which data structure among the follow has fastest sequential access ?
i) vector
ii) Singly linked list
iii) Doubly linked list
it won't be doubly linked list as it involves more pointer manipulations
than singly linked list...
--
Hello,
Algorithm--
1. Find out start and end index of contiguous subarray which has max sum
O(n)
2.Once u have start and end index calculate avg if it satisfies the
condition then done O(n)
2.1 other wise run a loop through start to end index and remove trailing
elements by increasing start
This question has been taken from codeforces.com. Any idea how to solve
this ?
Polycarpus has an array, consisting of *n* integers *a*1, *a*2, ..., *a**n*.
Polycarpus likes it when numbers in an array match. That's why he wants the
array to have as many equal numbers as possible. For that
Implementation
public class Kadanes {
static int maxAvgSum(int a[], float k) {
int max_so_far = 0, max_ending_here = 0, avg = 0, s = -1, e = -1, ts, te,
tsum;
boolean flag = false;
for (int i = 0; i a.length; i++) {
if (max_ending_here == 0)
s = e = i;
max_ending_here = max_ending_here +
Arnaldo and Bernaldo are playing within the classroom as follows:
they write initially under a positive integer n. then
alternately, beginning with Arnold, erase the number that is on the table
and
write a new number that can be:
• what has just been erased least the largest power of 2 (with
singly linked list
On Wed, Nov 21, 2012 at 8:21 PM, shady sinv...@gmail.com wrote:
which data structure among the follow has fastest sequential access ?
i) vector
ii) Singly linked list
iii) Doubly linked list
it won't be doubly linked list as it involves more pointer manipulations
@Ansum: Polycarpus should start by summing the numbers. If the sum is
divisible by n, then n numbers can be made equal. If the sum is not
divisible by n, then only n-1 numbers can be made equal.
Dave
On Wednesday, November 21, 2012 12:18:54 PM UTC-6, Ansum Baid wrote:
This question has
@Dave: Can you give a little insight on your approach?
On Wed, Nov 21, 2012 at 6:52 PM, Dave dave_and_da...@juno.com wrote:
@Ansum: Polycarpus should start by summing the numbers. If the sum is
divisible by n, then n numbers can be made equal. If the sum is not
divisible by n, then only n-1
@shady : as subject says fastest sequential access , then if i am not
getting it wrong.we only care of sequential access a value not modifying
the linked list.
so i guess double linked list would be helpful
1) bcozz it can move in both the direction , so if linked list is sorted
then it would be
