Related
The method is given NxN matrix always powers of 2 and a number,it will return true if the num is found example for 4x4 size:
this is what i wrote:
public class Search {
public static boolean Search (int [][] matrix, int num)
{
int value = matrix.length / 2;
int first_quarter_pivot = matrix[value-1][0]; // represents highest number in first quarter
int second_quarter_pivot = matrix[value-1][value]; // represents highest number in second quarter
int third_quarter_pivot = matrix[matrix.length-1][value]; // represents highest number in third quarter
int fourth_quarter_pivot = matrix[matrix.length-1][0]; // represents highest number in fourth quarter
boolean isBoolean = false;
int i=0;
int j;
// if the num is not in the range of biggest smallest number it means he can`t be there.
if(!(num >= first_quarter_pivot) && (num <= fourth_quarter_pivot)) {
return false;
}
// if num is one of the pivots return true;
if((num == first_quarter_pivot || (num ==second_quarter_pivot))
|| (num == third_quarter_pivot) || (num == fourth_quarter_pivot ))
return true;
// if num is smaller than first pivot it means num is the first quarter,we limit the search to first quarter.
// if not smaller move to the next quarter pivot
if(num < first_quarter_pivot){{
j =0;
do
if(matrix[i][j] == num) {
isBoolean = true;
break;
}
else if((j == value)) {
j = 0;
i++;
}
else if(matrix[i][j] != num){
j++;
}
while(isBoolean != true) ;
}
return isBoolean;
}
// if num is smaller than second pivot it means num is the second quarter,we limit the search to second quarter.
// if not smaller move to the next quarter pivot
if(num < second_quarter_pivot){{
j = value;// start (0,value) j++ till j=value
do
if(matrix[i][j] == num) {
isBoolean = true;
break;
}
else if((j == matrix.length-1)) {
j = value;
i++;
}
else if(matrix[i][j] != num){
j++;
}
while(isBoolean != true) ;
}
return isBoolean;
}
// if num is smaller than third pivot it means num is the third quarter,we limit the search to third quarter.
// if not smaller move to the next quarter pivot
if(num < third_quarter_pivot){{
i = value;
j = value;// start (0,value) j++ till j=value
do
if(matrix[i][j] == num) {
isBoolean = true;
break;
}
else if((j == matrix.length-1)) {
j = value;
i++;
}
else if(matrix[i][j] != num){
j++;
}
while(isBoolean != true) ;
}
return isBoolean;
}
// if num is smaller than fourth pivot it means num is the fourth quarter,we limit the search to fourth quarter.
// number must be here because we verfied his existence in the start.
if(num < fourth_quarter_pivot){
i = value;
j = 0;// start (0,value) j++ till j=value
do
if(matrix[i][j] == num) {
isBoolean = true;
break;
}
else if((j == value)) {
j = 0;
i++;
}
else if(matrix[i][j] != num){
j++;
}
while(isBoolean != true) ;
}
return isBoolean;
}
}
What i tried to do:
find in which quarter the wanted number is in,after that check
the same quarter by moving j++ until it hits the limit,than i++
until found
with the limits changing for each quarter,i cant understand if run time complexity is O(n^2) or lower? and will it be better do create one dimensional array and and move on the quarter this way: move right until limit,one down,move left until limit and il have a sorted array and just binear search
If you can map an array to a matrix, you can use a normal binary search.
You can define the translation table to achieve that like this:
X = [0, 0, 1, 1, 0, 0, 1, 1, 2, 2, 3, 3, 2, 2, 3, 3, ...]
Y = [0, 1, 1, 0, 2, 3, 3, 2, 2, 3, 3, 2, 0, 1, 1, 0, ...]
The final program looks like this.
