DashDash
BAN USER- -2of 2 votes
AnswersHow to print a variable 1000 times without using loops and recurssion
- DashDash in India| Report Duplicate | Flag | PURGE
Software Engineer / Developer - 0of 0 votes
AnswersWe have
- DashDash in India for Development
char *p = "abc";
I know we cant do p[0] = 'a'. What is the reason behind it?| Report Duplicate | Flag | PURGE
Software Engineer / Developer C++ - 0of 0 votes
AnswersWhat is the next number in the series
- DashDash in India
2,4,8,16,24...| Report Duplicate | Flag | PURGE
Goldman Sachs Applications Developer Brain Teasers - 0of 0 votes
Answersvoid copystring(char* dest, char *source)
- DashDash
{
while(*source != NULL)
{
*dest = *source;
dest++;
source++;
}
}
int _tmain(int argc, _TCHAR* argv[])
{
char input[10] = "hello";
char *dest;
dest = &input[1];
copystring(dest, input);
return 0;
}
What is the output of the program...| Report Duplicate | Flag | PURGE
Microsoft Software Engineer / Developer C++
void SumDiff(Tree *root, int num, int *Sum1, int *Sum2)
{
if(root != NULL)
{
if(num % 2 == 0)
{
*Sum2+= root->data;
SumDiff(root->left, num +1, Sum1, Sum2);
SumDiff(root->right, num +1, Sum1, Sum2);
}
else
{
*Sum1+= root->data;
SumDiff(root->left, num +1, Sum1, Sum2);
SumDiff(root->right, num +1, Sum1, Sum2);
}
}
}
Here is what I thought
Last element will be the largest
The diagonal just above the last element will have elements smaller to largest and greater than others
Eg.
I am taking a 3x3 matrix
4 9 11
5 13 15
6 16 25
Now 25 will be largest
Next diagonal elements are 16 and 15
Other next are 6 13 11
Hence we can calculate the kth max element by finding the diagonal where the kth number can be and then finding the kth element from that diagonal
void PrintDdistance(Tree *node, int d, const int D)
{
if(node == NULL)
return;
if(d == D)
Print the node
else
{
PrintDdistance(node->left, d+1, D);
PrintDdistance(node->right, d+1, D);
}
}
I doubt we can find the distance from the parent since only the node pointer is given
Scan the matrix and store the cell positions whose values are 0 in a separate array.
Now every year scan the array and for every element in the array check the adjacent values and if it is all 1 then remove the element from the array and change the matrix.
after k yrs the left elements in the array are the forests left
lets number every sequence like 1->3->8 = 138, so every sequence will be unique
Now let say the sequence with time is 1->3->6->3->5->4....
So the sequence combinations can be 136,363,635,354,........
Now counting the sequence which occurs most can be the answer
Please do let me know if I am wrong here.
We can divide the numbers in blocks and assign it to threads. Let say we get first 10 numbers we assign it to a thread which will give me a random number between 1-10. Now every thread will give me a random number between than range and the main thread then actually get the random number from the range of numbers provided by the threads.
- DashDash June 02, 2012Here is what I thought
Let say he starts from root and index = 0. Reduce index by 1 if he goes to left and increase index by 1 when goes to right. and enter into a hash with index as the key.
Now print the hash taking the min index and traversing to the max index.
int SumofRightTree(Tree *tree, bool isRight)
{
if(tree != NULL)
{
SumofRightTree(tree->left, false);
tree->data = tree->data + SumofRightTree(tree->right, true);
return tree->data;
}
else
return 0;
}
int _tmain(int argc, _TCHAR* argv[])
{
int arr[] = {30,10,5,20,50,40,60};
Tree *root= NULL;
for(int i=0;i<7; i++)
root = CreateTree(root, arr[i]);
SumofRightTree(root, true);
return 0;
}
int SumofRightTree(Tree *tree, bool isRight)
{
if(tree != NULL)
{
SumofRightTree(tree->left, false);
tree->data = tree->data + SumofRightTree(tree->right, true);
return tree->data;
}
else
return 0;
}
int _tmain(int argc, _TCHAR* argv[])
{
int arr[] = {30,10,5,20,50,40,60};
Tree *root= NULL;
for(int i=0;i<7; i++)
root = CreateTree(root, arr[i]);
SumofRightTree(root, true);
return 0;
}
int SumofRightTree(Tree *tree, bool isRight)
{
if(tree != NULL)
{
SumofRightTree(tree->left, false);
tree->data = tree->data + SumofRightTree(tree->right, true);
return tree->data;
}
else
return 0;
}
int _tmain(int argc, _TCHAR* argv[])
{
int arr[] = {30,10,5,20,50,40,60};
Tree *root= NULL;
for(int i=0;i<7; i++)
root = CreateTree(root, arr[i]);
SumofRightTree(root, true);
return 0;
}
Can we have DP here
- DashDash February 17, 2013SP(m,n) = min(SP(m-1, n-1), SP(m-1,n), SP(m,n-1), SP(m+1,n+1), SP(m+1,n), SP(m,n+1)) + 1 if x[m,n] != 0;
if x[m,n] = 0 then return 0