So I recently saw this puzzle posted by the British GCHQ:
It involves solving a 25x25 nonogram:
A nonogram is picture logic puzzles in which cells in a grid must be colored or left blank according to numbers at the side of the grid to reveal a hidden picture. In this puzzle type, the numbers are a form of discrete tomography that measures how many unbroken lines of filled-in squares there are in any given row or column. For example, a clue of "4 8 3" would mean there are sets of four, eight, and three filled squares, in that order, with at least one blank square between successive groups."
Naturally I had the inclination to try and write a program that would solve it for me. I was thinking of a recursive backtracking algorithm that starts at row 0, and for each possible arrangement of that row given the information from the row clue, it places a possible combination of the next row and verifies whether it is a valid placement given the column clues. If it is, it continues, if not it backtracks, until all the rows are placed in a valid configuration, or all possible row combinations have been exhausted.
I tested it on a few 5x5 puzzles and it works perfectly. The issue is that it takes too long to compute the 25x25 GCHQ puzzle. I need ways to make this algorithm more efficient - enough so that it can solve the puzzle linked above. Any ideas?
Here is my code for generating a set of the row possibilities for each row as well as the code for the solver (Note* it uses some non standard libraries but this shouldn't detract from the point):
// The Vector<int> input is a list of the row clues eg. for row 1, input = {7,3,1,1,7}. The
// int currentElemIndex keeps track of what block of the input clue we are dealing with i.e
// it starts at input[0] which is the 7 sized block and for all possible places it can be
// placed, places the next block from the clue recursively.
// The Vector<bool> rowState is the state of the row at the current time. True indicates a
// colored in square, false indicates empty.
// The Set< Vector<bool> >& result is just the set that stores all the possible valid row
// configurations.
// The int startIndex and endIndex are the bounds on the start point and end point between
// which the function will try to place the current block. The endIndex is calculated by
// subtracting the width of the board from the sum of the remaining block sizes + number
// of blocks remaining. Ie. if for row 1 with the input {7,3,1,1,7} we were placing the
// first block, the endIndex would be (3+1+1+7)+4=16 because if the first block was placed
// further than this, it would be impossible for the other blocks to fit.
// BOARD_WIDTH = 25;
// The containsPresets funtion makes sure that the row configuration is only added to the
// result set if it contains the preset values of the puzzle (the given squares
// at the start of the puzzle).
void Nonogram::rowPossibilitiesHelper(int currentElemIndex, Vector<bool>& rowState,
Vector<int>& input, Set< Vector<bool> >& result,
int startIndex, int rowIndex) {
if(currentElemIndex == input.size()) {
if(containsPresets(rowState, rowIndex)) {
result += rowState;
}
} else {
int endIndex = BOARD_WIDTH - rowSum(currentElemIndex+1, input);
int blockSize = input[currentElemIndex];
for(int i=startIndex; i<=endIndex-blockSize; i++) {
for(int j=0; j<blockSize; j++) {
rowState[i+j] = true; // set block
}
rowPossibilitiesHelper(currentElemIndex+1, rowState, input, result, i+blockSize+1, rowIndex); // explore
for(int j=0; j<blockSize; j++) {
rowState[i+j] = false; // unchoose
}
}
}
}
// The function is initally passed in 0 for the rowIndex. It gets a set of all possible
// valid arrangements of the board and for each one of them, sets the board row at rowIndex
// to the current rowConfig. Is then checks if the current configuration so far is valid in
// regards to the column clues. If it is, it solves the next row, if not, it unmarks the
// current configuration from the board row at rowIndex.
void Nonogram::solveHelper(int rowIndex) {
if(rowIndex == BOARD_HEIGHT) {
printBoard();
} else {
for(Vector<bool> rowConfig : rowPossisbilities(rowIndex)) {
setBoardRow(rowConfig, rowIndex);
if(isValidConfig(rowIndex)) { // set row
solveHelper(rowIndex+1); // explore
}
unsetBoardRow(rowIndex); // unset row
}
}
}
I've made a solution in Java, that for your example puzzle (25x25) solves it in about 50ms.
Full code and input examples: Github
Prerequisites
Notions of Java (understanding the samples)
Bitwise operations: I depend a lot on them, so read about it if you are not very familiar with it.
Graph traversal algorithms: DFS
Given:
R, C // number of rows, columns
int[][] rows; // for each row the block size from left to right (ex: rows[2][0] = first blocksize of 3 row)
int[][] cols; // for each column the block size from top to bottom
long[] grid; // bitwise representation of the board with the initially given painted blocks
Precalculate all permutations per row.
A permutation is also stored in a bitwise representation. Where the first bit is set to true if it fill the first column etc..
This is both time and space efficient. For the calculation we first count the number of extra spaces that can be added.
This is number_of_columns - sum_of_blocksize - (number_of_blocks-1)
Dfs over all possible permutations of placing extra spaces. See calcPerms and add them to the list of possible permutations if it's a match with the initially given painted blocks.
rowPerms = new long[R][];
for(int r=0;r<R;r++){
LinkedList<Long> res = new LinkedList<Long>();
int spaces = C - (rows[r].length-1);
for(int i=0;i<rows[r].length;i++){
spaces -= rows[r][i];
}
calcPerms(r, 0, spaces, 0, 0,res);
rowPerms[r] = new long[res.size()];
while(!res.isEmpty()){
rowPerms[r][res.size()-1]=res.pollLast();
}
}
...
// row, current block in row, extra spaces left to add, current permutation, current number of bits to shift
static void calcPerms(int r, int cur, int spaces, long perm, int shift, LinkedList<Long> res){
if(cur == rows[r].length){
if((grid[r]&perm)==grid[r]){
res.add(perm);
}
return;
}
while(spaces>=0){
calcPerms(r, cur+1, spaces, perm|(bits(rows[r][cur])<<shift), shift+rows[r][cur]+1,res);
shift++;
spaces--;
}
}
static long bits(int b){
return (1L<<b)-1; // 1 => 1, 2 => 11, 3 => 111, ...