static final int MAX_N = 64;
static final int MAX_NN = MAX_N * MAX_N;
static final int[] DX = {0, 0, 1, 1};
static final int[] DY = {0, 1, 1, 0};
static final int[] X = new int[MAX_NN];
static final int[] Y = new int[MAX_NN];
static { // initialize X and Y
for (int i = 0; i < MAX_NN; ++i) {
int x = 0, y = 0;
for (int t = i, f = 0; t > 0; ++f) {
int mod = t & 3;
x += DX[mod] << f; y += DY[mod] << f;
t >>= 2;
}
X[i] = x; Y[i] = y;
}
}
public static boolean Search(int [][] matrix, int num) {
int n = matrix.length, nn = n * n;
int lower = 0;
int upper = nn - 1;
while (lower <= upper) {
int mid = (lower + upper) / 2;
int value = matrix[X[mid]][Y[mid]];
if (value == num)
return true;
else if (value < num)
lower = mid + 1;
else
upper = mid - 1;
}
return false;
}
and
public static void main(String[] args) {
int[][] matrix = {
{1, 3, 7, 9},
{6, 4, 15, 11},
{36, 50, 21, 22},
{60, 55, 30, 26},
};
// case: exists
System.out.println(Search(matrix, 1));
System.out.println(Search(matrix, 60));
System.out.println(Search(matrix, 11));
// case: not exists
System.out.println(Search(matrix, 0));
System.out.println(Search(matrix, 70));
System.out.println(Search(matrix, 20));
}
output:
true
true
true
false
false
false
I want to create a function that lets me see if there is a break in a circuit which I've stored in a 2D Array by returning either true or false. For simplicity sake i have modeled the circuit using 1's and 0's. With 1's being components and 0's being spaces. I have tried messing with recursive functions however to no avail and am pretty much stuck at this point.
For example:
1 1 1
1 0 1
1 1 1
I would like this to return true as all the 1's are connected in series. This 2D array would be visualized as seen here.
1 1 1
0 0 1
1 1 1
I would like this to return false as there is a break in the circuit as
as shown here.
Any solutions or guidance would be much appreciated!
My current code is shown below. It returns incomplete when I used the completeCircuit as an input however returns complete when I have the incompleteCircuit as an input.
public class TraversalTest
{
boolean complete = false;
int runs = 0;
public static void main(String[] args)
{
TraversalTest traversal = new TraversalTest();
}
public TraversalTest()
{
Cell[][] completeCircuit =
{
{ new Cell( 0, 0, 1), new Cell( 1, 0, 1), new Cell( 2, 0, 1) },
{ new Cell( 0, 1, 1), new Cell( 1, 1, 0), new Cell( 2, 1, 1) },
{ new Cell( 0, 2, 1), new Cell( 1, 2, 1), new Cell( 2, 2, 1) }
};
Cell[][] incompleteCircuit =
{
{ new Cell( 0, 0, 1), new Cell( 1, 0, 1), new Cell( 2, 0, 1) },
{ new Cell( 0, 1, 0), new Cell( 1, 1, 0), new Cell( 2, 1, 1) },
{ new Cell( 0, 2, 1), new Cell( 1, 2, 1), new Cell( 2, 2, 1) }
};
completeCircuit[1][0].connected = true;
int cellsLeft = (numOfPositiveCells(completeCircuit));
checkComplete(completeCircuit, completeCircuit[1][0], cellsLeft);
incompleteCircuit[1][0].connected = true;
int cellsLeft1 = (numOfPositiveCells(incompleteCircuit));
checkComplete(incompleteCircuit, incompleteCircuit[1][0], cellsLeft1);
}
void checkComplete(Cell[][] circuit, Cell currentCell, int cellsLeft)
{
currentCell.connected = true;
if(cellsLeft > 0)
{
if(currentCell.x != 0 && circuit[currentCell.x-1][currentCell.y].value == 1 &&
circuit[currentCell.x-1][currentCell.y].connected == false)
{
cellsLeft--;
checkComplete(circuit, circuit[currentCell.x-1][currentCell.y], cellsLeft);
}
else if(currentCell.x != 2 &&circuit[currentCell.x+1][currentCell.y].value == 1 &&
circuit[currentCell.x+1][currentCell.y].connected == false)
{
cellsLeft--;
checkComplete(circuit, circuit[currentCell.x+1][currentCell.y], cellsLeft);
}
else if(currentCell.y != 0 && circuit[currentCell.x][currentCell.y-1].value == 1 &&
circuit[currentCell.x][currentCell.y-1].connected == false)
{
cellsLeft--;
checkComplete(circuit, circuit[currentCell.x][currentCell.y-1], cellsLeft);
}
else if(currentCell.y != 2 && circuit[currentCell.x][currentCell.y+1].value == 1 &&
circuit[currentCell.x][currentCell.y+1].connected == false)
{
cellsLeft--;
checkComplete(circuit, circuit[currentCell.x][currentCell.y+1], cellsLeft);
}
else
{
complete = false;
System.out.println("Incomplete");
}
}
else
{
complete = true;
System.out.println("Complete");
}
}
int numOfPositiveCells(Cell[][] circuit)
{
int num = 0;
for(int x=0; x < 3; x++)
for(int y=0; y < 3; y++)
if(circuit[x][y].value == 1)
num++;
return num;
}
}
class Cell
{
public boolean connected;
public int value;
public int x;
public int y;
public Cell(int x, int y, int value)
{
this.x = x;
this.y = y;
this.value = value;
}
}
I think checking that every cell has 2 connections should work for your use cases as we only look for series connections.
