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:
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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 4struct Point{ int x, y;};// Function to find maximum length Snake sequence path// (i, j) corresponds to tail of the snakelist<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 sequencevoid 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 coordinates 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 codeint 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 itimport 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 snakestatic 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 sequencestatic 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 coordinates 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 codepublic 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 |
C#
// C# program to find maximum length// Snake sequence and print itusing 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 coordinates 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 |
Python3
def snakesequence(S, m, n): sequence = {} DP = [[1 for x in range(m+1)] for x in range(n+1)] a, b, maximum = 0, 0, 0 position = [0, 0] for i in range(0, n+1): for j in range(0, m+1): a, b = 0, 0 p = "initial" if(i > 0 and abs(S[i][j] - S[i-1][j]) == 1): a = DP[i-1][j] if(j > 0 and abs(S[i][j] - S[i][j-1]) == 1): b = DP[i][j-1] if a != 0 and a >= b: p = str(i-1) + " " + str(j) elif b != 0: p = str(i) + " " + str(j-1) q = str(i) + " " + str(j) sequence[q] = p DP[i][j] = DP[i][j] + max(a, b) if DP[i][j] >= maximum: maximum = DP[i][j] position[0] = i position[1] = j snakeValues = [] snakePositions = [] snakeValues.append(S[position[0]][position[1]]) check = 'found' str_next = str(position[0]) + " " + str(position[1]) findingIndices = sequence[str_next].split() while(check == 'found'): if sequence[str_next] == 'initial': snakePositions.insert(0, str_next) check = 'end' continue findingIndices = sequence[str_next].split() g = int(findingIndices[0]) h = int(findingIndices[1]) snakeValues.insert(0, S[g][h]) snake_position = str(g) + " " + str(h) snakePositions.insert(0, str_next) str_next = sequence[str_next] return [snakeValues, snakePositions]S = [[9, 6, 5, 2], [8, 7, 6, 5], [7, 3, 1, 6], [1, 1, 10, 7]]m = 3n = 3seq = snakesequence(S, m, n)for i in range(len(seq[0])): print(seq[0][i], ",", seq[1][i].split()) |
Javascript
function snakesequence(S, m, n){ let sequence = {} let DP = new Array(n + 1) for (var i = 0; i <= n; i++) DP[i] = new Array(m + 1).fill(1) let a = 0, b = 0, maximum = 0 let position = [0, 0] for (var i = 0; i <= n; i++) { for (var j = 0; j <= m; j++) { a = 0 b = 0 let p = "initial" if(i > 0 && Math.abs(S[i][j] - S[i-1][j]) == 1) a = DP[i-1][j] if(j > 0 && Math.abs(S[i][j] - S[i][j-1]) == 1) b = DP[i][j-1] if (a != 0 && a >= b) p = String(i-1) + " " + String(j) else if (b != 0) p = String(i) + " " + String(j-1) let q = String(i) + " " + String(j) sequence[q] = p DP[i][j] = DP[i][j] + Math.max(a, b) if (DP[i][j] >= maximum) { maximum = DP[i][j] position[0] = i position[1] = j } } } let snakeValues = [] let snakePositions = [] snakeValues.push(S[position[0]][position[1]]) let check = 'found' let String_next = String(position[0]) + " " + String(position[1]) let findingIndices = sequence[String_next].split(" ") while(check == 'found') { if (sequence[String_next] == 'initial') { snakePositions.unshift(String_next) check = 'end' continue } findingIndices = sequence[String_next].split(" ") let g = parseInt(findingIndices[0]) let h = parseInt(findingIndices[1]) snakeValues.unshift(S[g][h]) let snake_position = String(g) + " " + String(h) snakePositions.unshift(String_next) String_next = sequence[String_next] } return [snakeValues, snakePositions]}// Driver Code let S = [[9, 6, 5, 2], [8, 7, 6, 5], [7, 3, 1, 6], [1, 1, 10, 7]]let m = 3let n = 3let seq = snakesequence(S, m, n)for (var i = 0; i < seq[0].length; i++) console.log(seq[0][i] + ",", seq[1][i].split(" ")) |
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.
This article is contributed by Aditya Goel. If you like neveropen and would like to contribute, you can also write an article and mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
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