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An array(10^5 size) of 32-bit binary numbers is given, we're required to count the no. of ones for every bit of those numbers.
For example:
Array : {10101,1011,1010,1}
Counts : {1's place: 3, 2's place: 2, 3's place: 1, 4's place: 2, 5's place: 1}
No bit manipulation technique seems to satisfy the constraints to me.
Well, this should be solveable with two loops: one going over the array the other one masking the right bits. Running time should be not too bad for your constraints.
Here is a rust implementation (out of my head, not throughtfully tested):
fn main() {
let mut v = vec!();
for i in 1..50*1000 {
v.push(i);
}
let r = bitcount_arr(v);
r.iter().enumerate().for_each( |(i,x)| print!("index {}:{} ",i+1,x));
}
fn bitcount_arr(input:Vec<u32>) -> [u32;32] {
let mut res = [0;32];
for num in input {
for i in 0..31 {
let mask = 1 << i;
if num & mask != 0 {
res[i] += 1;
}
}
}
res
}
This can be done with transposed addition, though the array is a bit long for it.
To transpose addition, use an array of counters, but instead of using one counter for every position we'll use one counter for every bit of the count. So a counter that tracks for each position whether the count is even/odd, a counter that tracks for each position whether the count has a 2 in it, etc.
To add an element of the array into this, only half-add operations (& to find the new carry, ^ to update) are needed, since it's only a conditional increment: (not tested)
uint32_t counters[17];
for (uint32_t elem : array) {
uint32_t c = elem;
for (int i = 0; i < 17; i++) {
uint32_t nextcarry = counters[i] & c;
counters[i] ^= c;
c = nextcarry;
}
}
I chose 17 counters because log2(10^5) is just less than 17. So even if all bits are 1, the counters won't wrap.
To read off the result for bit k, take the k'th bit of every counter.
There are slightly more efficient ways that can add several elements of the array into the counters at once using some full-adds and duplicated counters.
I want to traverse all the elements in the set Q = [0, 2^16) in a non sequential manner. To do so I need a function f(x) Q --> Q which gives the order in which the set will be sorted. for example:
f(0) = 2345
f(1) = 4364
f(2) = 24
(...)
To recover the order I would need the inverse function f'(x) Q --> Q which would output:
f(2345) = 0
f(4364) = 1
f(24) = 2
(...)
The function must be bijective, for each element of Q the function uniquely maps to another element of Q.
How can I generate such a function or are there any know functions that do this?
EDIT: In the following answer, f(x) is "what comes after x", not "what goes in position x". For example, if your first number is 5, then f(5) is the next element, not f(1). In retrospect, you probably thought of f(x) as "what goes in position x". The function defined in this answer is much weaker if used as "what goes in position x".
Linear congruential generators fit your needs.
A linear congruential generator is defined by the equation
f(x) = a*x+c (mod m)
for some constants a, c, and m. In this case, m = 65536.
An LCG has full period (the property you want) if the following properties hold:
c and m are relatively prime.
a-1 is divisible by all prime factors of m.
If m is a multiple of 4, a-1 is a multiple of 4.
We'll go with a = 5, c = 1.
To invert an LCG, we solve for f(x) in terms of x:
x = (a^-1)*(f(x) - c) (mod m)
We can find the inverse of 5 mod 65536 by the extended Euclidean algorithm, or since we just need this one computation, we can plug it into Wolfram Alpha. The result is 52429.
Thus, we have
f(x) = (5*x + 1) % 65536
f^-1(x) = (52429 * (x - 1)) % 65536
There's many approaches to solving this.
Since your set size is small, the requirement for generating the function and its inverse can simply be done via memory lookup. So once you choose your permutation, you can store the forward and reverse directions in lookup tables.
One approach to creating a permutation is mapping out all elements in an array and then randomly swapping them "enough" times. C code:
int f[PERM_SIZE], inv_f[PERM_SIZE];
int i;
// start out with identity permutation
for (i=0; i < PERM_SIZE; ++i) {
f[i] = i;
inv_f[i] = i;
}
// seed your random number generator
srand(SEED);
// look "enough" times, where we choose "enough" = size of array
for (i=0; i < PERM_SIZE; ++i) {
int j, k;
j = rand()%PERM_SIZE;
k = rand()%PERM_SIZE;
swap( &f[i], &f[j] );
}
// create inverse of f
for (i=0; i < PERM_SIZE; ++i)
inv_f[f[i]] = i;
Enjoy
I have an assignment to write an algorithm (not in any particular language, just pseudo-code) that receives a matrix [size: M x N] that is sorted in a way that all of it's rows are sorted and all of it's columns are sorted individually, and finds a certain value within this matrix. I need to write the most time-efficient algorithm I can think of.