}
Implement validations per row
Validating the rows:
[Trivial:] We are going to use precalculated permutations so we don't need any extra validation per row.
Validating the columns:
Herefor I'm keeping for each row and column the index of the current blocksize colIx, and the position in that size colVal.
This is calculated by the value and index of the previous row:
The value increased by 1 if the column is painted in the current row.
the value reset to 0 and index increased by 1 if the column was painted in the previous row and is not in the current row.
Sample:
static void updateCols(int row){
long ixc = 1L;
for(int c=0;c<C;c++,ixc<<=1){
// copy from previous
colVal[row][c]=row==0 ? 0 : colVal[row-1][c];
colIx[row][c]=row==0 ? 0 : colIx[row-1][c];
if((grid[row]&ixc)==0){
if(row > 0 && colVal[row-1][c] > 0){
// bit not set and col is not empty at previous row => close blocksize
colVal[row][c]=0;
colIx[row][c]++;
}
}else{
colVal[row][c]++; // increase value for set bit
}
}
}
Now we can use these index/values to determine which bits are expected to be false/true in the next row.
Used datastructure for validation:
static long[] mask; // per row bitmask, bit is set to true if the bit has to be validated by the val bitmask
static long[] val; // per row bitmask with bit set to false/true for as expected for the current row
When bit in previous row is set, we expect the bit in current row to be set to true if and only if the current size is still smaller than the expected size for the current index. Else it has to be 0 because you want to cut it off at the current row.
Or when the last blocksize is already used for the column we can not start a new block. Hence bit has to be 0.
static void rowMask(int row){
mask[row]=val[row]=0;
if(row==0){
return;
}
long ixc=1L;
for(int c=0;c<C;c++,ixc<<=1){
if(colVal[row-1][c] > 0){
// when column at previous row is set, we know for sure what has to be the next bit according to the current size and the expected size
mask[row] |= ixc;
if(cols[c][colIx[row-1][c]] > colVal[row-1][c]){
val[row] |= ixc; // must set
}
}else if(colVal[row-1][c] == 0 && colIx[row-1][c]==cols[c].length){
// can not add anymore since out of indices
mask[row] |= ixc;
}
}
}
Dfs all rows and check if still valid
This makes the actual dfs part as easy as your own.
If the rowmask fits with the current configuration we can update the column indices/values and traverse to the next row and eventually end up at row R.
static boolean dfs(int row){
if(row==R){
return true;
}
rowMask(row); // calculate mask to stay valid in the next row
for(int i=0;i<rowPerms[row].length;i++){
if((rowPerms[row][i]&mask[row])!=val[row]){
continue;
}
grid[row] = rowPerms[row][i];
updateCols(row);
if(dfs(row+1)){
return true;
}
}
return false;
}
Related
I am trying to fill a matrix with data read from a text file. The matrix is simply a 2D array that has already been created with parameters. I have check and the array is being created to the correct size based on parameters. For some reason when I scan the next value (value) and then try to insert it at a certain point in the array it is successful once and then quits without filling the rest of the array. If the insertAt method is removed than the loops work perfectly going through all the needed cells. For some reason when the method is added and called it works fine for the first cell but then quits after that.
While loop that inserts the value at a specific point.
while(scan.hasNext()) {
for(int i = 0; i < m1.row; i++) {
for(int j = 0; j < m1.column; j++) {
value = scan.nextInt();
m1.insertAt(i, j, value);
}
}
}
Method for matrix that control the insertion of the value.
public void insertAt(int row, int column, int value) {
if(row >= 0 && column >= 0) {
matrix[row][column] = value;
} else {
System.out.println("Error inserting value. Row: " + row + " Column: " + column);
}
}
Discovered it does this because you cannot put a variable in at a specific point at one time, even if you try to put them in in order. If the matrix is initialized and filled with 0's or some other digit it works fine. It has to be filled creating an empty matrix with a specific size does not work. Otherwise you have to restart the entire method each time to add 1 variable.
Suppose you need to count the number of islands on a matrix
{1, 1, 0, 0, 0},
{0, 1, 0, 0, 1},
{1, 0, 0, 1, 1},
{0, 0, 0, 0, 0},
{1, 0, 1, 0, 1}
We could simply use DFS or BFS when the input matrix size can be fitting into the memory.
However, what do we do if the input matrix is really large which could not be fitting into the memory?
I could chunk/split the input matrix into different small files and read them respectively.
But how to merge them?
I got stuck at how to merge them. I have the idea that when merging them we have to read some overlapped portion. But what is a concrete way to do so?
Trying to understand Matt's solution.
When I drew the below sample on the whiteboard and process it row by row.
Merge left then merge top and it seems won't work.
From Matt's solution.
not sure what are topidx, botidx meaning
int topidx = col * 2;
int botidx = topidx + 1;
Using union-find, the basic algorithm (without worrying about memory) is:
Create a set for every 1
Merge the sets for every pair of adjacent 1s. It doesn't matter what order you find them in, so reading order is usually fine.
Count the number of root sets -- there will be one for every island.
Easy, and with a little care, you can do this using sequential access to the matrix and only 2 rows worth of memory:
Initialize the island count to 0
Read the first row, create a set for each 1, and merge sets in adjacent columns.
For each additional row:
Read the row, create a set for each 1, and merge sets in adjacent columns;
Merge sets in the new row with adjacent sets in the previous row. ALWAYS POINT THE LINKS DOWNWARD, so that you never end up with a set in the new row linked to a parent in the old row.
Count the remaining root sets in the previous row, and add the number to your island count. These will never be able to merge with anything else.
Discard all the sets in the previous row -- you're never going to need them again, because you already counted them and nothing links to them.
Finally, count the root sets in the last row and add them to your island count.
The key to this, of course, is always pointing the links downward whenever you link sets in different rows. This will not hurt the complexity of the algorithm, and if you're using your own union-find, then it is easy to accomplish. If you're using a library data structure then you can use it just for each row, and keep track of the links between root sets in different rows yourself.
