Length of longest connected 1’s in a Binary Grid - GeeksforGeeks

12 Feb, 2022
Given a grid of size N*M consists of 0 and 1 only, the task is to find the length of longest connected 1s in the given grid. We can only move to left, right, up or down from any current cell of the grid.
Examples: 
Input: N = 3, M = 3, grid[][] = { {0, 0, 0}, {0, 1, 0}, {0, 0, 0} } Output: 1 Explanation: The longest possible route is 1 as there cant be any movement from (1, 1) position of the matrix.
Input: N = 6, M = 7, grid[][] = { {0, 0, 0, 0, 0, 0, 0}, {0, 1, 0, 1, 0, 0, 0}, {0, 1, 0, 1, 0, 0, 0}, {0, 1, 0, 1, 0, 1, 0}, {0, 1, 1, 1, 1, 1, 0}, {0, 0, 0, 0, 0, 0, 0}} Output: 9 Explanation: The longest possible route is 9 starting from (1, 1) -> (2, 1) -> (3, 1) -> (4, 1) -> (4, 2) -> (4, 3) -> (4, 4) -> (4, 5) -> (3, 5). 
Approach: The idea is to do DFS Traversal on grid where the value of current cell is 1 and recursively call for all the four direction of the current cell where value is 1 and updated the maximum length of connected 1.
Below is the implementation of the above approach:  
C++
Java
Python3
C#
Javascript
// C++ program for the above approach #include <bits/stdc++.h>#define row 6#define col 7using namespace std; int vis[row + 1][col + 1], id;int diameter = 0, length = 0; // Keeps a track of directions// that is up, down, left, rightint dx[] = { -1, 1, 0, 0 };int dy[] = { 0, 0, -1, 1 }; // Function to perform the dfs traversalvoid dfs(int a, int b, int lis[][col],         int& x, int& y){     // Mark the current node as visited    vis[a][b] = id;     // Increment length from this node    length++;     // Update the diameter length    if (length > diameter) {        x = a;        y = b;        diameter = length;    }    for (int j = 0; j < 4; j++) {         // Move to next cell in x-direction        int cx = a + dx[j];         // Move to next cell in y-direction        int cy = b + dy[j];         // Check if cell is invalid        // then continue        if (cx < 0 || cy < 0 || cx >= row            || cy >= col || lis[cx][cy] == 0            || vis[cx][cy]) {            continue;        }         // Perform DFS on new cell        dfs(cx, cy, lis, x, y);    }     vis[a][b] = 0;     // Decrement the length    length--;} // Function to find the maximum length of// connected 1s in the given gridvoid findMaximumLength(int lis[][col]){     int x, y;     // Increment the id    id++;    length = 0;    diameter = 0;     // Traverse the grid[]    for (int i = 0; i < row; i++) {         for (int j = 0; j < col; j++) {             if (lis[i][j] != 0) {                 // Find start point of                // start dfs call                dfs(i, j, lis, x, y);                i = row;                break;            }        }    }     id++;    length = 0;    diameter = 0;     // DFS Traversal from cell (x, y)    dfs(x, y, lis, x, y);     // Print the maximum length    cout << diameter;} // Driver Codeint main(){    // Given grid[][]    int grid[][col] = { { 0, 0, 0, 0, 0, 0, 0 },                        { 0, 1, 0, 1, 0, 0, 0 },                        { 0, 1, 0, 1, 0, 0, 0 },                        { 0, 1, 0, 1, 0, 1, 0 },                        { 0, 1, 1, 1, 1, 1, 0 },                        { 0, 0, 0, 1, 0, 0, 0 } };     // Function Call    findMaximumLength(grid);     return 0;}
// Java program for the above approachimport java.util.*; class GFG{     static final int row = 6;static final int col = 7;static int [][]vis = new int[row + 1][col + 1];static int id;static int diameter = 0, length = 0;static int x = 0, y = 0; // Keeps a track of directions// that is up, down, left, rightstatic int dx[] = { -1, 1, 0, 0 };static int dy[] = { 0, 0, -1, 1 }; // Function to perform the dfs traversalstatic void dfs(int a, int b, int lis[][]){         // Mark the current node as visited    vis[a][b] = id;     // Increment length from this node    length++;     // Update the diameter length    if (length > diameter)    {        x = a;        y = b;        diameter = length;    }     for(int j = 0; j < 4; j++)    {                // Move to next cell in x-direction       int cx = a + dx[j];               // Move to next cell in y-direction       int cy = b + dy[j];               // Check if cell is invalid       // then continue       if (cx < 0 || cy < 0 ||           cx >= row || cy >= col ||           lis[cx][cy] == 0 || vis[cx][cy] > 0)       {           continue;       }               // Perform DFS on new cell       dfs(cx, cy, lis);    }         vis[a][b] = 0;     // Decrement the length    length--;} // Function to find the maximum length of// connected 1s in the given gridstatic void findMaximumLength(int lis[][]){         // Increment the id    id++;    length = 0;    diameter = 0;     // Traverse the grid[]    for(int i = 0; i < row; i++)    {       for(int j = 0; j < col; j++)       {          if (lis[i][j] != 0)          {                             // Find start point of              // start dfs call              dfs(i, j, lis);              i = row;              break;          }       }    }     id++;    length = 0;    diameter = 0;     // DFS Traversal from cell (x, y)    dfs(x, y, lis);     // Print the maximum length    System.out.print(diameter);} // Driver Codepublic static void main(String[] args){         // Given grid[][]    int grid[][] = { { 0, 0, 0, 0, 0, 0, 0 },                     { 0, 1, 0, 1, 0, 0, 0 },                     { 0, 1, 0, 1, 0, 0, 0 },                     { 0, 1, 0, 1, 0, 1, 0 },                     { 0, 1, 1, 1, 1, 1, 0 },                     { 0, 0, 0, 1, 0, 0, 0 } };     // Function Call    findMaximumLength(grid);}} // This code is contributed by amal kumar choubey