The matrix looks something like:
1 3 5
4 6 8
7 9 10
My idea is to start at the first row and last column and simply check the value, if it's bigger go down and if it's smaller than go left and keep doing so until the value is found or until the indexes are out of bounds (in case the value does not exist). This algorithm works at linear complexity O(m+n). I've been told that it's possible to do so with a logarithmic complexity. Is it possible? and if so, how?
Your matrix looks like this:
a ..... b ..... c
. . . . .
. 1 . 2 .
. . . . .
d ..... e ..... f
. . . . .
. 3 . 4 .
. . . . .
g ..... h ..... i
and has following properties:
a,c,g < i
a,b,d < e
b,c,e < f
d,e,g < h
e,f,h < i
So value in lowest-rigth most corner (eg. i) is always the biggest in whole matrix
and this property is recursive if you divide matrix into 4 equal pieces.
So we could try to use binary search:
probe for value,
divide into pieces,
choose correct piece (somehow),
goto 1 with new piece.
Hence algorithm could look like this:
input: X - value to be searched
until found
divide matrix into 4 equal pieces
get e,f,h,i as shown on picture
if (e or f or h or i) equals X then
return found
if X < e then quarter := 1
if X < f then quarter := 2
if X < h then quarter := 3
if X < i then quarter := 4
if no quarter assigned then
return not_found
make smaller matrix from chosen quarter
This looks for me like a O(log n) where n is number of elements in matrix. It is kind of binary search but in two dimensions. I cannot prove it formally but resembles typical binary search.
and that's how the sample input looks? Sorted by diagonals? That's an interesting sort, to be sure.
Since the following row may have a value that's lower than any value on this row, you can't assume anything in particular about a given row of data.
I would (if asked to do this over a large input) read the matrix into a list-struct that took the data as one pair of a tuple, and the mxn coord as the part of the tuple, and then quicksort the matrix once, then find it by value.
Alternately, if the value of each individual location is unique, toss the MxN data into a dictionary keyed on the value, then jump to the dictionary entry of the MxN based on the key of the input (or the hash of the key of the input).
EDIT:
Notice that the answer I give above is valid if you're going to look through the matrix more than once. If you only need to parse it once, then this is as fast as you can do it:
for (int i = 0; i<M; i++)
for (int j=0; j<N; j++)
if (mat[i][j] == value) return tuple(i,j);
Apparently my comment on the question should go down here too :|
#sagar but that's not the example given by the professor. otherwise he had the fastest method above (check the end of the row first, then proceed) additionally, checking the end of the middlest row first would be faster, a bit of a binary search.
Checking the end of each row (and starting on the end of the middle row) to find a number higher than the checked for number on an in memory array would be fastest, then doing a binary search on each matching row till you find it.
in log M you can get a range of rows able to contain the target (binary search on the first value of rows, binary search on last value of rows, keep only those rows whose first <= target and last >= target) two binary searches is still O(log M)
then in O(log N) you can explore each of these rows, with again, a binary search!
that makes it O(logM x logN)
tadaaaa
public static boolean find(int a[][],int rows,int cols,int x){
int m=0;
int n=cols-1;
while(m<rows&&n>=0){
if(a[m][n]==x)
return1;
else if(a[m][n]>x)
n--;
else m++;
}
}
what about getting the diagonal out, then binary search over the diagonal, start bottom right check if it is above, if yes take the diagonal array position as the column it is in, if not then check if it is below. each time running a binary search on the column once you have a hit on the diagonal (using the array position of the diagonal as the column index). I think this is what was stated by #user942640
you could get the running time of the above and when required (at some point) swap the algo to do a binary search on the initial diagonal array (this is taking into consideration its n * n elements and getting x or y length is O(1) as x.length = y.length. even on a million * million binary search the diagonal if it is less then half step back up the diagonal, if it is not less then binary search back towards where you where (this is a slight change to the algo when doing a binary search along the diagonal). I think the diagonal is better than the binary search down the rows, Im just to tired at the moment to look at the maths :)
by the way I believe running time is slightly different to analysis which you would describe in terms of best/worst/avg case, and time against memory size etc. so the question would be better stated as in 'what is the best running time in worst case analysis', because in best case you could do a brute linear scan and the item could be in the first position and this would be a better 'running time' than binary search...
Here is a lower bound of n. Start with an unsorted array A of length n. Construct a new matrix M according to the following rule: the secondary diagonal contains the array A, everything above it is minus infinity, everything below it is plus infinity. The rows and columns are sorted, and looking for an entry in M is the same as looking for an entry in A.
