celeritas
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AnswersGiven a matrix, you need to create another matrix such that the value (i,j) is either -1, 0 or 1.
- celeritas in United States
1 - if multiplication of all values in ith row and jth column is greater than 0.
-1 - if multiplication of all values in ith row and jth column is less than 0.
0 - if multiplication of all the values in ith row and jth column is 0.
e.g.
1 2 3 1
1 0 -1 2
-1 1 1 1
o/p
-1 0 -1 1
0 0 0 0
1 0 1 -1| Report Duplicate | Flag | PURGE
Amazon Software Engineer Intern Algorithm
They have just asked us to write a code to test the reverse function written to reverse a linked list.
Assume
Node * reverse(Node * head)
returns a pointer to the linked list which is reversed.
So something like, should work.
Node * reverseReverse = reverse(reverse(orig));
boolean compare(Node * orig, Node * reverseReverse)
{
Node * temp = orig; Node * temp1 = reverseReverse
if(temp == NULL & temp1 == NULL)
return true;
while(temp != NULL && temp1 != NULL)
{
if(temp->value != temp1->value)
return false;
}
if((temp != NULL && temp1 == NULL) || (temp == NULL && temp != NULL))
return false;
return true;
}
int[] reverse(int[] A, int shifts)
{
int[] temp = new int[A.length * 2];
System.arrayCopy(A, 0, temp, 0, A.length);
System.arrayCopy(A, 0, temp, A.length, A.length);
shifts = shifts % A.length;
int[] result = new int[A.length];
System.arrayCopy(temp, shifts - 1, result, 0, A.length);
}
Explanation: temp array is A + A. And temp.length = 2*A.length.
e.g. A = [4,3,2,5] then temp = [4,3,2,5,4,3,2,5].
Now select the array between (including) (shifts - 1) and including shifts - 1 + A.length - 1.
Repwilliamland1990, Associate at Advent
Gifted in donating shaving cream worldwide. Crossed the country getting my feet wet with the elderly in Salisbury, MD. Have ...
We should start decomposing the problem:
- celeritas August 07, 20181. we will only store data for last 30 days in a rolling window.
2. we will have 3 (or one?) max heap data structures (one for 10 frequent words in a day, one for 10 frequent words in an hour and one for 10 frequent in a month).
3. there will be a system that maintains and stores that data.