Ms Questions Iit D

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<x<20000, how do u sort it. U can load only 1000 numbers at a time in memory. 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 There were many more...

Solution for Qn 4 4) Given an array having 16000 unique integers, each lying within the range 1<x<20000, how do u sort it. U can load only 1000 numbers at a time in memory. 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 <stdlib.h> #include <string.h> #include <stdio.h> 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