Meta (Facebook) Software Engineer Interview Questions

Yomna H. - "Evication Strategy: LFU Access Pattern: Write Around"See full answer

Shuchi A. - "A few months ago I joined a micro-services platform engineering team as their manager, at that time my team was struggling to deliver towards an upcoming production deadline for a customer facing product. Production date had been moved 5 times already and there were about 40% of product features which were remaining to be tested and signed off to move to production . I was made responsible to deliver the release of this product within the deadline and turnaround the software delivery throughput."See full answer

Rjj - "I tried solving this question and here is the recorded video for the entire solution - https://youtu.be/G_LIbTp58XA Feel free to comment here or on the video for further discussion."See full answer

Tiago R. - "const ops = { '+': (a, b) => a+b, '-': (a, b) => a-b, '/': (a, b) => a/b, '': (a, b) => ab, }; function calc(expr) { // Search for + or - for (let i=expr.length-1; i >= 0; i--) { const char = expr.charAt(i); if (['+', '-'].includes(char)) { return opschar), calc(expr.slice(i+1))); } } // Search for / or * for (let i=expr.length-1; i >= 0; i--) { const char = expr.charAt(i); if"See full answer

Komal S. - "a. Sort the array elements. b. take two pointers at index 0 and index Len-1; c. if the sum at the two pointers is target; break and return the pair d. if the sum is smaller, then move left pointer by 1 e. else move right pointer by 1; run the logic till the target is met or right pointer crosses the left pointer."See full answer

Anonymous Salamander - "Microservices are small parts of a application we can deploy them seprately and use them as a application feature."See full answer

Srihitha J. - "my answer: void* memcpy(void* dest, const void* src, size_t n) { unsigned char* uDest = static_cast(dest); const unsigned char* ucSrc = static_cast(src); for(size_t i= 0; i(dest); const unsigned c"See full answer

Mahaboob P. - "int[] sqSorted(int[] nums) { int i = 0, j = nums.length-1; int k = nums.length-1; int[] sqs = new int[nums.length]; while(i n1) { sqs[k--] = n2; j--; } else { sqs[k--] = n1; i++; } } for(int n: sqs) System.out.println(n); return sqs; }"See full answer

Guilherme F. - "A much better solution than the one in the article, below: It looks like the ones writing articles here in Javascript do not understand the time/space complexity of javascript methods. shift, splice, sort, etc... In the solution article you have a shift and a sort being done inside a while, that is, the multiplication of Ns. My solution, below, iterates through the list once and then sorts it, separately. It´s O(N+Log(N)) class ListNode { constructor(val = 0, next = null) { th"See full answer

Ds S. - "Initially I asked clarifying questions like whether the tree can be empty or not and asked the interviewer to explain what is meant by left view and the explanation for the sample inputs. Then I came up with the level order traversal approach where we visit each level in the binary tree at once using a queue and at each level print the value of the first node. Interviewer seemed satisfied with the approach and asked me to code it up. Finally gave the time and space complexity of the solution."See full answer

Techzen I. - "Implemented the Java code to find the largest island. It is similar to count the island. But in this we need to keep track of max island and compute its perimeter."See full answer

Alvaro R. - "bool isValidBST(TreeNode* root, long min = LONGMIN, long max = LONGMAX){ if (root == NULL) return true; if (root->val val >= max) return false; return isValidBST(root->left, min, root->val) && isValidBST(root->right, root->val, max); } `"See full answer

Bhaskar B. - "This could be done using two-pointer approach assuming array is sorted: left and right pointers. We need track two sums (left and right) as we move pointers. For moving pointers we will move left to right by 1 (increment) when right sum is greater. We will move right pointer to left by 1 (decrement) when left sum is greater. at some point we will either get the sum same and that's when we exit from the loop. 0-left will be one array and right-(n-1) will be another array. We are not going to mo"See full answer

Aryan D. - "int main() { int a1[7]={1,2,3,4,5,6,7}; int a2[7]={1,9,10,11,12,13,14}; vectorv; v.insert(v.begin(),begin(a1),end(a1)); v.insert(v.begin(),begin(a2),end(a2)); int a3[v.size()]; sort(v.begin(),v.end()); for(int i=0;i<v.size();i++) { a3[i]=v[i]; } } `"See full answer

Moshe S. - "Law is my passion. Traveling all over the world in 5 years"See full answer