This is in the vein of Michal's answer (from which I will steal the nice graphic).
Matrix:
min ..... b ..... c
. . .
. II . I .
. . .
d .... mid .... f
. . .
. III . IV .
. . .
g ..... h ..... max
Min and max are the smallest and largest values, respectively. "mid" is not necessarily the average/median/whatever value.
We know that the value at mid is >= all values in quadrant II, and <= all values in quadrant IV. We cannot make such claims for quadrants I and III. If we recurse, we can eliminate one quadrant at each level.
Thus, if the target value is less than mid, we must search quadrants I, II, and III. If the target value is greater than mid, we must search quadrants I, III, and IV.
The space reduces to 3/4 its previous at each step:
n * (3/4)x = 1
n = (4/3)x
x = log4/3(n)
Logarithms differ by a constant factor, so this is O(log(n)).
find(min, max, target)
if min is max
if target == min
return min
else
return not found
else if target < min or target > max
return not found
else
set mid to average of min and max
if target == mid
return mid
else
find(b, f, target), return if found
find(d, h, target), return if found
if target < mid
return find(min, mid, target)
else
return find(mid, max, target)
JavaScript solution:
//start from the top right corner
//if value = el, element is found
//if value < el, move to the next row, element can't be in that row since row is sorted
//if value > el, move to the previous column, element can't be in that column since column is sorted
function find(matrix, el) {
//some error checking
if (!matrix[0] || !matrix[0].length){
return false;
}
if (!el || isNaN(el)){
return false;
}
var row = 0; //first row
var col = matrix[0].length - 1; //last column
while (row < matrix.length && col >= 0) {
if (matrix[row][col] === el) { //element is found
return true;
} else if (matrix[row][col] < el) {
row++; //move to the next row
} else {
col--; //move to the previous column
}
}
return false;
}
this is wrong answer
I am really not sure if any of the answers are the optimal answers. I am going at it.
binary search first row, and first column and find out the row and column where "x" could be. you will get 0,j and i,0. x will be on i row or j column if x is not found in this step.
binary search on the row i and the column j you found in step 1.
I think the time complexity is 2* (log m + log n).
You can reduce the constant, if the input array is a square (n * n), by binary searching along the diagonal.
As part of a recent job application I was asked to code a solution to this problem.
Given,
n = number of people standing in a circle.
k = number of people to count over each time
Each person is given a unique (incrementing) id. Starting with the first person (the lowest id), they begin counting from 1 to k.
The person at k is then removed and the circle closes up. The next remaining person (following the eliminated person) resumes counting at 1. This process repeats until only one person is left, the winner.
The solution must provide:
the id of each person in the order they are removed from the circle
the id of the winner.
Performance constraints:
Use as little memory as possible.
Make the solution run as fast as possible.
I remembered doing something similar in my CS course from years ago but could not recall the details at the time of this test. I now realize it is a well known, classic problem with multiple solutions. (I will not mention it by name yet as some may just 'wikipedia' an answer).
I've already submitted my solution so I'm absolutely not looking for people to answer it for me. I will provide it a bit later once/if others have provided some answers.
My main goal for asking this question is to see how my solution compares to others given the requirements and constraints.
(Note the requirements carefully as I think they may invalidate some of the 'classic' solutions.)
Manuel Gonzalez noticed correctly that this is the general form of the famous Josephus problem.
If we are only interested in the survivor f(N,K) of a circle of size N and jumps of size K, then we can solve this with a very simple dynamic programming loop (In linear time and constant memory). Note that the ids start from 0:
int remaining(int n, int k) {
int r = 0;
for (int i = 2; i <= n; i++)
r = (r + k) % i;
return r;
}
It is based on the following recurrence relation:
f(N,K) = (f(N-1,K) + K) mod N
This relation can be explained by simulating the process of elimination, and after each elimination re-assigning new ids starting from 0. The old indices are the new ones with a circular shift of k positions. For a more detailed explanation of this formula, see http://blue.butler.edu/~phenders/InRoads/MathCounts8.pdf.
I know that the OP asks for all the indices of the eliminated items in their correct order. However, I believe that the above insight can be used for solving this as well.
You can do it using a boolean array.
Here is a pseudo code:
Let alive be a boolean array of size N. If alive[i] is true then ith person is alive else dead. Initially it is true for every 1>=i<=N
Let numAlive be the number of persons alive. So numAlive = N at start.
i = 1 # Counting starts from 1st person.
count = 0;
# keep looping till we've more than 1 persons.
while numAlive > 1 do
if alive[i]
count++
end-if
# time to kill ?
if count == K
print Person i killed
numAlive --
alive[i] = false
count = 0
end-if
i = (i%N)+1 # Counting starts from next person.
end-while
# Find the only alive person who is the winner.
while alive[i] != true do
i = (i%N)+1
end-while
print Person i is the winner
The above solution is space efficient but not time efficient as the dead persons are being checked.