Since this is actually one of my favorite algorithms, here is an implementation in Java. This is not the most readable implementation since it involves some low-level tricks, but is super-efficient and short -- the kind of thing I'd write where performance is very important:
import java.util.Arrays;
public class Islands
{
private static final String[] matrix=new String[] {
" ############# ### ",
" # ##### ## ",
" # ## ## # # ",
" ### ## # # ",
" # ######### ## ## ",
" ## ## ",
" ########## ",
};
// find with path compression.
// If sets[s] < 0 then it is a link to ~sets[s]. Otherwise it is size of set
static int find(int[] sets, int s)
{
int parent = ~sets[s];
if (parent>=0)
{
int root = find(sets, parent);
if (root != parent)
{
sets[s] = ~root;
}
return root;
}
return s;
}
// union-by-size
// If sets[s] < 0 then it is a link to ~sets[s]. Otherwise it is size of set
static boolean union(int[] sets, int x, int y)
{
x = find(sets,x);
y = find(sets,y);
if (x!=y)
{
if ((sets[x] < sets[y]))
{
sets[y] += sets[x];
sets[x] = ~y;
}
else
{
sets[x] += sets[y];
sets[y] = ~x;
}
return true;
}
return false;
}
// Count islands in matrix
public static void main(String[] args)
{
// two rows of union-find sets.
// top row is at even indexes, bottom row is at odd indexes. This arrangemnt is chosen just
// to make resizing this array easier.
// For each value x:
// x==0 => no set. x>0 => root set of size x. x<0 => link to ~x
int cols=4;
int[] setrows= new int[cols*2];
int islandCount = 0;
for (String s : matrix)
{
System.out.println(s);
//Make sure our rows are big enough
if (s.length() > cols)
{
cols=s.length();
if (setrows.length < cols*2)
{
int newlen = Math.max(cols,setrows.length)*2;
setrows = Arrays.copyOf(setrows, newlen);
}
}
//Create sets for land in bottom row, merging left
for (int col=0; col<s.length(); ++col)
{
if (!Character.isWhitespace(s.charAt(col)))
{
int idx = col*2+1;
setrows[idx]=1; //set of size 1
if (idx>=2 && setrows[idx-2]!=0)
{
union(setrows, idx, idx-2);
}
}
}
//merge up
for (int col=0; col<cols; ++col)
{
int topidx = col*2;
int botidx = topidx+1;
if (setrows[topidx]!=0 && setrows[botidx]!=0)
{
int toproot=find(setrows,topidx);
if ((toproot&1)!=0)
{
//top set is already linked down
union(setrows, toproot, botidx);
}
else
{
//link top root down. It does not matter that we aren't counting its size, since
//we will shortly throw it aaway
setrows[toproot] = ~botidx;
}
}
}
//count root sets, discard top row, and move bottom row up while fixing links
for (int col=0; col<cols; ++col)
{
int topidx = col * 2;
int botidx = topidx + 1;
if (setrows[topidx]>0)
{
++islandCount;
}
int v = setrows[botidx];
setrows[topidx] = (v>=0 ? v : v|1); //fix up link if necessary
setrows[botidx] = 0;
}
}
//count remaining root sets in top row
for (int col=0; col<cols; ++col)
{
if (setrows[col*2]>0)
{
++islandCount;
}
}
System.out.println("\nThere are "+islandCount+" islands there");
}
}
I have the following question and I am unsure how to approach. I would like some help/hints with designing an efficient algorithm for the following requirements
Input
The first line of the input contains an integer N, which is length of the series.
This is followed by N lines, each of which contains a string containing only lower case characters.
1<=N<=100000.
Length of each string is between 1 and 10 (inclusive).
Output
Output the minimum length of the consecutive sub-series which contains all distinct strings.
Sample Input
6
letitbe
mihon
mihon
omi
omi
letitbe
Sample Output
18
Explanation: the last 4 consecutive strings contains all unique strings with the minimum length (smallest number of characters)
If I understood this correctly, you want the subseries which:
Contains at least 1 instance of "letitbe, "mihon" and "omi"
Has the lowest possible sum of string lengths
Here is how to do this efficiently, code in C#, algorithm explained in comments:
static void Main(string[] args)
{
// Input
var elements = new List<string> { "letitbe", "mihon", "mihon", "omi", "omi", "letitbe" };
// Find distinct elements
var distinctElements = elements.Distinct().ToList();
// Create a dictionary that tells us how many copies of each element we have in the current subseries, initialize all values to 0
var copiesOfElementInCurrentSubseries = distinctElements.ToDictionary(key => key, value => 0);
// The sum of lengths of strings in the current subseries
// Our goal is to minimize this
var lengthOfCurrentSubseries = 0;
// How many distinct elements are covered by the current subseries
// The condition under which we minimize lengthOfCurrentSubseries is that numberOfElementsCoveredByCurrentSubseries equals distinctElements
var numberOfElementsCoveredByCurrentSubseries = 0;
// We remember the solution in these
var bestStartIndex = 0;
var bestLength = elements.Sum(e => e.Length);
var bestNum = elements.Count;
// Start with startIndex and endIndex at 0, increase endIndex until we cover all distinct elements
// The subseries from startIndex to endIndex (inclusive) is our current subseries
for (int startIndex = 0, endIndex = 0; endIndex < elements.Count; endIndex++)
{
// We add the element at endIndex to our current subseries:
// If we found an element that previously wasn't covered, increase the count of covered elements
// Note that we never decrease this, because once we find a solution that covers all elements, we never make a change which "loses" some element
if (copiesOfElementInCurrentSubseries[elements[endIndex]] == 0)
{
numberOfElementsCoveredByCurrentSubseries++;
}
// Increase the number of copies of the element we added
copiesOfElementInCurrentSubseries[elements[endIndex]]++;
// Increase the total length of subseries by this element's length
lengthOfCurrentSubseries += elements[endIndex].Length;
// Initially, we will just loop increasing endIndex until all elements are covered
// Once we are covering all elements, try to improve the solution
if (numberOfElementsCoveredByCurrentSubseries == distinctElements.Count)
{
// Move startIndex to the right as far as possible while still covering all elements
while (copiesOfElementInCurrentSubseries[elements[startIndex]] > 1)
{
lengthOfCurrentSubseries -= elements[startIndex].Length;
copiesOfElementInCurrentSubseries[elements[startIndex]]--;
startIndex++;
}
// If the new solution is better, remember it
if (lengthOfCurrentSubseries < bestLength)
{
bestLength = lengthOfCurrentSubseries;
bestStartIndex = startIndex;
bestNum = endIndex - startIndex + 1;
}
}
// Now we add another element by moving endIndex one place to the right, then try improving the solution by moving startIndex to the right, and we repeat this process...
