Microsoft Interview Questions
These Are one of the Microsoft Interview Questions Answer:I answered hash tables but he said suggest a better Round 2: Prologue: Well in my case he started with how they require not only a programmer but a designer and coder who writes perfect code. write it perfectly as if it should ready to be shipped after you code it. Flip 1 to 0 and 0 remains as they are. for example results in 0 0 0 0 0 Round 3: Microsoft Interview Questions and Answers 1. How could you determine if a linked list contains a cycle in it, and, at what node the cycle starts? There are a number of approaches. The approach I shared is in time N (where N is the number of nodes in your linked list). Assume that the node definition contains a boolean flag, bVisited. Then, to determine whether a node has a loop, you could first set this flag to false for all of the nodes: A much better approach was submitted by 4Guys visitor George R., a Microsoft interviewer/employee. He recommended using the following technique, which is in time O(N) and space O(1). Use two pointers. // error checking and checking for NULL at end of list omitted do { p2 is moving through the list twice as fast as p1. If the list is circular, (i.e. a cycle exists) it will eventually get around to that sluggard, p1. 2. How would you reverse a doubly-linked list? This problem isn't too hard. You just need to start at the head of the list, and iterate to the end. At each node, swap the values of pNext and pPrev. Finally, set pHead to the last node in the list. Node * pCurrent = pHead, *pTemp; pHead = pCurrent; pCurrent = temp; 3. Assume you have an array that contains a number of strings (perhaps char * a[100]). Each string is a word from the dictionary. Your task, described in high-level terms, is to devise a way to determine and display all of the anagrams within the array (two words are anagrams if they contain the same characters; for example, tales and slate are anagrams.) Begin by sorting each element in the array in alphabetical order. So, if one element of your array was slate, it would be rearranged to form aelst (use some mechanism to know that the particular instance of aelst maps to slate). At this point, you slate and tales would be identical: aelst. 4. Given the following prototype: write a function that will take a sorted array, possibly with duplicates, and compact the array, returning the new length of the array. That is, if p points to an array containing: 1, 3, 7, 7, 8, 9, 9, 9, 10, when the function returns, the contents of p should be: 1, 3, 7, 8, 9, 10, with a length of 5 returned. A single loop will accomplish this. int compact(int * p, int size)
one.He said suggest a better one and then gave me one
Tree sort of DS and then asked me to compare the two.
Well i gave a the some weird eg where the struct was not a Binary tree but if passed through the test will give positive results.then he asked me to solve for that too.
The interviewer gets a bit serious with each stage. He will test ur work for all possible set of inputs.
"aabbccdef" and o/p "a2b2c2def" or "a4bd2g4" for "aaaabddgggg"
1 0 1 1 0
0 1 1 1 0
1 1 1 1 1
1 0 1 1 1
1 1 1 1 1
0 0 0 0 0
0 0 1 1 0
0 0 0 0 0
0 0 1 1 0
For me some Qs on DB Lock Manager?
(dont even think of sorting the two arrays in a third array , though u can sort them. Try something better than order N ..order LogN )
struct Node
{
...
bool bVisited;
};
// Detect cycle
// Note: pHead points to the head of the list (assume already exists)
Node *pCurrent = pHead;
while (pCurrent)
{
pCurrent->bVisited = false;
pCurrent = pCurrent->pNext;
}
p1 = p2 = head;
p1 = p1-gt;next;
p2 = p2-gt;next->next;
} while (p1 != p2);
while (pCurrent)
{ pTemp = pCurrent-gt;pNext;
pCurrent-gt;pNext = pCurrent->pPrev;
pCurrent-gt;pPrev = temp;
}
Next, sort the entire array of these modified dictionary words. Now, all of the anagrams are grouped together. Finally, step through the array and display duplicate terms, mapping the sorted letters (aelst) back to the word (slate or tales).
int compact(int * p, int size);
{
int current, insert = 1;
for (current=1; current < size; current++)
if (p[current] != p[insert-1])
{
p[insert] = p[current];
current++;
insert++;
} else
current++;
}
0 comments: