Software Engineer Interview Questions

There are a lot of online solutions posted about this.

Graph implemented by matrix: import java.util.LinkedList; import java.util.List; import java.util.Queue; import java.util.Stack; public class GraphMatrix { int[][] mat; public GraphMatrix(int[][] mat){ this.mat = mat; } public int[] bfs(int i){ int[] dis = new int[mat.length]; Queue q = new LinkedList(); int[] couple = {i,0}; q.add(couple); while (!q.isEmpty()){ int[] curr = q.poll(); dis[curr[0]] = curr[1]; for (int j=0; j q = new Stack(); int[] couple = {i,0}; q.add(couple); while (!q.isEmpty()){ int[] curr = q.peek(); dis[curr[0]] = curr[1]; boolean foundNeigh=false; for (int j=0; j

void rotateLeft( Node p) { Node r = p.right; p.right = r.left; if (r.left != null) r.left.parent = p; r.parent = p.parent; if (p.parent == null) root = r; else if (p.parent.left == p) p.parent.left = r; else p.parent.right = r; r.left = p; p.parent = r; } void rotateRight(Node p) { Node l = p.left; p.left = l.right; if (l.right != null) l.right.parent = p; l.parent = p.parent; if (p.parent == null) root = l; else if (p.parent.right == p) p.parent.right = l; else p.parent.left = l; l.right = p; p.parent = l; }

To help visualise, right rotation is shown below. Consider a given tree: R /\ / \ P C /\ / \ A B Imagine nodes as metal coins suspended by strings (branches). For right rotation: 1. Pick up node P and pull it up so that node R falls down to the right of P. 2. Take B (right child of P) and make it left child of node R. The tree has been right rotated now and looks like this: P /\ / \ A R /\ / \ B C Similar process can be applied to this tree to rotate it left. This time start by pulling up R above P. Hope it helps.

For sql , w3schools tutorial is enough to brush up the things

Second highest number in an array public class SecondHighest { public static int secondhighest(int[] arr){ if (arr == null){ return -1; } if (arr.length ==1){ return -1; } int max = arr[0] > arr[1] ? arr[0] : arr[1]; int max2 = arr[0] max){ max2 = max; max = arr[i]; } else if (arr[i] > max2){ max2 = arr[i]; } } return max2; } public static void main (String[] args){ int[] arr = {1,-2,4,7,-9,45,34,-92}; System.out.println(secondhighest(arr)); } }

They are shared amongst same types.

For each instantiation of the template class the static members are shared only between the objects of that class and are completely independent from static members of other instantiations of the template class. E.g. this program prints "50": #include template class A { public: static int x; }; template int A::x = 0; int main() { A a; A b; a.x = 5; std::cout << a.x << b.x << std::endl; return 0; }

if(false)

Its just to check logic, finally will not run if system.exit(0) is called.

function lca(root, a, b){ var res = null; (function(subroot){ var ret = 0; if(subroot == a) ret += 1; if(subroot == b) ret += 2; // a, b can be the same // lca can be either a or b as well if( ret == 3 ) { res = subroot; return 0; } if(subroot.left != null){ ret += arguments.callee(subroot.left); } if(subroot.right != null){ ret += arguments.callee(subroot.right); } if (ret === 3) { res = subroot; return 0; } return ret; })(root); return res; }

class Node { int value; int height Node next; Node right; Node parent; } public int height(Node root) { int retValue = 0; while((root=parent(root)) != NULL) { retValue+ } return retValue; } public Node LCA(Node A, Node B) { Node temp; A.height = height(A); B.height = height(B); if(A.height B.height; i--, A=parent(A)); do { A = parent(A); B = parent(B); } while (A != B); return A; }

// NOT the best possible solution but an alternative: // This code gets parents of node1 and puts in an array. // Then gets parents of node2 and puts in another array. // Then goes through the list to see if it contains the first node. // NOTE: Not tried in editor, may not compile. Please use as pseudo code. public static void main(String[] args){ // assumption: Nodes are initialized and known already Node node1; Node node2; Node lca = getLCA(node1, node2); // TODO: print lcs System.out.println("LCA is: " + lca.toString(); } static Node getLCA(Node node1, Node node2){ List list1 = new ArrayList(); getParents(list1, node1); List list2 = new ArrayList(); getParents(list2, node2); for(Node n : list2){ if(list1.contains(n){ return n; } } } static void getParents(List list, Node node){ if(list == null){ return; } if(node == null){ return; } list.add(node); getParents(list, node.parent); } public class Node{ String value; Node parent; @Override public String toString() { return value; } }

memmove function Interviewer was really trying to pull this one out of me, but I was just stumped. He was like: "you can copy it, or you can - what?" I had no clue, never heard of it. I think this one finally destroyed my chances.

i guess it can be moved by using pointer? Below is an answer on the stackoverflow http://stackoverflow.com/questions/6118432/how-to-reverse-words-one-by-one-in-c-using-pointers-and-arrays

I'm not sure how it works in the UK, but in USA, it's illegal to ask all three of those questions.

I think its pretty common in Europe. I have been asked few times about whether I have a partner or not.