Sorted the array and stored the minimum difference in a variable and then traversed the array for the pairs having minimum difference"See full answer

Sachin R. - "If 0's aren't a concern, couldn't we just multiply all numbers. and then divide product by each number in the list ? if there's more than one zero, then we just return an array of 0s if there's one zero, then we just replace 0 with product and rest 0s. what am i missing?"See full answer

Taheia S. - "at first I want to know number of cluster I will put random number if I don't know and I will use method called Elbow method or Silhouette Score ,Gap Statistic and Davies–Bouldin Index to know the best number of cluster and I will use scikit-learn library to import kmeans from sklearn.cluster import KMeans kmeans = KMeans(nclusters=2, randomstate=0) kmeans.fit(X) and X this my data "See full answer

Praveen D. - "Let me try to explain it with simple life analogy You're cooking dinner in the kitchen. Multithreading is when you've got a bunch of friends helping out. Each friend does a different job—like one chops veggies while another stirs a sauce. Everyone focuses on their task, and together, you all make the meal faster. In a computer, it's like different jobs happening all at once, making stuff happen quicker, just like having lots of friends helping makes dinner ready faster."See full answer

Yashasvi V. - "WITH RECURSIVE fibonacci_series AS ( SELECT 1 AS n, 0 AS fib1, 1 AS fib2 UNION ALL SELECT n + 1 AS n, fib2 AS fib1, fib1 + fib2 AS fib2 FROM fibonacci_series WHERE n < 20 -- Limit the series to 20 numbers ) SELECT n, fib1 AS fib FROM fibonacci_series ORDER BY n; `"See full answer

Sue G. - "Abstract class A class that can have Abstract methods - without implementations and Concerete Methods i.e with implementation. Can have private, protected and public access modifiers. Supports Single inheritance i.e a class can extend only 1 abstract class Can have constructors Mainly used when sharing common behaviors Interface Class A collection of abstract methods ( can have static and default methods also - onwards of java 8) Public, static, final are the access"See full answer

Stanley Y. - "First of all, stack and heap memory are abstraction on top of the hardware by the compiler. The hardware is not aware of stack and heap memory. There is only a single piece of memory that a program has access to. The compiler creates the concepts of stack and heap memory to run the programs efficiently. Programs use stack memory to store local variables and a few important register values such as frame pointer and return address for program counter. This makes it easier for the compiler to gene"See full answer

Nilay B. - "int reverse(int x) { int rev = 0; bool isNegative = false; if(x 0){ int r = x % 10; rev = rev * 10 + r; x = x / 10; } if(isNegative){ return -rev; } return rev; } `"See full answer

An D. - "First find the node that we need to delete. After it's found, think about ways to keep the tree BST after deleting the node. a. If there's no left or right subtree, we found the leaf. Delete this node without any further traversing. b. If it's not a leaf node, what node we can use from the subtree that can replace the delete node and still maintain the BST property? We can either replace the delete node with the minimum from the right subtree (if right exists) or we can replace the delete"See full answer

Anonymous Wombat - "Was the statement very similar to the leetcode or was it changed and only the main idea remained?"See full answer

Divya R. - "static TreeNode buildTree(int] preorder, int[] inorder) {//[2, 4, 5 if(preorder.length==0 || preorder.length!=inorder.length) { isNull=true; return null; } int parentElem=preorder[0];//2 int index=-1; index=binarySearch(inorder, parentElem);//3 //1 if(index==-1) { isNull=true; return null; } TreeNode root=new TreeNode(parentElem); int preOrder=0; int ind=-1; if(index>0) { int[] leftInOrder=Arrays.copyOfRange(inorder, 0, index); //[4, 2, 5]//[4] ind=binarySearch(preorder, inorder[index"See full answer

GalacticInterviewer - "Approach 1: Use sorting and return the kth largest element from the sorted list. Time complexity: O(nlogn) Approach 2: Use max heap and then select the kth largest element. time complexity: O(n+logn) Approach 3: Quickselect. Time complexity O(n) I explained my interviewer the 3 approaches. He told me to solve in a naive manner. Used Approach 1 had some time left so coded approach 3 also The average time complexity of Quickselect is O(n), making it very efficient for its purpose. However, in"See full answer

Basil A. - " import java.io.*; import java.util.*; class Solution { static int findRotations(int[] arr) { // your code goes here int n = arr.length; if(n == 0) { return 0; } int left = 0; int right = n-1; if(arr[left] arr[mid+1]) { return mid+1; } if(arr[left] <= arr[mid]) { left = mid + 1; } e"See full answer

Manu G. - "class Solution { public: vector topKFrequent(vector& sentences , int k) { unordered_map mp; auto cmp={ if(a.second==b.second) return a.first>b.first; return a.second,vector>,decltype(cmp)> pq(cmp); for(string s : sentences) { stringstream ss(s); string word; while(ss >> word) { mp[word]++; } } for(auto it=mp.begin();it!=mp.end();++it) pq.push({it->first,it->secon"See full answer