To make it more efficient time wise you can make use of a circular linked list. Every time you kill a person you delete a node from the list. You continue till a single node is left in the list.
The problem of determining the 'kth' person is called the Josephus Problem.
Armin Shams-Baragh from Ferdowsi University of Mashhad published some formulas for the Josephus Problem and the extended version of it.
The paper is available at: http://www.cs.man.ac.uk/~shamsbaa/Josephus.pdf
This is my solution, coded in C#. What could be improved?
public class Person
{
public Person(int n)
{
Number = n;
}
public int Number { get; private set; }
}
static void Main(string[] args)
{
int n = 10; int k = 4;
var circle = new List<Person>();
for (int i = 1; i <= n; i++)
{
circle.Add(new Person(i));
}
var index = 0;
while (circle.Count > 1)
{
index = (index + k - 1) % circle.Count;
var person = circle[index];
circle.RemoveAt(index);
Console.WriteLine("Removed {0}", person.Number);
}
Console.ReadLine();
}
Console.WriteLine("Winner is {0}", circle[0].Number);
Essentially the same as Ash's answer, but with a custom linked list:
using System;
using System.Linq;
namespace Circle
{
class Program
{
static void Main(string[] args)
{
Circle(20, 3);
}
static void Circle(int k, int n)
{
// circle is a linked list representing the circle.
// Each element contains the index of the next member
// of the circle.
int[] circle = Enumerable.Range(1, k).ToArray();
circle[k - 1] = 0; // Member 0 follows member k-1
int prev = -1; // Used for tracking the previous member so we can delete a member from the list
int curr = 0; // The member we're currently inspecting
for (int i = 0; i < k; i++) // There are k members to remove from the circle
{
// Skip over n members
for (int j = 0; j < n; j++)
{
prev = curr;
curr = circle[curr];
}
Console.WriteLine(curr);
circle[prev] = circle[curr]; // Delete the nth member
curr = prev; // Start counting again from the previous member
}
}
}
}
Here is a solution in Clojure:
(ns kthperson.core
(:use clojure.set))
(defn get-winner-and-losers [number-of-people hops]
(loop [people (range 1 (inc number-of-people))
losers []
last-scan-start-index (dec hops)]
(if (= 1 (count people))
{:winner (first people) :losers losers}
(let [people-to-filter (subvec (vec people) last-scan-start-index)
additional-losers (take-nth hops people-to-filter)
remaining-people (difference (set people)
(set additional-losers))
new-losers (concat losers additional-losers)
index-of-last-removed-person (* hops (count additional-losers))]
(recur remaining-people
new-losers
(mod last-scan-start-index (count people-to-filter)))))))
Explanation:
start a loop, with a collection of people 1..n
if there is only one person left, they are the winner and we return their ID, as well as the IDs of the losers (in order of them losing)
we calculate additional losers in each loop/recur by grabbing every N people in the remaining list of potential winners
a new, shorter list of potential winners is determined by removing the additional losers from the previously-calculated potential winners.
rinse & repeat (using modulus to determine where in the list of remaining people to start counting the next time round)
This is a variant of the Josephus problem.
General solutions are described here.
Solutions in Perl, Ruby, and Python are provided here. A simple solution in C using a circular doubly-linked list to represent the ring of people is provided below. None of these solutions identify each person's position as they are removed, however.
#include <stdio.h>
#include <stdlib.h>
/* remove every k-th soldier from a circle of n */
#define n 40
#define k 3
struct man {
int pos;
struct man *next;
struct man *prev;
};
int main(int argc, char *argv[])
{
/* initialize the circle of n soldiers */
struct man *head = (struct man *) malloc(sizeof(struct man));
struct man *curr;
int i;
curr = head;
for (i = 1; i < n; ++i) {
curr->pos = i;
curr->next = (struct man *) malloc(sizeof(struct man));
curr->next->prev = curr;
curr = curr->next;
}
curr->pos = n;
curr->next = head;
curr->next->prev = curr;
/* remove every k-th */
while (curr->next != curr) {
for (i = 0; i < k; ++i) {
curr = curr->next;
}
curr->prev->next = curr->next;
curr->next->prev = curr->prev;
}
/* announce last person standing */
printf("Last person standing: #%d.\n", curr->pos);
return 0;
}
Here's my answer in C#, as submitted. Feel free to criticize, laugh at, ridicule etc ;)
public static IEnumerable<int> Move(int n, int k)
{
// Use an Iterator block to 'yield return' one item at a time.