You just need to go through your array and make sure that all live cells have 2 and only 2 live cells connected to it
I'd suggest you make a dedicated class, this way you don't have to keep passing the circuit around in all the methods:
class CircuitChecker {
private final Cell[][] circuit;
private final int nbRows, nbCols;
public CircuitChecker(Cell[][] circuit) {
this.circuit = circuit;
this.nbRows = circuit.length;
this.nbCols = circuit[0].length;
}
public boolean isCircuitComplete() {
boolean isComplete = true;
for(int col = 0; col<nbCols; col++) {
for (int row = 0; row < nbRows; row++) {
if(cellIsLive(col, row) && !cellHas2LiveConnections(col, row)) {
isComplete = false;
break;
}
}
}
return isComplete;
}
private boolean cellIsLive(int col, int row) {
return circuit[row][col].value == 1;
}
private boolean cellHas2LiveConnections(int col, int row) {
Cell left = col > 0 ? circuit[col-1][row] : null;
Cell right = col < nbCols-1 ? circuit[col+1][row] : null;
Cell up = row > 0 ? circuit[col][row-1] : null;
Cell down = row < nbRows-1 ? circuit[col][row+1] : null;
int nbConnections = Stream.of(left, right, up, down)
.filter(Objects::nonNull)
.mapToInt(c -> c.value)
.sum();
return nbConnections == 2;
}
}
you call it this way:
new CircuitChecker(completeCircuit).isCircuitComplete();
new CircuitChecker(incompleteCircuit).isCircuitComplete();
One more thing, the fields in your class Cell should be private (and even maybe final) and you should access them via getters.
I'm studying the 3 Sum to implement it on my own, and came across the following implementation with the rules:
Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.
Note: Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)
The solution set must not contain duplicate triplets.
For example, given array S = {-1 0 1 2 -1 -4},
A solution set is:
(-1, 0, 1)
(-1, -1, 2)
And implementation (sorts the array, iterates through the list, and uses another two pointers to approach the target):
import java.util.*;
public class ThreeSum {
List<List<Integer>> threeSum(int[] num) {
Arrays.sort(num);
List<List<Integer>> res = new LinkedList<>();
for (int i=0; i<num.length-2; i++) {
if (i==0 || (i>0 && num[i] != num[i-1])) { //HERE
int lo = i+1;
int hi = num.length-1;
int sum = 0 - num[i];
while (lo < hi) {
if (num[lo] + num[hi] == sum) {
res.add(Arrays.asList(num[i], num[lo], num[hi]));
while (lo < hi && num[lo] == num[lo+1]) lo++; //HERE
while (lo < hi && num[hi] == num[hi-1]) hi--; //HERE
lo++; hi--;
} else if (num[lo] + num[hi] < sum) lo++;
else hi--;
}
}
}
return res;
}
//Driver
public static void main(String args[]) {
ThreeSum ts = new ThreeSum();
int[] sum = {-1, 0, 1, 2, -1, -4};
System.out.println(ts.threeSum(sum));
}
}
And my question is (located where commented: //HERE), what's the reason for checking num[i] != num[i-1], num[lo] == num[lo+1], and num[hi] == num[hi-1]? Supposedly they are supposed to skip the same result, but what does that mean? Examples would really help.
Thank you in advance and will accept answer/up vote.
Imagine you have {-1,-1,0,1,2,4} and considering triplet num[0], num[2], num[3] (-1,0,1).
lo=0 here. To exclude triplet num[1], num[2], num[3] with the same values, we should increment lo and pass over duplicate
This will prevent the list to have duplicate triplet.
For example, with you test :
int[] sum = {-1, 0, 1, 2, -1, -4};
will be sorted like :
sum = {-4, -1, -1, 0, 1, 2};
You see that you have -1 twice. Without these test, you would test twice if -1 = 0 + 1. This is not usefull so the algo simply search the next different value.
You could remove duplicate in the sorted List to prevent these test.
Thanks to MBo, we can't remove duplicate since we can have triplet with same value (but with different index)
All the three sentences is used to avoid the duplicate output.
Consider a sorted list {-2, -2 , 1, 1}
If there is no checking for num[i] != num[i-1], the output of the program would be(-2, 1, 1)and(-2, 1, 1), which are two duplicate triplets.
The checking for num[lo] != num[lo + 1]and num[hi] != num[hi - 1] are for the same reason.
Consider a sorted list
{-2,-1,-1,0,3}
If there is no checking for num[lo], you will get (-2,-1,3) and (-2,-1,3) as the output.
Still, I want to recommend a better solution for this problem. You can numerate the sum of two numbers in the list and find the 3rd number by hash or binary search. It will helps you to gain a O(n^2logn) time complexity rather than O(n^3). (I was wrong, the time complexity of this algorithm is O(n^2), sorry for that.)
Following program finds pairs of three integer with O(N*2)
Sort the input Array
and iterate each element in for loop and check for sum in program which is developed for Two sum.
Two sum in linear time after sorting ->
https://stackoverflow.com/a/49650614/4723446
public class ThreeSum {
private static int countThreeSum(int[] numbers) {
int count = 0;
for (int i = 0; i < numbers.length; i++) {
int front = 0, rear = numbers.length - 1;
while (front < rear) {
if (numbers[front] + numbers[rear] + numbers[i] == 0) {
System.out.printf(String.format("Front : {%d} Rear : {%d} I : {%d} \n", numbers[front],
numbers[rear], numbers[i]));
front++;
rear--;
count++;
} else {
if (Math.abs(numbers[front]) > Math.abs(numbers[rear])) {
front++;
} else {
rear--;
}
}
}
}
return count;
}
public static void main(String[] args) {
int[] numbers = { 1, 3, 5, 7, 12, 16, 19, 15, 11, 8, -1, -3, -7, -8, -11, -17, -15 };
Arrays.sort(numbers);
System.out.println(countThreeSum(numbers));
}
}
It's worked with any NSum (3Sum, 4Sum, 5Sum, ...) and quite fast.
public class ThreeSum {
private static final int RANDOM_RANGE = 20;
private Integer array[];
private Integer arrayIndex[];
private int result[];
private int bagLength;
private int resultIndex = 0;
private void generateData(int size) {
array = new Integer[size];
Random random = new Random();
for (int i = 0; i < size; i++) {
array[i] = random.nextInt(RANDOM_RANGE) - (RANDOM_RANGE/2);
}
}
private void markArrayIndex(int size) {
arrayIndex = new Integer[size];
for (int i = 0; i < size; i++) {
arrayIndex[i] = i;
}
}
private void prepareBeforeCalculate(int size, int sumExpected, int bagLength) {
this.bagLength = bagLength;
result = new int[bagLength];
generateData(size);
markArrayIndex(size);
}
void calculate(int size, int sumExpected, int bagLength) {
prepareBeforeCalculate(size, sumExpected, bagLength);
Arrays.sort(arrayIndex, (l, r) -> array[l].compareTo(array[r]));
System.out.println(Arrays.toString(array));
long startAt = System.currentTimeMillis();
if (sumExpected > 0) findLeft(sumExpected, 0, 0, array.length);
else findRight(sumExpected, 0, 0 - 1, array.length - 1);
System.out.println("Calculating in " + ((System.currentTimeMillis() - startAt) / 1000));
}
private void findLeft(int total, int indexBag, int left, int right) {
while (left < array.length && array[arrayIndex[left]] < 0 && indexBag < bagLength) {
navigating(total, arrayIndex[left], indexBag, left, right);
left++;
}
}
private void findRight(int total, int indexBag, int left, int right) {
while (right >= 0 && array[arrayIndex[right]] >= 0 && indexBag < bagLength) {
navigating(total, arrayIndex[right], indexBag, left, right);
right--;
}
}
private void navigating(int total, int index, int indexBag, int left, int right) {
result[indexBag] = index;
total += array[index];
if (total == 0 && indexBag == bagLength - 1) {
System.out.println(String.format("R[%d] %s", resultIndex++, toResultString()));
return;
}
if (total > 0) findLeft(total, indexBag + 1, left + 1, right);
else findRight(total, indexBag + 1, left, right - 1);
}
private String toResultString() {
int [] copyResult = Arrays.copyOf(result, result.length);
Arrays.sort(copyResult);
int iMax = copyResult.length - 1;
StringBuilder b = new StringBuilder();
b.append('[');
for (int i = 0; ; i++) {
b.append(array[copyResult[i]]);
if (i == iMax)
return b.append(']').toString();
b.append(", ");
}
}
}
public class ThreeSumTest {
#Test
public void test() {
ThreeSum test = new ThreeSum();
test.calculate(100, 0, 3);
Assert.assertTrue(true);
}
}
I have 2-D array of 1 and 0's.where top left corner and bottom right corner is filled with 1 and rest of index is filled with 0 and 1.
