## Saturday, June 4, 2016

### Find maximum length Snake sequence - GeeksforGeeks

Find maximum length Snake sequence - GeeksforGeeks
Given a grid of numbers, find maximum length Snake sequence and print it. If multiple snake sequences exists with the maximum length, print any one of them.
A snake sequence is made up of adjacent numbers in the grid such that for each number, the number on the right or the number below it is +1 or -1 its value. For example, if you are at location (x, y) in the grid, you can either move right i.e. (x, y+1) if that number is ± 1 or move down i.e. (x+1, y) if that number is ± 1.

The idea is to use Dynamic Programming. For each cell of the matrix, we keep maximum length of a snake which ends in current cell. The maximum length snake sequence will have maximum value. The maximum value cell will correspond to tail of the snake. In order to print the snake, we need to backtrack from tail all the way back to snake’s head.
Let T[i][i] represent maximum length of a snake which ends at cell (i, j), then for given matrix M, the DP relation is defined as –
T[0][0] = 0
T[i][j] = max(T[i][j], T[i][j – 1] + 1) if M[i][j] = M[i][j – 1] ± 1
T[i][j] = max(T[i][j], T[i – 1][j] + 1) if M[i][j] = M[i – 1][j] ± 1
Time complexity of above solution is O(M*N). Auxiliary space used by above solution is O(M*N). If we are not required to print the snake, space  can be further reduced to O(N) as we only uses the result from last row.

`// Function to find maximum length Snake sequence path`
`// (i, j) corresponds to tail of the snake`
`list<Point> findPath(``int` `grid[M][N], ``int` `mat[M][N],`
`                     ``int` `i, ``int` `j)`
`{`
`    ``list<Point> path;`
`    ``Point pt = {i, j};`
`    ``path.push_front(pt);`
`    ``while` `(grid[i][j] != 0)`
`    ``{`
`       ``if` `(i > 0 &&`
`           ``grid[i][j] - 1 == grid[i - 1][j])`
`       ``{`
`           ``pt = {i - 1, j};`
`           ``path.push_front(pt);`
`           ``i--;`
`       ``}`
`       ``else` `if` `(j > 0 &&`
`                ``grid[i][j] - 1 == grid[i][j - 1])`
`       ``{`
`           ``pt = {i, j - 1};`
`           ``path.push_front(pt);`
`           ``j--;`
`       ``}`
`    ``}`
`    ``return` `path;`
`}`
`// Function to find maximum length Snake sequence`
`void` `findSnakeSequence(``int` `mat[M][N])`
`{`
`    ``// table to store results of subproblems`
`    ``int` `lookup[M][N];`
`    ``// initialize by 0`
`    ``memset``(lookup, 0, ``sizeof` `lookup);`
`    ``// stores maximum length of Snake sequence`
`    ``int` `max_len = 0;`
`    ``// store cordinates to snake's tail`
`    ``int` `max_row = 0;`
`    ``int` `max_col = 0;`
`    ``// fill the table in bottom-up fashion`
`    ``for` `(``int` `i = 0; i < M; i++)`
`    ``{`
`        ``for` `(``int` `j = 0; j < N; j++)`
`        ``{`
`            ``// do except for (0, 0) cell`
`            ``if` `(i || j)`
`            ``{`
`                ``// look above`
`                ``if` `(i > 0 &&`
`                    ``abs``(mat[i - 1][j] - mat[i][j]) == 1)`
`                ``{`
`                    ``lookup[i][j] = max(lookup[i][j],`
`                               ``lookup[i - 1][j] + 1);`
`                    ``if` `(max_len < lookup[i][j])`
`                    ``{`
`                        ``max_len = lookup[i][j];`
`                        ``max_row = i, max_col = j;`
`                    ``}`
`                ``}`
`                ``// look left`
`                ``if` `(j > 0 &&`
`                    ``abs``(mat[i][j - 1] - mat[i][j]) == 1)`
`                ``{`
`                    ``lookup[i][j] = max(lookup[i][j],`
`                                       ``lookup[i][j - 1] + 1);`
`                    ``if` `(max_len < lookup[i][j])`
`                    ``{`
`                        ``max_len = lookup[i][j];`
`                        ``max_row = i, max_col = j;`
`                    ``}`
`                ``}`
`            ``}`
`        ``}`
`    ``}`
`    ``cout << ``"Maximum length of Snake sequence is: "`
`         ``<< max_len << endl;`
`    ``// find maximum length Snake sequence path`
`    ``list<Point> path = findPath(lookup, mat, max_row,`
`                                             ``max_col);`
`    ``cout << ``"Snake sequence is:"``;`
`    ``for` `(``auto` `it = path.begin(); it != path.end(); it++)`
`        ``cout << endl << mat[it->x][it->y] << ``" ("`
`             ``<< it->x << ``", "` `<< it->y << ``")"` `;`
`}`

http://algorithms.tutorialhorizon.com/find-longest-snake-sequence-in-a-given-matrix/
public int getMaxSequence(int [][] matrix){

int rows = matrix.length;
int cols = matrix[0].length;
int maxLenth =1;
int maxRow = 0;
int maxCol = 0;

//create result matrix
int [][] result = new int [rows][cols];

//if no sequence is found then every cell itself is a sequence of length 1
for (int i = 0; i <rows ; i++) {
for (int j = 0; j <cols ; j++) {
result[i][j] =1;
}
}

for (int i = 0; i <rows ; i++) {
for (int j = 0; j <cols ; j++) {
if(i!=0 || j!=0){
//check from left
if(i>0 && Math.abs(matrix[i][j]-matrix[i-1][j])==1){
result[i][j] = Math.max(result[i][j],
result[i-1][j]+1);
if(maxLenth<result[i][j]){
maxLenth = result[i][j];
maxRow = i;
maxCol = j;
}
}

//check from top
if(j>0 && Math.abs(matrix[i][j]-matrix[i][j-1])==1){
result[i][j] = Math.max(result[i][j],
result[i][j-1]+1);
if(maxLenth<result[i][j]){
maxLenth = result[i][j];
maxRow = i;
maxCol = j;
}
}
}
}
}

//Now we will check the max entry in the result[][].
System.out.println("Max Snake Sequence : " + maxLenth);
printPath(matrix, result, maxLenth, maxRow, maxCol);
return 0;
}

public void printPath(int [][] matrix, int [][] result, int maxLength, int maxRow, int maxCol){
int len =  maxLength;
while(maxLength>=1){
System.out.print(" - " + matrix[maxRow][maxCol]);
if(maxRow>0 && Math.abs(result[maxRow-1][maxCol]-result[maxRow][maxCol])==1){
maxRow--;
}else if(maxCol>0 && Math.abs(result[maxRow][maxCol-1]-result[maxRow][maxCol])==1){
maxCol--;
}
maxLength--;
}
}

http://www.techiedelight.com/maximum-length-snake-sequence/