How to select k sample nodes from a unknown sized linked list.?

http://en.wikipedia.org/wiki/Reservoir_sampling can be applied here. Is there any better solution for this problem?

Map reduce implementation of reservoir sampling is discussed in this blog post.
http://had00b.blogspot.in/2013/07/random-subset-in-mapreduce.html

Finding number of on bits (1 bits) given an unsigned integer?

Sol 1:
Simple Iterative solution.(Very slow)

int countbits (unsigned int number) {

int count = 0;

while (number) {

count += number & 0x1u;

number >>= 1;

}

return count;

}

Sol 2:
Faster Solution for 32 bit machines.
compute 16bits then add then to return full number of ones in 32 bit number.

static char ones_in_16bit_num [0x1u << 16] ;
int countbits (unsigned int number)
{
return ones_in_16bit_num [number & 0xffffu] + ones_in_16bit_num [(number >> 16) & 0xffffu] ;

}

Design a stack which have O(1) time complexity for finding the minimum element?

This problem can be solved using simple encoding technique. We should use the minimum element to get the next minimum element.Here is a solution which uses O(1) space.PUSH(elem) Operation: In this process we are keeping minimum element on the top of the stack at any given time. And encoding(i.e elem-min) the actual data with minimum and storing. Whenever there is a change in cur minimum data stored in the stack will be negative. This property will be used while poping the elements to find the next minimum. We can define it as shown below. if stack is empty. Store the element twice Else CurMin = POP(); Take difference between elem and Curmin and push it on to stack. Push the least of elem and CurMin on to stack.
POP() Operation: POP process is bit tricky. If minimum is not changing (i.e elements are positive on stack) then return the element + curMin(sitting on stack). If you see a negative stack element then it means there was a shift of curMin (or minimum element is the one which is being popped). In this case we take the current minimum and return it as popped element. We find the new minimum (as current minimum is popped out) by computing curMin-element and push it on top of stack as minimum element. We can summarize it as shown below.curMin = pop();element = pop(); If element is negative then return curMin and push curMin-element on to stack. Else push curMin on to stack and return element + curMin.
FindMin():
This is trivial operation as we are storing minimum element on the top of the stack at any given time. So we pop it and return to the caller. FindMin() steps will be...min=pop();push(min);return min;

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Given a string s1 and a string s2, write a snippet to say whether s2 is a rotation of s1 using only one call to strstr routine?

(eg given s1 = ABCD and s2 = CDAB, return true)
(given s1 = ABCD, and s2 = ACBD , return false)

Sol:
This is most frequently asked question. It can be solved with simple logic as given below.

1 . Check if both s1, s2 are of same length.
2. if( strstr(strcat(s1,s1), s2) !=NULL) return TRUE
else
return FALSE

Example:
if s1 = ABCD, the concatination will make it ABCDABCD and hence any rotation will fall within the concatenated string. s2 is italicized in conctenized in above concatenated string.

Reverse each individual word in sentence?

void reversestr(char* str, char* strend)
{
char tmp;
while(strend>str)
{
tmp = *strend;
*strend = *str;
*str = tmp;
str++;
strend--;
}
return;
}
void reverseSen(char *sentence)
{
char *wordstart=sentence, *wordstop=sentence;
char *pSend=sentence,*pSstart =sentence ;
int size=0;
while(*pSend != '\0'){
pSend++;
size++;
}
pSend--;
// reverese stentence
reversestr(pSstart,pSend);

while(wordstop < pSend)
{
//find the word
while((*wordstop!='\0')&&(*wordstop!=' ')) wordstop++;
wordstop--;
//reverese the word
reversestr(wordstart,wordstop);
wordstart = wordstop +2;
wordstop = wordstart;
}
}

Find out if a string is a palindrome?

bool Palindrome(char *str)

{
int end = strlen(str)-1;
for (int i = 0; i <>
if (*(str+i) != *(str+end))
return false;
return true;
}

Find a repeated integer in array of size n?

int DuplicatedNumber(int *a, int size)
{
map hash;
bool founddup = false;
for(int i=0; i 0)
{
founddup= true;
break;
}
else
hash[a[i]]++;
}

return founddup ? a[i] : -1; // return the duplicated value or -1 if not found
}

WAP to reverse a linked list?

void reverse(NODE **head) {
if (!*head) return;
NODE *cur = *head;
NODE *prev = NULL;
NODE *next = NULL;
while (cur) {
next = cur->next;
cur->next = prev;
prev = cur;
cur = next;
}
*head = prev;
return;

}

Recusive way call with original = list, parent = NULL

Node* reverselinkedlist(Node* original, Node*parent)
{
Node* reverse;

if(original == NULL)
{
reverse = parent;

}
else
{
reverse = reverselinkedlist(original->link,original);
original->link = parent;
}

return reverse;

}