}
Console.WriteLine(string.Join(" ", elements.Skip(bestStartIndex).Take(bestNum)));
}
Note that even though this has nested loops, the inner while loop can have at most length of input steps total in all passes of the inner loop combined, as startIndex keeps its value and always moves to the right.
In case you are unfamiliar with C# - Dictionary is basically a hashtable - it can efficiently look up values based on keys (as long as the keys have a good hash function, which strings do).
It's a Google interview question. There's a list of "T" and "F" only. All denotes a position such that T means position is occupied by a flower pot and F means pot is not there, so you can put another pot at this position. Find the number of pots that can be placed in a given arrangement such that no two pots are adjacent to each other(they can be adjacent in the given arrangement). If a position at the beginning is unoccupied then a pot can be placed if second position is also unoccupied and if the last position is unoccupied than a pot can be placed if second last position is also unoccupied. For ex.
TFFFTFFTFFFFT - returns 2
FFTTFFFFFTTFF - returns 4
I tried solving it by looking at adjacent values for every position with value F. Increased the counter if both adjacent positions were F and set this position as T. I need a better solution or any other solution(if any).
Let's analyse what has to be done.
So first we probably need to visit and examine each place. That suggests loop of some sort. E.g.:
for (int i = 0; i < myPlaces.Length; ++i)
When we are at a spot we have to check if it's occupied
if (place[i] == 'F')
but that's not enough to place the flower pot there. We have to check if the next and previous place is free
place[i-1]
place[i+1]
If all tree contain F you can put the flower pot there and move to next field
Now, we also have some exceptions from the rule. Beginning and end of the list. So you have to deal with them separately. E.g
if (i == 0)
{
// only check current position and next position
}
if (i == myPlaces.Length - 1) // minus 1 because indexing usually starts from 0
{
// only check current position and previous position
}
After that you can perform the checks mentioned previously.
Now let's think of the input data. Generally, it's a good habit not to modify the input data but make a copy and work on the copy. Also some data structures work better than the others for different tasks. Here you can use simple string to keep entry values. But I would say an array of chars would be a better option because then, when you find a place where you can put a flower pot you can actually replace the F with the T in an array. Then when you move to new spot your data structers knows that there is already a pot in the previous position so your algorithm won't put an adjacent one.
You would not be able to do that with string as strings are immutable and you would need to generate a new string each time.
Note that it's only a naive algorithm with a lot of scope for improvement and optimization. But my goal was rather to give some idea how to approach this kind of problems in general. I'll leave implementing of the details to you as an afternoon exercise before targeting a job at Google.
You may be able to do this with a modified Mergesort. Consider the flowerpots that can be placed in the singletons, then the flowerpots that can be placed in the doubleton merges of those singletons, up the tree to the full arrangement. It would complete in O(n lg n) for a list of n flowerpots.
There is certainly a way to do this with a modified Rod Cutting algorithm with complexity O(n^2). The subproblem is whether or not an open "false set" exists in the substring being considered. The "closed false sets" already have some maximum value computed for them. So, when a new character is added, it either increases the amount of flowerpots that can be inserted, or "locks in" the maximum quantity of available flowerpots for the substring.
Also, you know that the maximum flowerpots that can be placed in a set of n open positions bound by closed positions is n - 2 (else n-1 if only bracketed on one side, i.e. the string begins or ends with a "false set". The base condition (the first position is open, or the first position is closed) can calculated upon reaching the second flowerpot.
So, we can build up to the total number of flowerpots that can be inserted into the whole arrangement in terms of the maximum number of flowerpots that can be inserted into smaller subarrangements that have been previously calculated. By storing our previous calculations in an array, we reduce the amount of time necessary to calculate the maximum for the next subarrangement to a single array lookup and some constant-time calculations. This is the essence of dynamic programming!
EDIT: I updated the answer to provide a description of the Dynamic Programming approach. Please consider working through the interactive textbook I mentioned in the comments! http://interactivepython.org/runestone/static/pythonds/index.html
I would approach the problem like this. You need FFF to have one more pot, FFFFF for two pots, etc. To handle the end cases, add an F at each end.
Because this is very similar to a 16-bit integer, the algorithm should use tricks like binary arithmetic operations.
Here is an implementation in Python that uses bit masking (value & 1), bit shifting (value >>= 1) and math ((zeros - 1) / 2) to count empty slots and calculate how many flower pots could fit.
#value = 0b1000100100001
value = 0b0011000001100
width = 13
print bin(value)
pots = 0 # number of flower pots possible
zeros = 1 # number of zero bits in a row, start with one leading zero
for i in range(width):
if value & 1: # bit is one, count the number of zeros
if zeros > 0:
pots += (zeros - 1) / 2
zeros = 0
else: # bit is zero, increment the number found
zeros += 1
value >>= 1 # shift the bits to the right
zeros += 1 # add one trailing zero
pots += (zeros - 1) / 2
print pots, "flower pots"
The solution is really simple, check the previous and current value of the position and mark the position as plantable (or puttable) and increment the count. Read the next value, if it is already is planted, (backtrack and) change the previous value and decrement the count. The complexity is O(n). What we really want to check is the occurrence of 1001. Following is the implementation of the algorithm in Java.
