📢 แจ้งเตือน: web//https://www.geeksforgeeks.org/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 the implementation of the idea –

C++

// 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; 
} 

Java

// Java program to find maximum length 
// Snake sequence and print it 
import java.util.*; 
  
class GFG  
{ 
  
static int M = 4; 
static int N = 4; 
  
static class Point 
{ 
    int x, y; 
  
    public Point(int x, int y)  
    { 
        this.x = x; 
        this.y = y; 
    } 
}; 
  
// Function to find maximum length Snake sequence path 
// (i, j) corresponds to tail of the snake 
static List<Point> findPath(int grid[][],    
                            int mat[][],  
                            int i, int j) 
{ 
    List<Point> path = new LinkedList<>(); 
  
    Point pt = new Point(i, j); 
    path.add(0, pt); 
  
    while (grid[i][j] != 0) 
    { 
        if (i > 0 && 
            grid[i][j] - 1 == grid[i - 1][j]) 
        { 
            pt = new Point(i - 1, j); 
            path.add(0, pt); 
            i--; 
        } 
        else if (j > 0 && grid[i][j] - 1 ==  
                          grid[i][j - 1]) 
        { 
            pt = new Point(i, j - 1); 
            path.add(0, pt); 
            j--; 
        } 
    } 
    return path; 
} 
  
// Function to find maximum length Snake sequence 
static void findSnakeSequence(int mat[][]) 
{ 
    // table to store results of subproblems 
    int [][]lookup = new int[M][N]; 
  
    // initialize by 0 
  
    // 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 != 0 || j != 0) 
            { 
                // look above 
                if (i > 0 && 
                    Math.abs(mat[i - 1][j] -  
                             mat[i][j]) == 1) 
                { 
                    lookup[i][j] = Math.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 && 
                    Math.abs(mat[i][j - 1] -  
                             mat[i][j]) == 1) 
                { 
                    lookup[i][j] = Math.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; 
                    } 
                } 
            } 
        } 
    } 
    System.out.print("Maximum length of Snake " +  
                     "sequence is: " + max_len + "\n"); 
  
    // find maximum length Snake sequence path 
    List<Point> path = findPath(lookup, mat, max_row, 
                                             max_col); 
  
    System.out.print("Snake sequence is:"); 
    for (Point it : path) 
        System.out.print("\n" + mat[it.x][it.y] + " (" +  
                                    it.x + ", " + it.y + ")"); 
} 
  
// Driver code 
public static void main(String[] args) 
{ 
    int mat[][] = {{9, 6, 5, 2}, 
                   {8, 7, 6, 5}, 
                   {7, 3, 1, 6}, 
                   {1, 1, 1, 7}}; 
  
    findSnakeSequence(mat); 
} 
} 
  
// This code is contributed by 29AjayKumar 

Python3

# Python program to find maximum length 
# Snake sequence and print it 
  
M = 4
N = 4
  
class Point: 
    def __init__(self, x, y): 
        self.x = x 
        self.y = y 
  
# Function to find maximum length Snake sequence path 
# (i, j) corresponds to tail of the snake 
def findPath(grid, mat, i, j): 
    path = list() 
  
    pt = Point(i, j) 
    path.append(pt) 
  
    while (grid[i][j] != 0): 
        if (i > 0 and grid[i][j]-1 == grid[i-1][j]): 
            pt = Point(i-1, j) 
            path.append(pt) 
            i -= 1
        elif (j > 0 and grid[i][j]-1 == grid[i][j-1]): 
            pt = Point(i, j-1) 
            path.append(pt) 
            j -= 1
    return path 
  
# Function to find maximum length Snake sequence 
def findSnakeSequence(mat): 
  
    # table to store results of subproblems 
    # initialize by 0 
    lookup = [[0 for i in range(N)] for j in range(M)] 
  
    # stores maximum length of Snake sequence 
    max_len = 0
  
    # store cordinates to snake's tail 
    max_row = 0
    max_col = 0
  
    # fill the table in bottom-up fashion 
    for i in range(M): 
        for j in range(N): 
            # do except for (0, 0) cell 
            if (i or j): 
                # look above 
                if (i > 0 and
                    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 and
                    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 
  
    print("Maximum length of Snake sequence is:", max_len) 
  
    # find maximum length Snake sequence path 
    path = findPath(lookup, mat, max_row, max_col) 
  
    print("Snake sequence is:") 
    for ele in reversed(path): 
        print(mat[ele.x][ele.y], 
              " (", ele.x, ", ", ele.y, ")", sep = "") 
  
# Driver code 
mat = [[9, 6, 5, 2], 
       [8, 7, 6, 5], 
       [7, 3, 1, 6], 
       [1, 1, 1, 7]] 
  
findSnakeSequence(mat) 
  
# This code is contributed 
# by Soumen Ghosh 

C#

// C# program to find maximum length 
// Snake sequence and print it 
using System; 
using System.Collections.Generic; 
      
class GFG  
{ 
static int M = 4; 
static int N = 4; 
  
public class Point 
{ 
    public int x, y; 
  
    public Point(int x, int y)  
    { 
        this.x = x; 
        this.y = y; 
    } 
}; 
  
// Function to find maximum length Snake sequence path 
// (i, j) corresponds to tail of the snake 
static List<Point> findPath(int [,]grid,  
                            int [,]mat,  
                            int i, int j) 
{ 
    List<Point> path = new List<Point>(); 
  
    Point pt = new Point(i, j); 
    path.Insert(0, pt); 
  
    while (grid[i, j] != 0) 
    { 
        if (i > 0 && 
            grid[i, j] - 1 == grid[i - 1, j]) 
        { 
            pt = new Point(i - 1, j); 
            path.Insert(0, pt); 
            i--; 
        } 
        else if (j > 0 && grid[i, j] - 1 ==  
                          grid[i, j - 1]) 
        { 
            pt = new Point(i, j - 1); 
            path.Insert(0, pt); 
            j--; 
        } 
    } 
    return path; 
} 
  
// Function to find maximum length Snake sequence 
static void findSnakeSequence(int [,]mat) 
{ 
    // table to store results of subproblems 
    int [,]lookup = new int[M, N]; 
  
    // initialize by 0 
  
    // 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 != 0 || j != 0) 
            { 
                // look above 
                if (i > 0 && 
                    Math.Abs(mat[i - 1, j] -  
                             mat[i, j]) == 1) 
                { 
                    lookup[i, j] = Math.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 && 
                    Math.Abs(mat[i, j - 1] -  
                             mat[i, j]) == 1) 
                { 
                    lookup[i, j] = Math.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; 
                    } 
                } 
            } 
        } 
    } 
    Console.Write("Maximum length of Snake " +  
                  "sequence is: " + max_len + "\n"); 
  
    // find maximum length Snake sequence path 
    List<Point> path = findPath(lookup, mat, max_row, 
                                             max_col); 
  
    Console.Write("Snake sequence is:"); 
    foreach (Point it in path) 
        Console.Write("\n" + mat[it.x,it.y] +  
                      " (" + it.x + ", " + it.y + ")"); 
} 
  
// Driver code 
public static void Main(String[] args) 
{ 
    int [,]mat = {{9, 6, 5, 2}, 
                  {8, 7, 6, 5}, 
                  {7, 3, 1, 6}, 
                  {1, 1, 1, 7}}; 
  
    findSnakeSequence(mat); 
} 
} 
  
// This code is contributed by Princi Singh 

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

This article is contributed by Aditya Goel. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Find maximum length Snake sequence

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