int children = n;
int childrenToSkip = k - 1;
LinkedList<int> linkedList = new LinkedList<int>();
// Set up the linked list with children IDs
for (int i = 0; i < children; i++)
{
linkedList.AddLast(i);
}
LinkedListNode<int> currentNode = linkedList.First;
while (true)
{
// Skip over children by traversing forward
for (int skipped = 0; skipped < childrenToSkip; skipped++)
{
currentNode = currentNode.Next;
if (currentNode == null) currentNode = linkedList.First;
}
// Store the next node of the node to be removed.
LinkedListNode<int> nextNode = currentNode.Next;
// Return ID of the removed child to caller
yield return currentNode.Value;
linkedList.Remove(currentNode);
// Start again from the next node
currentNode = nextNode;
if (currentNode== null) currentNode = linkedList.First;
// Only one node left, the winner
if (linkedList.Count == 1) break;
}
// Finally return the ID of the winner
yield return currentNode.Value;
}
I came across this problem in an interview website. The problem asks for efficiently implement three stacks in a single array, such that no stack overflows until there is no space left in the entire array space.
For implementing 2 stacks in an array, it's pretty obvious: 1st stack grows from LEFT to RIGHT, and 2nd stack grows from RIGHT to LEFT; and when the stackTopIndex crosses, it signals an overflow.
Thanks in advance for your insightful answer.
You can implement three stacks with a linked list:
You need a pointer pointing to the next free element. Three more pointers return the last element of each stack (or null, if the stack is empty).
When a stack gets another element added, it has to use the first free element and set the free pointer to the next free element (or an overflow error will be raised). Its own pointer has to point to the new element, from there back to the next element in the stack.
When a stack gets an element removed it will hand it back into the list of free elements. The own pointer of the stack will be redirected to the next element in the stack.
A linked list can be implemented within an array.
How (space) efficent is this?
It is no problem to build a linked list by using two cells of an array for each list element (value + pointer). Depending on the specification you could even get pointer and value into one array element (e.g. the array is long, value and pointer are only int).
Compare this to the solution of kgiannakakis ... where you lose up to 50% (only in the worst case). But I think that my solution is a bit cleaner (and maybe more academic, which should be no disadvantage for an interview question ^^).
See Knuth, The Art of Computer Programming, Volume 1, Section 2.2.2. titled "Sequential allocation". Discusses allocating multiple queues/stacks in a single array, with algorithms dealing with overflows, etc.
We can use long bit array representing to which stack the i-th array cell belongs to.
We can take values by modulo 3 (00 - empty, 01 - A, 10 - B, 11 - C). It would take N/2 bits or N/4 bytes of additional memory for N sized array.
For example for 1024 long int elements (4096 bytes) it would take only 256 bytes or 6%.
This bit array map can be placed in the same array at the beginning or at the end, just shrinking the size of the given array by constant 6%!
First stack grows from left to right.
Second stack grows from right to left.
Third stack starts from the middle. Suppose odd sized array for simplicity. Then third stack grows like this:
* * * * * * * * * * *
5 3 1 2 4
First and second stacks are allowed to grow maximum at the half size of array. The third stack can grow to fill in the whole array at a maximum.
Worst case scenario is when one of the first two arrays grows at 50% of the array. Then there is a 50% waste of the array. To optimise the efficiency the third array must be selected to be the one that grows quicker than the other two.
This is an interesting conundrum, and I don't have a real answer but thinking slightly outside the box...
it could depend on what each element in the stack consists of. If it's three stacks of true/false flags, then you could use the first three bits of integer elements. Ie. bit 0 is the value for the first stack, bit 1 is the value for the second stack, bit 2 is the value for the third stack. Then each stack can grow independently until the whole array is full for that stack. This is even better as the other stacks can also continue to grow even when the first stack is full.
I know this is cheating a bit and doesn't really answer the question but it does work for a very specific case and no entries in the stack are wasted. I am watching with interest to see whether anyone can come up with a proper answer that works for more generic elements.
Split array in any 3 parts (no matter if you'll split it sequentially or interleaved). If one stack grows greater than 1/3 of array you start filling ends of rest two stacks from the end.
aaa bbb ccc
1 2 3
145 2 3
145 2 6 3
145 2 6 3 7
145 286 3 7
145 286 397
The worse case is when two stacks grows up to 1/3 boundary and then you have 30% of space waste.
Assuming that all array positions should be used to store values - I guess it depends on your definition of efficiency.