public boolean canPlaceFlowers(List<Boolean> flowerbed, int numberToPlace) {
Boolean previous = false;
boolean puttable = false;
boolean prevChanged = false;
int planted = 0;
for (Boolean current : flowerbed) {
if (previous == false && current == false) {
puttable = true;
}
if (prevChanged == true && current == true) {
planted--;
}
if (puttable) {
previous = true;
prevChanged = true;
planted++;
puttable = false;
} else {
previous = current;
prevChanged = false;
}
}
if (planted >= numberToPlace) {
return true;
}
return false;
}
private static void canPlaceOneFlower(List<Boolean> flowerbed, FlowerBed fb) {
boolean result;
result = fb.canPlaceFlowers(flowerbed, 1);
System.out.println("Can place 1 flower");
if (result) {
System.out.println("-->Yes");
} else {
System.out.println("-->No");
}
}
private static void canPlaceTwoFlowers(List<Boolean> flowerbed, FlowerBed fb) {
boolean result;
result = fb.canPlaceFlowers(flowerbed, 2);
System.out.println("Can place 2 flowers");
if (result) {
System.out.println("-->Yes");
} else {
System.out.println("-->No");
}
}
private static void canPlaceThreeFlowers(List<Boolean> flowerbed, FlowerBed fb) {
boolean result;
result = fb.canPlaceFlowers(flowerbed, 3);
System.out.println("Can place 3 flowers");
if (result) {
System.out.println("-->Yes");
} else {
System.out.println("-->No");
}
}
private static void canPlaceFourFlowers(List<Boolean> flowerbed, FlowerBed fb) {
boolean result;
result = fb.canPlaceFlowers(flowerbed, 4);
System.out.println("Can place 4 flowers");
if (result) {
System.out.println("-->Yes");
} else {
System.out.println("-->No");
}
}
public static void main(String[] args) {
List<Boolean> flowerbed = makeBed(new int[] { 0, 0, 0, 0, 0, 0, 0 });
FlowerBed fb = new FlowerBed();
canPlaceFourFlowers(flowerbed, fb);
canPlaceThreeFlowers(flowerbed, fb);
flowerbed = makeBed(new int[] { 0, 0, 0, 1, 0, 0, 0 });
canPlaceFourFlowers(flowerbed, fb);
canPlaceThreeFlowers(flowerbed, fb);
canPlaceTwoFlowers(flowerbed, fb);
flowerbed = makeBed(new int[] { 1, 0, 0, 1, 0, 0, 0, 1 });
canPlaceFourFlowers(flowerbed, fb);
canPlaceThreeFlowers(flowerbed, fb);
canPlaceTwoFlowers(flowerbed, fb);
canPlaceOneFlower(flowerbed, fb);
}
My solution using dynamic programming.
ar is array in the form of ['F','T','F'].
import numpy as np
def pot(ar):
s = len(ar)
rt = np.zeros((s,s))
for k in range(0,s):
for i in range(s-k):
for j in range(i,i+k+1):
left = 0
right = 0
if ar[j] != 'F':
continue
if j-1 >= i and ar[j-1] == 'T':
continue
else:
left = 0
if j+1 <= i+k and ar[j+1] == 'T':
continue
else:
right = 0
if j-2 >= i:
left = rt[i][j-2]
if j+2 <= i+k:
right = rt[j+2][i+k]
rt[i][i+k] = max(rt[i][i+k], left+right+1)
return rt[0][len(ar)-1]
My solution written in C#
private static int CheckAvailableSlots(string str)
{
int counter = 0;
char[] chrs = str.ToCharArray();
if (chrs.FirstOrDefault().Equals('F'))
if (chrs.Length == 1)
counter++;
else if (chrs.Skip(1).FirstOrDefault().Equals('F'))
counter++;
if (chrs.LastOrDefault().Equals('F') && chrs.Reverse().Skip(1).FirstOrDefault().Equals('F'))
counter++;
for (int i = 1; i < chrs.Length - 2; i++)
{
if (chrs[i - 1].Equals('T'))
continue;
else if (chrs[i].Equals('F') && chrs[i + 1].Equals('F'))
{
chrs[i] = 'T';
counter++;
i++;
}
else
i++;
}
return counter;
}
// 1='T'
// 0='F'
int[] flowerbed = new int[] {1,0,0,0,0,1};
public boolean canPlaceFlowers(int[] flowerbed, int n) {
int tg = 0;
for (int i = 0, g = 1; i < flowerbed.length && tg < n; i++) {
g += flowerbed[i] == 0 ? flowerbed.length - 1 == i ? 2 : 1 : 0;
if (flowerbed[i] == 1 || i == flowerbed.length - 1) {
tg += g / 2 - (g % 2 == 0 ? 1 : 0);
g = 0;
}
}
return tg >= n;
}
Most of these answers (unless they alter the array or traverse and a copy) dont consider the situation where the first 3 (or last 3) pots are empty. These solutions will incorrectly determine that FFFT will contain 2 spaces, rather than just one. We therefore need to start at the third element (rather than then second) and end at index length - 3 (rather than length - 2). Also, while looping through the array, if an eligible index is found, the index just be incremented by 2, otherwise TTFFFFT would give 2 available plots instead of one. This is true unless you alter the array while looping or use a copy of the array and alter it.
Edit: this holds true unless the question is how many spaces are available for planting, rather than how many total plants can be added
There are ones, zeroes and ‘U’s in a particular order. (E.g. “1001UU0011”) The number of ones and zeroes are the same, and there’s always two ‘U’s next to each other. You can swap the pair of ‘U’s with any pair of adjacent digits. Here’s a sample move:
__
/ \
1100UU0011 --> 11001100UU
The task is to put all the zeroes before the ones.
Here's a sample solution:
First step:
__
/ \
1100UU0011
Second step:
____
/ \
UU00110011
000011UU11 --> DONE
It’s pretty easy to create a brute-force algorithm. But with that it takes hundreds or even thousands of moves to solve a simple one like my example. So I’m looking for something more “clever” algorithm.
It's not homework; it was a task in a competition. The contest is over but I can’t find the solution for this.
Edit: The task here is the create an algorithm that can sort those 0s and 1s - not just output N 0s and N 1s and 2 Us. You have to show the steps somehow, like in my example.