a[][]= {{1,1,0,0,1}{0,1,1,0,0}{0,1,0,1,0}{0,1,1,1,1}} .
write a program/algorithm that will return the other 2-D array of same size, contain the path from start to end.
output[][]={{1,1,0,0,0}{0,1,0,0,0}{0,1,0,0,0}{0,1,1,1,1}}.
please help me out for above issue. Also the program should work for other sequence of input.
Maybe I can provide you with a cleaner version of the backtracking algorithm. Her you go:
import java.util.Arrays;
public class Backtrack {
public static void main(String... args) {
int[][] inputArray = {
{ 1, 1, 1, 0, 0 },
{ 0, 1, 1, 0, 0 },
{ 0, 1, 0, 1, 0 },
{ 0, 1, 1, 1, 1 } };
int[][] findPath = findPath(inputArray);
System.out.println(Arrays.deepToString(findPath));
}
public static int[][] findPath(int[][] maze) {
int[][] solution = new int[maze.length][];
for (int i = 0; i < maze.length; i++) {
solution[i] = new int[maze[i].length];
}
if (!findPath(maze, solution, 0, 0)) {
System.out.println("Didn't find a solution.");
}
return solution;
}
private static boolean findPath(int[][] maze, int[][] solution, int x, int y) {
if (0 <= y && y < maze.length && 0 <= x && x < maze[y].length) {
if (y == maze.length - 1 && x == maze[y].length - 1) {
solution[y][x] = 1;
return true;
} else if (solution[y][x] != 1 && maze[y][x] == 1) {
solution[y][x] = 1;
if (findPath(maze, solution, x, y + 1)
|| findPath(maze, solution, x + 1, y)
|| findPath(maze, solution, x - 1, y)
|| findPath(maze, solution, x, y - 1)) {
return true;
}
solution[y][x] = 0;
}
}
return false;
}
}
I am working on a problem to try to find what is called a Snake Sequence in a typical XY graph (aka grid). A Snake sequence is defined as a sequence of numbers where each new number, which can only be located to the right or down of the current number, is either plus or minus one. For example, if you are in the center of the graph, you can either move right (if that number is + or - 1) or move down (if that number is + or - 1). The goal of the problem is to find the longest path (aka Snake Sequence) through the graph (keeping in mind you can only chart a path to a new cell whose value is +- 1 with the current cell).
So, for the following XY graph, the longest snake sequence is: 9, 8, 7, 6, 5, 6, 7
9, 6, 5, 2
8, 7, 6, 5
7, 3, 1, 6
1, 1, 1, 7
Below is my code, which does not seem to work.
Question: How would you solve this problem above? (I offer my code showing what I have thus far, but it does not work)
import java.util.ArrayList;
public class SnakeSequence {
private final int maxX = 3;
private final int maxY = 3;
private final int[][] board = new int[][]{
{1, 2, 3, 4},
{2, 1, -1, 5},
{3, 0, -1, 6},
{6, 2, 1, 7}
};
private ArrayList<Integer> findSequence(int xPos,
int yPos, ArrayList<Integer> currentPath) {
currentPath.add(board[yPos][xPos]);
ArrayList<Integer> pathRight = new ArrayList<Integer>(currentPath);
ArrayList<Integer> pathDown = new ArrayList<Integer>(currentPath);
if (xPos < maxX || yPos < maxY) {
if (yPos < maxY && (board[yPos + 1][xPos] + 1 == board[yPos][xPos] ||
board[yPos + 1][xPos] - 1 == board[yPos][xPos])) {
pathDown = findSequence(xPos, yPos + 1, currentPath);
}
if (xPos < maxX && (board[yPos][xPos + 1] + 1 == board[yPos][xPos] ||
board[yPos][xPos + 1] - 1 == board[yPos][xPos])) {
pathRight = findSequence(xPos + 1, yPos, currentPath);
}
if (pathDown.size() > pathRight.size()) {
return pathDown;
} else {
return pathRight;
}
}
return currentPath;
}
private void getSequence() {
ArrayList<Integer> currentPath = new ArrayList<Integer>();
ArrayList<Integer> result;
result = findSequence(0, 0, currentPath);
for (int i = 0; i < result.size(); i++) {
System.out.println(result.get(i));
}
}
public static void main(String[] args) {
SnakeSequence sequence = new SnakeSequence();
sequence.getSequence();
}
}
You can imagine your table as an oriented graph, then you problem is just to find the longest path.