If you do the two stack solution, place the third stack somewhere in the middle, and track both its bottom and top, then most operations will continue to be efficient, at a penalty of an expensive Move operation (of the third stack towards wherever free space remains, moving to the half way point of free space) whenever a collision occurs.
It's certainly going to be quick to code and understand. What are our efficiency targets?
A rather silly but effective solution could be:
Store the first stack elements at i*3 positions: 0,3,6,...
Store the second stack elements at i*3+1 positions: 1,4,7...
And third stack elements at i*3+2 positions.
The problem with this solution is that the used memory will be always three times the size of the deepest stack and that you can overflow even when there are available positions at the array.
Make a HashMap with keys to the begin and end positions e.g. < "B1" , 0 >, <"E1" , n/3 >
for each Push(value) add a condition to check if position of Bx is previous to Ex or there is some other "By" in between. -- lets call it condition (2)
with above condition in mind,
if above (2) is true // if B1 and E1 are in order
{ if ( S1.Push()), then E1 ++ ;
else // condition of overflow ,
{ start pushing at end of E2 or E3 (whichever has a space) and update E1 to be E2-- or E3-- ; }
}
if above (2) is false
{ if ( S1.Push()), then E1 -- ;
else // condition of overflow ,
{ start pushing at end of E2 or E3 (whichever has a space) and update E1 to be E2-- or E3-- ; }
}
Assume you only has integer index. if it's treated using FILO (First In Last Out) and not referencing individual, and only using an array as data. Using it's 6 space as stack reference should help:
[head-1, last-1, head-2, last-2, head-3, last-3, data, data, ... ,data]
you can simply using 4 space, because head-1 = 0 and last-3 = array length. If using FIFO (First In First Out) you need to re-indexing.
nb: I’m working on improving my English.
first stack grows at 3n,
second stack grows at 3n+1,
third grows at 3n+2
for n={0...N}
Yet another approach (as additional to linked-list) is to use map of stacks. In that case you'll have to use additional log(3^n)/log(2) bits for building map of data distribution in your n-length array. Each of 3-value part of map says which stack is owns next element.
Ex. a.push(1); b.push(2); c.push(3); a.push(4); a.push(5); will give you image
aacba
54321
appropriate value of map is calculated while elements is pushed onto stack (with shifting contents of array)
map0 = any
map1 = map0*3 + 0
map2 = map1*3 + 1
map3 = map2*3 + 2
map4 = map3*3 + 0
map5 = map4*3 + 0 = any*3^5 + 45
and length of stacks 3,1,1
Once you'll want to do c.pop() you'll have to reorganize your elements by finding actual position of c.top() in original array through walking in cell-map (i.e. divide by 3 while mod by 3 isn't 2) and then shift all contents in array back to cover that hole. While walking through cell-map you'll have to store all position you have passed (mapX) and after passing that one which points to stack "c" you'll have to divide by 3 yet another time and multiply it by 3^(amount positions passed-1) and add mapX to get new value of cells-map.
Overhead for that fixed and depends on size of stack element (bits_per_value):
(log(3n)/log(2)) / (nlog(bits_per_value)/log(2)) = log(3n) / (nlog(bits_per_value)) = log(3) / log(bits_per_value)
So for bits_per_value = 32 it will be 31.7% space overhead and with growing bits_per_value it will decay (i.e. for 64 bits it will be 26.4%).
In this approach, any stack can grow as long as there is any free space in the array.
We sequentially allocate space to the stacks and we link new blocks to the previous block. This means any new element in a stack keeps a pointer to the previous top element of that particular stack.
int stackSize = 300;
int indexUsed = 0;
int[] stackPointer = {-1,-1,-1};
StackNode[] buffer = new StackNode[stackSize * 3];
void push(int stackNum, int value) {
int lastIndex = stackPointer[stackNum];
stackPointer[stackNum] = indexUsed;
indexUsed++;
buffer[stackPointer[stackNum]]=new StackNode(lastIndex,value);
}
int pop(int stackNum) {
int value = buffer[stackPointer[stackNum]].value;
int lastIndex = stackPointer[stackNum];
stackPointer[stackNum] = buffer[stackPointer[stackNum]].previous;
buffer[lastIndex] = null;
indexUsed--;
return value;
}
int peek(int stack) { return buffer[stackPointer[stack]].value; }
boolean isEmpty(int stackNum) { return stackPointer[stackNum] == -1; }
class StackNode {
public int previous;
public int value;
public StackNode(int p, int v){
value = v;
previous = p;
}
}
This code implements 3 stacks in single array. It takes care of empty spaces and fills the empty spaces in between the data.