Edit 2: The task didn't ask for the result with the least moves or anything like that. But personally I would love the see an algorithm that provides that :)
I think this should work:
Iterate once to find the position of
the U's. If they don't occupy the last
two spots, move them there by
swapping with the last two.
Create a
variable to track the currently
sorted elements, initially set to
array.length - 1, meaning anything
after it is sorted.
Iterate
backwards. Every time you encounter a
1:
swap the the one and its element before it with the U's.
swap the U's back to the the currently sorted elements tracker -1, update variable
Continue until the beginning of the array.
This is quite an interesting problem - so let's try to solve it. I will start with an precise analysis of the problem and see what one can find out. I will add piece by piece to this answer over the next days. Any help is welcome.
A problem of size n is a problem with exactly exactly n zeros, n ones, and two Us, hence it consists of 2n+2 symbols.
There are
(2n)!
-----
(n!)²
different sequences of exactly n zeros and nones. Then there are 2n+1 possible positions to insert the two Us, hence there are
(2n)! (2n+1)!
-----(2n+1) = -------
(n!)² (n!)²
problem instances of size n.
Next I am looking for a way to assign a score to each problem instance and how this score changes under all possible moves hoping to find out what the minimal number of required moves is.
Instance of size one are either already sorted
--01 0--1 01--
(I think I will use hyphens instead of Us because they are easier to recognize) or cannot be sorted.
--10 ==only valid move==> 10--
-10- no valid move
10-- ==only valid move==> --10
In consequence I will assume n >= 2.
I am thinking about the inverse problem - what unordered sequences can be reached starting from an ordered sequence. The ordered sequences are determined up to the location of the both hyphens - so the next question is if it is possible to reach every ordered sequence from every other order sequence. Because a sequence of moves can be performed forward and backward it is sufficient to show that one specific ordered sequence is reachable from all other. I choose (0|n)(1|n)--. ((0|x) represents exactly x zeros. If x is not of the form n-m zero or more is assumed. There may be additional constraints like a+b+2=n not explicitly stated. ^^ indicates the swap position. The 0/1 border is obviously between the last zero and first one.)
// n >= 2, at least two zeros between -- and the 0/1 border
(0|a)--(0|b)00(1|n) => (0|n)--(1|n-2)11 => (0|n)(1|n)--
^^ ^^
// n >= 3, one zero between -- and 0/1 boarder
(0|n-1)--01(1|n-1) => (0|n)1--(1|n-3)11 => (0|n)(1|n)--
^^ ^^
// n >= 2, -- after last zero but at least two ones after --
(0|n)(1|a)--(1|b)11 => (0|n)(1|n)--
^^
// n >= 3, exactly one one after --
(0|n)(1|n-3)11--1 => (0|n)(1|n-3)--111 => (0|n)(1|n)--
^^ ^^
// n >= 0, nothing to move
(0|n)(1|n)--
For the remaining two problems of size two - 0--011 and 001--1 - it seems not to be possible to reach 0011--. So for n >= 3 it is possible to reach every ordered sequence from every other ordered sequence in at most four moves (Probably less in all cases because I think it would have been better to choose (0|n)--(1|n) but I leave this for tomorrow.). The preliminary goal is to find out at what rate and under what conditions one can create (and in consequence remove) 010 and 101 because they seem to be the hard cases as already mentioned by others.
If you use a WIDTH-first brute force, it's still brute force, but at least you are guaranteed to come up with the shortest sequence of moves, if there is an answer at all. Here's a quick Python solution using a width-first search.
from time import time
def generate(c):
sep = "UU"
c1, c2 = c.split(sep)
for a in range(len(c1)-1):
yield c1[0:a]+sep+c1[(a+2):]+c1[a:(a+2)]+c2
for a in range(len(c2)-1):
yield c1+c2[a:(a+2)]+c2[0:a]+sep+c2[(a+2):]
def solve(moves,used):
solved = [cl for cl in moves if cl[-1].rindex('0') < cl[-1].index('1')]
if len(solved) > 0: return solved[0]
return solve([cl+[d] for cl in moves for d in generate(cl[-1]) if d not in used and not used.add(d)],used)
code = raw_input('enter the code:')
a = time()
print solve([[code]],set())
print "elapsed time:",(time()-a),"seconds"
Well, the first thing that gets up to my mind is top-down dynamic programming approach. It's kind of easy to understand but could eat a lot of memory. While I'm trying to apply a bottom-up approach you can try this one:
Idea is simple - cache all of the results for the brute-force search. It will become something like this:
function findBestStep(currentArray, cache) {
if (!cache.contains(currentArray)) {
for (all possible moves) {
find best move recursively
}
cache.set(currentArray, bestMove);
}
return cache.get(currentArray);
}
This method complexity would be... O(2^n) which is creepy. However I see no logical way it can be smaller as any move is allowed.
If if find a way to apply bottom-up algorithm it could be a little faster (it does not need a cache) but it will still have O(2^n) complexity.
Added:
Ok, I've implemented this thing in Java. Code is long, as it always is in Java, so don't get scared of it's size. The main algorithm is pretty simple and can be found at the bottom. I don't think there can be any way faster than this (this is more of a mathematical question if it can be faster). It eats tonns of memory but still computes it all pretty fast.
This 0,1,0,1,0,1,0,1,0,1,0,1,0,1,2,2 computes in 1 second, eating ~60mb memory resulting in 7 step sorting.