Fortunatly for you, only moving down and right is allowed, so your graph is acyclic, so you can use algorithms like critical path method.
This is how your graph would look like:
However, you want to find longest path between any two cells. To do that, I would compute for each cell the longest path starting at that cell. It is simmilar to what you do, but you compute same thing more times. Consider this:
6 -> 5
| |
v v
7 -> 6
At both 5 and 7 you compute how long is the path from 6 at down right, and that is useless repeated computation. In worst case scenario, this could lead to exponential time consumption, while the problem can be resolved in linear time!
Moreover, there is no guarantee that the longest path will start at (0,0).
(one possible) Solution:
Compute longest path from each cell, starting from bottom-right to upper-left. At each cel.. remember how long the longest path from that cell is, and witch way from that cell the path leads. (I will measure path length by number of cells on it). So for example, for the only 8 in your grapth, we would remeber [length=8, direction=right].
Why so complicated? Because it is now extramly easy to compute longest path at a cell, if we know the longest path of the cells to the right and down. Example (I made up):
The correct data for 2 now would be [length=4, direction=down] because can't go from 2 to 4.
You can also keep globally longest path and it's start. After the computation is complete, just walk the longest path from that start through the direction and write the numbers, position or whatever you need.
Apologies for my Java (I am primarily a c# programmer) but here is one solution. I separated out the algorithm that discovers the snakes from the algorithm (implementing the interface ISnakeProcessor) that processes each one. That way you could enhance the code to, e.g., collect the snake with the largest sum of values, or collect all the longest snakes, in case there are more than one, by adding more ISnakeProcessor classes.
import java.util.*;
import java.lang.*;
class Rextester
{
public static void main(String args[])
{
SnakeSequence sequence = new SnakeSequence();
sequence.getSequence();
}
}
interface ISnakeProcessor
{
void process(List<Pair<Integer, Integer>> snake);
}
class SnakeSequence {
private final int[][] board;
public SnakeSequence()
{
this(new int[][]{
{1, 2, 3, 4},
{2, 1, -1, 5},
{3, 0, -1, 6},
{6, 2, 1, 7}
});
}
public SnakeSequence(int[][] board)
{
this.board = board;
}
public boolean isValid(int iRow, int iCol)
{
if (iRow < 0 || iRow >= board.length)
return false;
if (iCol < 0 || iCol >= board[iRow].length)
return false;
return true;
}
private boolean continuesInRow(int iRow, int iCol)
{
if (!isValid(iRow, iCol) || !isValid(iRow+1, iCol))
return false;
int myVal = board[iRow][iCol];
if (board[iRow+1][iCol] == myVal - 1 || board[iRow+1][iCol] == myVal + 1)
return true;
return false;
}
private boolean continuesInCol(int iRow, int iCol)
{
if (!isValid(iRow, iCol) || !isValid(iRow, iCol+1))
return false;
int myVal = board[iRow][iCol];
if (board[iRow][iCol+1] == myVal - 1 || board[iRow][iCol+1] == myVal + 1)
return true;
return false;
}
private boolean isHead(int iRow, int iCol)
{
if (!isValid(iRow, iCol))
return false;
if (isValid(iRow-1, iCol) && continuesInRow(iRow-1, iCol))
return false;
if (isValid(iRow, iCol-1) && continuesInRow(iRow, iCol-1))
return false;
return true;
}
private boolean isTail(int iRow, int iCol)
{
if (!isValid(iRow, iCol))
return false;
if (continuesInRow(iRow, iCol))
return false;
if (continuesInCol(iRow, iCol))
return false;
return true;
}
private void testHead()
{
System.out.println("Dumping list of heads");
for (int iRow = 0; iRow < board.length; iRow++)
{
for (int iCol = 0; iCol < board[iRow].length; iCol++)
{
boolean head = isHead(iRow, iCol);
boolean tail = isTail(iRow, iCol);
if (head && tail)
System.out.print(" B");
else if (head)
System.out.print(" H");
else if (tail)
System.out.print(" T");
else
System.out.print(" -");
}
System.out.println("");
}
}