#include <stdio.h>
struct stacknode {
int value;
int prev;
};
struct stacknode stacklist[50];
int top[3] = {-1, -1, -1};
int freelist[50];
int stackindex=0;
int freeindex=-1;
void push(int stackno, int value) {
int index;
if(freeindex >= 0) {
index = freelist[freeindex];
freeindex--;
} else {
index = stackindex;
stackindex++;
}
stacklist[index].value = value;
if(top[stackno-1] != -1) {
stacklist[index].prev = top[stackno-1];
} else {
stacklist[index].prev = -1;
}
top[stackno-1] = index;
printf("%d is pushed in stack %d at %d\n", value, stackno, index);
}
int pop(int stackno) {
int index, value;
if(top[stackno-1] == -1) {
printf("No elements in the stack %d\n", value, stackno);
return -1;
}
index = top[stackno-1];
freeindex++;
freelist[freeindex] = index;
value = stacklist[index].value;
top[stackno-1] = stacklist[index].prev;
printf("%d is popped put from stack %d at %d\n", value, stackno, index);
return value;
}
int main() {
push(1,1);
push(1,2);
push(3,3);
push(2,4);
pop(3);
pop(3);
push(3,3);
push(2,3);
}
Another solution in PYTHON, please let me know if that works as what you think.
class Stack(object):
def __init__(self):
self.stack = list()
self.first_length = 0
self.second_length = 0
self.third_length = 0
self.first_pointer = 0
self.second_pointer = 1
def push(self, stack_num, item):
if stack_num == 1:
self.first_pointer += 1
self.second_pointer += 1
self.first_length += 1
self.stack.insert(0, item)
elif stack_num == 2:
self.second_length += 1
self.second_pointer += 1
self.stack.insert(self.first_pointer, item)
elif stack_num == 3:
self.third_length += 1
self.stack.insert(self.second_pointer - 1, item)
else:
raise Exception('Push failed, stack number %d is not allowd' % stack_num)
def pop(self, stack_num):
if stack_num == 1:
if self.first_length == 0:
raise Exception('No more element in first stack')
self.first_pointer -= 1
self.first_length -= 1
self.second_pointer -= 1
return self.stack.pop(0)
elif stack_num == 2:
if self.second_length == 0:
raise Exception('No more element in second stack')
self.second_length -= 1
self.second_pointer -= 1
return self.stack.pop(self.first_pointer)
elif stack_num == 3:
if self.third_length == 0:
raise Exception('No more element in third stack')
self.third_length -= 1
return self.stack.pop(self.second_pointer - 1)
def peek(self, stack_num):
if stack_num == 1:
return self.stack[0]
elif stack_num == 2:
return self.stack[self.first_pointer]
elif stack_num == 3:
return self.stack[self.second_pointer]
else:
raise Exception('Peek failed, stack number %d is not allowd' % stack_num)
def size(self):
return len(self.items)
s = Stack()
# push item into stack 1
s.push(1, '1st_stack_1')
s.push(1, '2nd_stack_1')
s.push(1, '3rd_stack_1')
#
## push item into stack 2
s.push(2, 'first_stack_2')
s.push(2, 'second_stack_2')
s.push(2, 'third_stack_2')
#
## push item into stack 3
s.push(3, 'FIRST_stack_3')
s.push(3, 'SECOND_stack_3')
s.push(3, 'THIRD_stack_3')
#
print 'Before pop out: '
for i, elm in enumerate(s.stack):
print '\t\t%d)' % i, elm
#
s.pop(1)
s.pop(1)
#s.pop(1)
s.pop(2)
s.pop(2)
#s.pop(2)
#s.pop(3)
s.pop(3)
s.pop(3)
#s.pop(3)
#
print 'After pop out: '
#
for i, elm in enumerate(s.stack):
print '\t\t%d)' % i, elm
May be this can help you a bit...i wrote it by myself
:)
// by ashakiran bhatter
// compile: g++ -std=c++11 test.cpp
// run : ./a.out
// sample output as below
// adding: 1 2 3 4 5 6 7 8 9
// array contents: 9 8 7 6 5 4 3 2 1
// popping now...