public class Main {
public static final int UU_CODE = 2;
public static void main(String[] args) {
new Main();
}
private static class NumberSet {
private final int uuPosition;
private final int[] numberSet;
private final NumberSet parent;
public NumberSet(int[] numberSet) {
this(numberSet, null, findUUPosition(numberSet));
}
public NumberSet(int[] numberSet, NumberSet parent, int uuPosition) {
this.numberSet = numberSet;
this.parent = parent;
this.uuPosition = uuPosition;
}
public static int findUUPosition(int[] numberSet) {
for (int i=0;i<numberSet.length;i++) {
if (numberSet[i] == UU_CODE) {
return i;
}
}
return -1;
}
protected NumberSet getNextNumberSet(int uuMovePos) {
final int[] nextNumberSet = new int[numberSet.length];
System.arraycopy(numberSet, 0, nextNumberSet, 0, numberSet.length);
System.arraycopy(this.getNumberSet(), uuMovePos, nextNumberSet, uuPosition, 2);
System.arraycopy(this.getNumberSet(), uuPosition, nextNumberSet, uuMovePos, 2);
return new NumberSet(nextNumberSet, this, uuMovePos);
}
public Collection<NumberSet> getNextPositionalSteps() {
final Collection<NumberSet> result = new LinkedList<NumberSet>();
for (int i=0;i<=numberSet.length;i++) {
final int[] nextNumberSet = new int[numberSet.length+2];
System.arraycopy(numberSet, 0, nextNumberSet, 0, i);
Arrays.fill(nextNumberSet, i, i+2, UU_CODE);
System.arraycopy(numberSet, i, nextNumberSet, i+2, numberSet.length-i);
result.add(new NumberSet(nextNumberSet, this, i));
}
return result;
}
public Collection<NumberSet> getNextSteps() {
final Collection<NumberSet> result = new LinkedList<NumberSet>();
for (int i=0;i<=uuPosition-2;i++) {
result.add(getNextNumberSet(i));
}
for (int i=uuPosition+2;i<numberSet.length-1;i++) {
result.add(getNextNumberSet(i));
}
return result;
}
public boolean isFinished() {
boolean ones = false;
for (int i=0;i<numberSet.length;i++) {
if (numberSet[i] == 1)
ones = true;
else if (numberSet[i] == 0 && ones)
return false;
}
return true;
}
#Override
public boolean equals(Object obj) {
if (obj == null) {
return false;
}
if (getClass() != obj.getClass()) {
return false;
}
final NumberSet other = (NumberSet) obj;
if (!Arrays.equals(this.numberSet, other.numberSet)) {
return false;
}
return true;
}
#Override
public int hashCode() {
int hash = 7;
hash = 83 * hash + Arrays.hashCode(this.numberSet);
return hash;
}
public int[] getNumberSet() {
return this.numberSet;
}
public NumberSet getParent() {
return parent;
}
public int getUUPosition() {
return uuPosition;
}
}
void precacheNumberMap(Map<NumberSet, NumberSet> setMap, int length, NumberSet endSet) {
int[] startArray = new int[length*2];
for (int i=0;i<length;i++) startArray[i]=0;
for (int i=length;i<length*2;i++) startArray[i]=1;
NumberSet currentSet = new NumberSet(startArray);
Collection<NumberSet> nextSteps = currentSet.getNextPositionalSteps();
List<NumberSet> nextNextSteps = new LinkedList<NumberSet>();
int depth = 1;
while (nextSteps.size() > 0) {
for (NumberSet nextSet : nextSteps) {
if (!setMap.containsKey(nextSet)) {
setMap.put(nextSet, nextSet);
nextNextSteps.addAll(nextSet.getNextSteps());
if (nextSet.equals(endSet)) {
return;
}
}
}
nextSteps = nextNextSteps;
nextNextSteps = new LinkedList<NumberSet>();
depth++;
}
}
public Main() {
final Map<NumberSet, NumberSet> cache = new HashMap<NumberSet, NumberSet>();
final NumberSet startSet = new NumberSet(new int[] {0,1,0,1,0,1,0,1,0,1,0,1,0,1,2,2});
precacheNumberMap(cache, (startSet.getNumberSet().length-2)/2, startSet);
if (cache.containsKey(startSet) == false) {
System.out.println("No solutions");
} else {
NumberSet cachedSet = cache.get(startSet).getParent();
while (cachedSet != null && cachedSet.parent != null) {
System.out.println(cachedSet.getUUPosition());
cachedSet = cachedSet.getParent();
}
}
}
}
Here's a try:
Start:
let c1 = the total number of 1s
let c0 = the total number of 0s
if the UU is at the right end of the string, goto StartFromLeft
StartFromRight
starting at the right end of the string, move left, counting 1s,
until you reach a 0 or the UU.
If you've reached the UU, goto StartFromLeft.
If the count of 1s equals c1, you are done.
Else, swap UU with the 0 and its left neighbor if possible.
If not, goto StartFromLeft.
StartFromLeft
starting at the left end of the string, move right, counting 0s,
until you reach a 1 or the UU.
If you've reached the UU, goto StartFromRight.
If the count of 0s equals c0, you are done.
Else, swap UU with the 1 and its right neighbor, if possible.
If not, goto StartFromRight
Then goto StartFromRight.
So, for the original 1100UU0011:
1100UU0011 - original
110000UU11 - start from right, swap UU with 00
UU00001111 - start from left, swap UU with 11
For the trickier 0101UU01
0101UU01 - original
0UU11001 - start from right, can't swap UU with U0, so start from left and swap UU with 10
00011UU1 - start from right, swap UU with 00
However, this won't solve something like 01UU0...but that could be fixed by a flag - if you've gone through the whole algorithm once, made no swaps and it isn't solved...do something.
About the question... It never asked for the optimal solution and these types of questions do not want that. You need to write a general purpose algorithm to handle this problem and a brute-force search to find the best solution is not feasible for strings that may be megabytes in length. Also I noticed late that there are guaranteed to be the same number of 0s and 1s, but I think it's more interesting to work with the general case where there may be different numbers of 0s and 1s. There actually isn't guaranteed to be a solution in every case if the length of the input string is less than 7, even in the case where you have 2 0s and 2 1s.
Size 3: Only one digit so it is sorted by definition (UU0 UU1 0UU 1UU)
Size 4: No way to alter the order. There are no moves if UU is in the middle, and only swap with both digits if it is at an end (1UU0 no moves, UU10->10UU->UU10, etc)
Size 5: UU in the middle can only move to the far end and not change the order of the 0s and 1s (1UU10->110UU). UU at an end can move to middle and not change order, but only move back to the same end so there is no use for it (UU110->11UU0->UU110). The only way to change digits is if the UU is at an end and to swap with the opposite end. (UUABC->BCAUU or ABCUU->UUCAB). This means that if UU is at positions 0 or 2 it can solve if 0 is in the middle (UU101->011UU or UU100->001UU) and if UU is at positions 1 or 3 it can solve if 1 is in the middle (010UU->UU001 or 110UU->UU011). Anything else is already solved or is unsolvable. If we need to handle this case, I would say hard-code it. If sorted, return result (no moves). If UU is in the middle somewhere, move it to the end. Swap from the end to the other end and that is the only possible swap whether it is now sorted or not.
