Microsoft Questions at IIT Delhi

1) Two linked lists intersect at one node (imagine Y shape)...after intersecting, remaining nodes are common to both the link lists. how do u find the point of intersection

2) Given a BST, how do u check whether it is a valid BST

3) You have n machines, each having n integers. Now u have to find median of these n^2 numbers, but u can load only 2n integers at a time in memory

4) Given an array having 16000 unique integers, each lying within the range 1
5) Remove alternate nodes from a link list

6) Write code for removing loop from a link list

7) You have a BST containing integers. now Given any two numbers x and y, how do u find the common ancestor of nodes which have these values in them. You are given pointet to root of the BST.

8) Code for printing all permutations of a string.

9) Code for reversing words of the string

Solution for Qn 4
4) Given an array having 16000 unique integers, each lying within the range 1

Solution:

Have an array of 2500 bytes (memory equaivalent of 625 ints) - Each bit in the array will show the presence/absence of a num betw 1-20,000. Lets call this the "presence-bit-map".

Since you can store the presence of 8 nos in a byte, you need (20,000/8) = 2500 bytes.

Initialize all bytes in the presence-bit-map to 0, meaning 'absent'.

Iterate the large integer list(be it in file / mem) once. For every number encountered, set the corresponding bit to 1(indicating present) in the presence-bit-map.

Now, by scanning the presence-bit-map once, you can write the present numbers in sorted order.

#include 

#include <string.h>

#include 

void swap(char* src, char* dst)

{

        char ch = *dst;

        *dst = *src;

        *src = ch;

}

/* permute [set[begin], set[end]) */

int permute(char* set, int begin, int end)

{

        int i;

        int range = end - begin;

        if (range == 1) {

                printf("set: %s\n", set);

        } else {

                for(i=0; i

                        swap(&set[begin], &set[begin+i]);

                        permute(set, begin+1, end);

                        swap(&set[begin], &set[begin+i]);       /* set back */

                }

        }

        return 0;

}

int main()

{

        char str[] = "abcd";

        permute(str, 0, strlen(str));

        return 0;

}