// array contents: 8 7 6 5 4 3 2 1
#include <iostream>
#include <cstdint>
#define MAX_LEN 9
#define LOWER 0
#define UPPER 1
#define FULL -1
#define NOT_SET -1
class CStack
{
private:
int8_t array[MAX_LEN];
int8_t stack1_range[2];
int8_t stack2_range[2];
int8_t stack3_range[2];
int8_t stack1_size;
int8_t stack2_size;
int8_t stack3_size;
int8_t stack1_cursize;
int8_t stack2_cursize;
int8_t stack3_cursize;
int8_t stack1_curpos;
int8_t stack2_curpos;
int8_t stack3_curpos;
public:
CStack();
~CStack();
void push(int8_t data);
void pop();
void print();
};
CStack::CStack()
{
stack1_range[LOWER] = 0;
stack1_range[UPPER] = MAX_LEN/3 - 1;
stack2_range[LOWER] = MAX_LEN/3;
stack2_range[UPPER] = (2 * (MAX_LEN/3)) - 1;
stack3_range[LOWER] = 2 * (MAX_LEN/3);
stack3_range[UPPER] = MAX_LEN - 1;
stack1_size = stack1_range[UPPER] - stack1_range[LOWER];
stack2_size = stack2_range[UPPER] - stack2_range[LOWER];
stack3_size = stack3_range[UPPER] - stack3_range[LOWER];
stack1_cursize = stack1_size;
stack2_cursize = stack2_size;
stack3_cursize = stack3_size;
stack1_curpos = stack1_cursize;
stack2_curpos = stack2_cursize;
stack3_curpos = stack3_cursize;
}
CStack::~CStack()
{
}
void CStack::push(int8_t data)
{
if(stack3_cursize != FULL)
{
array[stack3_range[LOWER] + stack3_curpos--] = data;
stack3_cursize--;
} else if(stack2_cursize != FULL) {
array[stack2_range[LOWER] + stack2_curpos--] = data;
stack2_cursize--;
} else if(stack1_cursize != FULL) {
array[stack1_range[LOWER] + stack1_curpos--] = data;
stack1_cursize--;
} else {
std::cout<<"\tstack is full...!"<<std::endl;
}
}
void CStack::pop()
{
std::cout<<"popping now..."<<std::endl;
if(stack1_cursize < stack1_size)
{
array[stack1_range[LOWER] + ++stack1_curpos] = 0;
stack1_cursize++;
} else if(stack2_cursize < stack2_size) {
array[stack2_range[LOWER] + ++stack2_curpos] = 0;
stack2_cursize++;
} else if(stack3_cursize < stack3_size) {
array[stack3_range[LOWER] + ++stack3_curpos] = 0;
stack3_cursize++;
} else {
std::cout<<"\tstack is empty...!"<<std::endl;
}
}
void CStack::print()
{
std::cout<<"array contents: ";
for(int8_t i = stack1_range[LOWER] + stack1_curpos + 1; i <= stack1_range[UPPER]; i++)
std::cout<<" "<<static_cast<int>(array[i]);
for(int8_t i = stack2_range[LOWER] + stack2_curpos + 1; i <= stack2_range[UPPER]; i++)
std::cout<<" "<<static_cast<int>(array[i]);
for(int8_t i = stack3_range[LOWER] + stack3_curpos + 1; i <= stack3_range[UPPER]; i++)
std::cout<<" "<<static_cast<int>(array[i]);
std::cout<<"\n";
}
int main()
{
CStack stack;
std::cout<<"adding: ";
for(uint8_t i = 1; i < 10; i++)
{
std::cout<<" "<<static_cast<int>(i);
stack.push(i);
}
std::cout<<"\n";
stack.print();
stack.pop();
stack.print();
return 0;
}
Perhaps you can implement N number of stacks or queues in the single array. My defination of using single array is that we are using single array to store all the data of all the stacks and queues in the single array, anyhow we can use other N array to keep track of indices of all elements of particular stack or queue.
solution :
store data sequentially to in the array during the time of insertion in any of the stack or queue. and store it's respective index to the index keeping array of that particular stack or queue.
for eg : you have 3 stacks (s1, s2, s3) and you want to implement this using a single array (dataArray[]). Hence we will make 3 other arrays (a1[], a2[], a3[]) for s1, s2 and s3 respectively which will keep track of all of their elements in dataArray[] by saving their respective index.
insert(s1, 10) at dataArray[0] a1[0] = 0;
insert(s2, 20) at dataArray[1] a2[0] = 1;
insert(s3, 30) at dataArray[2] a3[0] = 2;
insert(s1, 40) at dataArray[3] a1[1] = 3;
insert(s3, 50) at dataArray[4] a3[1] = 4;
insert(s3, 60) at dataArray[5] a3[2] = 5;
insert(s2, 30) at dataArray[6] a2[1] = 6;
and so on ...
now we will perform operation in dataArray[] by using a1, a2 and a3 for respective stacks and queues.
to pop an element from s1
return a1[0]
shift all elements to left
do similar approach for other operations too and you can implement any number of stacks and queues in the single array.