# Find maximum length Snake sequence

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.

For example,

9, 6, 5, 2
8, 7, 6, 5
7, 3, 1, 6
1, 1, 1, 7

In above grid, the longest snake sequence is: (9, 8, 7, 6, 5, 6, 7)

Below figure shows all possible paths –

We strongly recommend you to minimize your browser and try this yourself first.

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

Below is C++ implementation of the idea –

```// C++ program to find maximum length
// Snake sequence and print it
#include <bits/stdc++.h>
using namespace std;
#define M 4
#define N 4

struct Point
{
int x, y;
};

// 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 << ")" ;
}

// Driver code
int main()
{
int mat[M][N] =
{
{9, 6, 5, 2},
{8, 7, 6, 5},
{7, 3, 1, 6},
{1, 1, 1, 7},
};

findSnakeSequence(mat);

return 0;
}
```

Output :

```Maximum length of Snake sequence is: 6
Snake sequence is:
9 (0, 0)
8 (1, 0)
7 (1, 1)
6 (1, 2)
5 (1, 3)
6 (2, 3)
7 (3, 3)```

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.

Reference: Stack Overflow

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