Size 6: Now we get so a position where we can have a string specified according to the rules where we can make moves but where there can be no solution. This is the problem point with any algorithm, because I would think a condition of any solution should be that it will let you know if it cannot be solved. For instance 0010, 0100, 1000, 1011, 1100, 1101, and 1110 can be solved no matter where the UU is placed and the worst cases take 4 moves to solve. 0101 and 1010 can only be solved if UU is in an odd position. 0110 and 1001 can only be solved if UU is in an even position (either end or middle).
I think the best way will be something like the following, but I haven't written it yet. First, make sure you place a '1' at the end of the list. If the end is currently 0, move UU to the end then move it to the last '1' position - 1. After that you continually move UU to the first '1', then to the first '0' after the new UU. This will move all the 0s to the start of the list. I've seen a similar answer the other way, but it didn't take into account the final character on either end. This can run into issues with small values still (i.e. 001UU01, cannot move to first 1, move to end 00101UU lets us move to start but leaves 0 at end 00UU110).
My guess is that you can hard-code special cases like that. I'm thinking there may be a better algorithm though. For instance you could use the first two characters as a 'temporary swap variable. You would put UU there and then do combinations of operations on others to leave UY back at the start. For instance, UUABCDE can swap AB with CD or DE or BC WITH DE (BCAUUDE->BCADEUU->UUADEBC).
Another possible thing would be to treat the characters as two blocks of two base-3 bits
0101UU0101 will show up as 11C11 or 3593. Maybe also something like a combination of hard-coded swaps. For instance if you ever see 11UU, move UU left 2. If you ever see UU00, move UU right two. If you see UU100, or UU101, move UU right 2 to get 001UU or 011UU.
Maybe another possibility would be some algorithm to move 0s left of center and 1s right of center (if it is given that there are the same number of 0s and 1s.
Maybe it would be better to work on an a structure that contained only 0s and 1s with a position for UU.
Maybe look at the resulting condition better, allowing for UU to be anywhere in the string, these conditions MUST be satisfied:
No 0s after Length/2
No 1s before (Length/2-1)
Maybe there are more general rules, like it's really good to swap UU with 10 in this case '10111UU0' because a '0' is after UU now and that would let you move the new 00 back to where the 10 was (10111UU0->UU111100->001111UU).
Anyway, here's the brute force code in C#. The input is a string and an empty Dictionary. It fills the dictionary with every possible resulting string as the keys and the list of shortest steps to get there as the value:
Call:
m_Steps = new Dictionary<string, List<string>>();
DoSort("UU1010011101", new List<string>);
It includes DoTests() which calls DoSort for every possible string with the given number of digits (not including UU):
Dictionary<string, List<string>> m_Steps = new Dictionary<string, List<string>>();
public void DoStep(string state, List<string> moves) {
if (m_Steps.ContainsKey(state) && m_Steps[state].Count <= moves.Count + 1) // have better already
return;
// we have a better (or new) solution to get to this state, so set it to the moves we used to get here
List<string> newMoves = new List<string>(moves);
newMoves.Add(state);
m_Steps[state] = newMoves;
// if the state is a valid solution, stop here
if (state.IndexOf('1') > state.LastIndexOf('0'))
return;
// try all moves
int upos = state.IndexOf('U');
for (int i = 0; i < state.Length - 1; i++) {
// need to be at least 2 before or 2 after the UU position (00UU11 upos is 2, so can only move to 0 or 4)
if (i > upos - 2 && i < upos + 2)
continue;
char[] chars = state.ToCharArray();
chars[upos] = chars[i];
chars[upos + 1] = chars[i + 1];
chars[i] = chars[i + 1] = 'U';
DoStep(new String(chars), newMoves);
}
}
public void DoTests(int digits) { // try all combinations
char[] chars = new char[digits + 2];
for (int value = 0; value < (2 << digits); value++) {
for (int uupos = 0; uupos < chars.Length - 1; uupos++) {
for (int i = 0; i < chars.Length; i++) {
if (i < uupos)
chars[i] = ((value >> i) & 0x01) > 0 ? '1' : '0';
else if (i > uupos + 1)
chars[i] = ((value >> (i - 2)) & 0x01) > 0 ? '1' : '0';
else
chars[i] = 'U';
}
m_Steps = new Dictionary<string, List<string>>();
DoSort(new string(chars), new List<string>);
foreach (string key in m_Steps.AllKeys))
if (key.IndexOf('1') > key.LastIndexOf('0')) { // winner
foreach (string step in m_Steps[key])
Console.Write("{0}\t", step);
Console.WriteLine();
}
}
}
}
Counting sort.
If A is the number of 0s, A is also the number of 1s, and U is the number of Us:
for(int i=0; i<A; i++)
data[i] = '0';
for(int i=0; i<A; i++)
data[A+i] = '1';
for(int i=0; i<U; i++)
data[A+A+i] = 'U';
There are only 2 Us?
Why not just count the number of 0s and store the position of the us:
numberOfZeros = 0
uPosition = []
for i, value in enumerate(sample):
if value = 0:
numberOfZeros += 1
if value = U
uPosition.append(i)
result = []
for i in range(len(sample)):
if i = uPosition[0]
result.append('U')
uPosition.pop(0)
continue
if numberOfZeros > 0:
result.append('0')
numberOfZeros -= 1
continue
result.append('1')
Would result in a runtime of O(n)
Or even better:
result = []
numberOfZeros = (len(sample)-2)/2
for i, value in enumerate(sample):
if value = U
result.append('U')
continue
if numberOfZeros > 0:
result.append(0)
numberOfZeros -= 1
continue
result.append(1)