private void walkSnake(ISnakeProcessor processor, int iRow, int iCol, ArrayList<Pair<Integer, Integer>> snake)
{
snake.add(new Pair<Integer, Integer>(iRow, iCol));
boolean isTail = true;
if (continuesInRow(iRow, iCol))
{
walkSnake(processor, iRow+1, iCol, snake);
isTail = false;
}
if (continuesInCol(iRow, iCol))
{
walkSnake(processor, iRow, iCol+1, snake);
isTail = false;
}
if (isTail)
{
processor.process(snake);
}
snake.remove(snake.size() - 1);
}
private void walkSnakes(ISnakeProcessor processor)
{
ArrayList<Pair<Integer, Integer>> snake = new ArrayList<Pair<Integer, Integer>>();
for (int iRow = 0; iRow < board.length; iRow++)
for (int iCol = 0; iCol < board[iRow].length; iCol++)
if (isHead(iRow, iCol))
walkSnake(processor, iRow, iCol, snake);
}
class LongestSnakeFinder implements ISnakeProcessor
{
private final SnakeSequence parent;
ArrayList<Pair<Integer, Integer>> longest = new ArrayList<Pair<Integer, Integer>>();
public LongestSnakeFinder(SnakeSequence parent)
{
this.parent = parent;
}
public void process(List<Pair<Integer, Integer>> snake)
{
if (snake.size() > longest.size())
{
longest.clear();
longest.addAll(snake);
}
}
public void dumpLongest()
{
System.out.format("The first encountered longest snake has length %d:\n", longest.size());
for (int i = 0; i < longest.size(); i++)
{
int iRow = longest.get(i).getFirst();
int iCol = longest.get(i).getSecond();
System.out.format(" (%d,%d): %d\n", iRow, iCol, parent.getValue(iRow, iCol));
}
}
}
public int getNRows() { return board.length; }
public int getNCols(int iRow) { return board[iRow].length; }
public int getValue(int iRow, int iCol) { return board[iRow][iCol]; }
public void getSequence() {
testHead();
LongestSnakeFinder finder = new LongestSnakeFinder(this);
walkSnakes(finder);
finder.dumpLongest();
}
}
class Pair<F, S> {
private F first; //first member of pair
private S second; //second member of pair
public Pair(F first, S second) {
this.first = first;
this.second = second;
}
public F getFirst() {
return first;
}
public S getSecond() {
return second;
}
}
Example run here: http://rextester.com/AKUFNL43897
Update - cleaned code a little. New sample run here: http://rextester.com/AVOAIY11573
And, the output:
Dumping list of heads
H - - -
- - B -
T - T -
B H T T
The first encountered longest snake has length 7:
(0,0): 1
(0,1): 2
(0,2): 3
(0,3): 4
(1,3): 5
(2,3): 6
(3,3): 7
Is this what you want?
Here is one simple way to correct your solution and avoid copying of path on every step
import java.util.ArrayList;
import java.util.Collections;
public class SnakeSequence {
private final int maxX = 3;
private final int maxY = 3;
private final int[][] board = new int[][]{
{1, 2, 3, 4},
{2, 1, -1, 5},
{3, 0, -1, 6},
{6, 2, 1, 7}
};
private ArrayList<Integer> findSequence(int xPos,
int yPos) {
ArrayList<Integer> pathRight = new ArrayList<Integer>();
ArrayList<Integer> pathDown = new ArrayList<Integer>();
if (yPos < maxY && (board[yPos + 1][xPos] + 1 == board[yPos][xPos] ||
board[yPos + 1][xPos] - 1 == board[yPos][xPos])) {
pathDown = findSequence(xPos, yPos + 1);
}
if (xPos < maxX && (board[yPos][xPos + 1] + 1 == board[yPos][xPos] ||
board[yPos][xPos + 1] - 1 == board[yPos][xPos])) {
pathRight = findSequence(xPos + 1, yPos);
}
ArrayList<Integer> ans;
if (pathDown.size() > pathRight.size()) {
ans = pathDown;
} else {
ans = pathRight;
}
ans.add(board[yPos][xPos]);
return ans;
}
private void getSequence() {
ArrayList<Integer> result;
result = findSequence(0, 0);
Collections.reverse(result);
for (int i = 0; i < result.size(); i++) {
System.out.println(result.get(i));
}
}
public static void main(String[] args) {
SnakeSequence sequence = new SnakeSequence();
sequence.getSequence();
}
}
But this way it can work much faster for big arrays due to no recalculating the longest path every time you visiting the same number during recursion. Actually, in this version each number is visited at most twice. It's achieved through saving best solution for every node. Separate storage of path and it length allows not to copy path when it's not needed.
