Vamsi Pavan’s Place

void permute(char *aHead,char *aTail)
{
// If there are no characters remaining in Head, i.e. we got one permutation
// So print it.
if(strlen(aHead) == 0){
printf(”%s\n”,aTail);
return;
}
// Each time remove one character from Head and add it to the tail,
// and then recurse ……
for(int i=0;i< strlen(aHead);i++)
int j=0,insertAt;
char *NewTail = new char[100];
char *NewHead = new char[100];
// Create the arguments that need […]

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int insertAt=0,index;
for(index=insertAt+1;index
if(a[insertAt] == a[index])
index+=1;
else
a[insertAt++] = a[index];
}
insertAt–;
Identification of duplicant element in an array
Identify the element that is not repeating in an array of 2n+1 elements, in O(n) time
X-or all the elements, same elements cancel each other.

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If p is a prime number other than 2, then using formats thearom it follows the condition:
2^(p-1) mod p = 1
So In order to check whether a number is prime or not, we have to check the value 2^(p-1) mod p.
And 2^(p-1) can be calculated efficiently, by following a binary search strategy. (Refer to the […]

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We will see it with an example, Lets calculate pow(x,23)
x^23 = x^22 * x
x^22 = x^11 * x^11
x^11 = x^10 * x
x^10 = x^5 * x^5
x^5

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The pseudo random numbers can be generated using linear congruential method. Successive members of the linear congruential sequence {x} are generated using the expression:
X_n+1 = ( a*X_n + b) mod m

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DLNode *DelNode(DLNode *aHead, int DelValue)
{

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DLNode *revList(DLNode *aHead)
{

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a = a+b-(b=a)
OR
a = a^b^(b=a)
And both of them will work also for chars and stuff ….

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ptr1 = ptr2 = head;
do{
ptr1 = ptr1 -> next;
ptr2 = ptr2 -> next -> next;
}while(ptr1 != ptr2);
ptr2 is iterating along the list with double the speed of ptr1. If there is any cycle, then ptr1 and ptr2 will loop in the list. And since ptr1 is moving at one step and ptr2 is moving faster […]

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