import java.util.ArrayList;
import java.util.Collections;
public class SnakeSequence {
private final int maxX = 3;
private final int maxY = 3;
private final int[][] board = new int[][]{
{1, 2, 3, 4},
{2, 3, -1, 5},
{3, 2, -1, 6},
{6, 1, 2, 3}
};
int[][] pathLength;
ArrayList<ArrayList<ArrayList<Integer>>> paths;
private ArrayList<Integer> findSequence(int xPos,
int yPos) {
if(pathLength[yPos][xPos] >= 0)
{
ArrayList<Integer> ans = new ArrayList<Integer>();
int length = pathLength[yPos][xPos];
ArrayList<Integer> path = paths.get(yPos).get(xPos);
for(int i = 0; i < length; i++)
ans.add(path.get(i));
return ans;
}
ArrayList<Integer> pathRight = new ArrayList<Integer>();
ArrayList<Integer> pathDown = new ArrayList<Integer>();
if (yPos < maxY && (board[yPos + 1][xPos] + 1 == board[yPos][xPos] ||
board[yPos + 1][xPos] - 1 == board[yPos][xPos])) {
pathDown = findSequence(xPos, yPos + 1);
}
if (xPos < maxX && (board[yPos][xPos + 1] + 1 == board[yPos][xPos] ||
board[yPos][xPos + 1] - 1 == board[yPos][xPos])) {
pathRight = findSequence(xPos + 1, yPos);
}
ArrayList<Integer> ans;
if (pathDown.size() > pathRight.size()) {
ans = pathDown;
} else {
ans = pathRight;
}
ans.add(board[yPos][xPos]);
paths.get(yPos).set(xPos,ans);
pathLength[yPos][xPos] = ans.size();
return ans;
}
private void getSequence() {
ArrayList<Integer> result;
pathLength = new int[maxX + 1][maxY + 1];
paths = new ArrayList<ArrayList<ArrayList<Integer>>>();
for(int y = 0; y <= maxY; y++)
{
ArrayList<ArrayList<Integer>> line = new ArrayList<ArrayList<Integer>>();
for(int x = 0; x <= maxX; x++)
{
line.add(null);
pathLength[y][x] = -1;
}
paths.add(line);
}
result = findSequence(0, 0);
Collections.reverse(result);
for (int i = 0; i < result.size(); i++) {
System.out.println(result.get(i));
}
}
public static void main(String[] args) {
SnakeSequence sequence = new SnakeSequence();
sequence.getSequence();
}
}
Simple Recursive solution :
import java.util.ArrayList;
import java.util.List;
public class MaximumLengthSnakeSequence {
static int max = -1;
static List<Integer> maxListTemp = new ArrayList<>();
public static void main(String args[]) {
int count = 0;
int n = 4;
int m = 4;
int mat[][] = { { 9, 6, 5, 2 }, { 8, 7, 6, 5 }, { 7, 3, 1, 6 }, { 1, 1, 1, 7 }, };
List<Integer> maxList = new ArrayList<>();
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
List<Integer> list = new ArrayList<>();
list.add(mat[i][j]);
List<Integer> testList = recur(i, j, count, mat, n, m, list);
if (maxList.size() < testList.size()) {
maxList = new ArrayList<>(testList);
}
maxListTemp.clear();
}
}
System.out.println("max is " + maxList);
}
static List<Integer> recur(int i, int j, int count, int mat[][], int n, int m, List<Integer> list) {
int curData = mat[i][j];
int rightData = 0;
int downData = 0;
if (j + 1 < n && i < m) {
rightData = mat[i][j + 1];
if (Math.abs(curData - rightData) == 1) {
list.add(rightData);
recur(i, j + 1, count + 1, mat, n, m, list);
list.remove(list.size() - 1);
}
}
if (count > max) {
max = count;
}
if (maxListTemp.size() < list.size()) {
maxListTemp = new ArrayList<>(list);
}
if (i + 1 < m && j < n) {
downData = mat[i + 1][j];
if (Math.abs(curData - downData) == 1) {
list.add(downData);
recur(i + 1, j, count + 1, mat, n, m, list);
list.remove(list.size() - 1);
}
}
return maxListTemp;
}
}