# Python3 program for the above approachrow = 6col = 7  vis = [[0 for i in range(col + 1)]          for j in range(row + 1)]id = 0diameter = 0length = 0  # Keeps a track of directions# that is up, down, left, rightdx = [ -1, 1, 0, 0 ]dy = [ 0, 0, -1, 1 ]  # Function to perform the dfs traversaldef dfs(a, b, lis, x, y):         global id, length, diameter         # Mark the current node as visited    vis[a][b] = id      # Increment length from this node    length += 1      # Update the diameter length    if (length > diameter):        x = a        y = b        diameter = length         for j in range(4):          # Move to next cell in x-direction        cx = a + dx[j]          # Move to next cell in y-direction        cy = b + dy[j]          # Check if cell is invalid        # then continue        if (cx < 0 or cy < 0 or            cx >= row or cy >= col or            lis[cx][cy] == 0 or vis[cx][cy]):            continue             # Perform DFS on new cell        dfs(cx, cy, lis, x, y)         vis[a][b] = 0      # Decrement the length    length -= 1         return x, y # Function to find the maximum length of# connected 1s in the given griddef findMaximumLength(lis):         global id, length, diameter         x = 0    y = 0      # Increment the id    id += 1         length = 0    diameter = 0      # Traverse the grid[]    for i in range(row):        for j in range(col):            if (lis[i][j] != 0):                  # Find start point of                # start dfs call                x, y = dfs(i, j, lis, x, y)                i = row                break      id += 1    length = 0    diameter = 0      # DFS Traversal from cell (x, y)    x, y = dfs(x, y, lis, x, y)      # Print the maximum length    print(diameter) # Driver Codeif __name__=="__main__":         # Given grid[][]    grid = [ [ 0, 0, 0, 0, 0, 0, 0 ],             [ 0, 1, 0, 1, 0, 0, 0 ],             [ 0, 1, 0, 1, 0, 0, 0 ],             [ 0, 1, 0, 1, 0, 1, 0 ],             [ 0, 1, 1, 1, 1, 1, 0 ],             [ 0, 0, 0, 1, 0, 0, 0 ] ]      # Function Call    findMaximumLength(grid)  # This code is contributed by rutvik_56
// C# program for the above approachusing System; class GFG{     static readonly int row = 6;static readonly int col = 7;static int [,]vis = new int[row + 1, col + 1];static int id;static int diameter = 0, length = 0;static int x = 0, y = 0; // Keeps a track of directions// that is up, down, left, rightstatic int []dx = { -1, 1, 0, 0 };static int []dy = { 0, 0, -1, 1 }; // Function to perform the dfs traversalstatic void dfs(int a, int b, int [,]lis){         // Mark the current node as visited    vis[a, b] = id;     // Increment length from this node    length++;     // Update the diameter length    if (length > diameter)    {        x = a;        y = b;        diameter = length;    }     for(int j = 0; j < 4; j++)    {                 // Move to next cell in x-direction        int cx = a + dx[j];                 // Move to next cell in y-direction        int cy = b + dy[j];                 // Check if cell is invalid        // then continue        if (cx < 0 || cy < 0 ||            cx >= row || cy >= col ||            lis[cx, cy] == 0 || vis[cx, cy] > 0)        {            continue;        }                 // Perform DFS on new cell        dfs(cx, cy, lis);    }    vis[a, b] = 0;         // Decrement the length    length--;} // Function to find the maximum length of// connected 1s in the given gridstatic void findMaximumLength(int [,]lis){         // Increment the id    id++;    length = 0;    diameter = 0;     // Traverse the grid[]    for(int i = 0; i < row; i++)    {        for(int j = 0; j < col; j++)        {            if (lis[i, j] != 0)            {                                 // Find start point of                // start dfs call                dfs(i, j, lis);                i = row;                break;            }        }    }         id++;    length = 0;    diameter = 0;     // DFS Traversal from cell (x, y)    dfs(x, y, lis);     // Print the maximum length    Console.Write(diameter);} // Driver Codepublic static void Main(String[] args){         // Given grid[,]    int [,]grid = { { 0, 0, 0, 0, 0, 0, 0 },                    { 0, 1, 0, 1, 0, 0, 0 },                    { 0, 1, 0, 1, 0, 0, 0 },                    { 0, 1, 0, 1, 0, 1, 0 },                    { 0, 1, 1, 1, 1, 1, 0 },                    { 0, 0, 0, 1, 0, 0, 0 } };     // Function Call    findMaximumLength(grid);}} // This code is contributed by amal kumar choubey
<script> // JavaScript program to implement// the above approach let row = 6; let col = 7;let vis = new Array(row + 1);// Loop to create 2D array using 1D arrayfor (var i = 0; i < vis.length; i++) {    vis[i] = new Array(2);} let id = 0;let diameter = 0, length = 0;let x = 0, y = 0;  // Keeps a track of directions// that is up, down, left, rightlet dx = [ -1, 1, 0, 0 ];let dy = [ 0, 0, -1, 1 ];  // Function to perform the dfs traversalfunction dfs(a, b, lis){          // Mark the current node as visited    vis[a][b] = id;      // Increment length from this node    length++;      // Update the diameter length    if (length > diameter)    {        x = a;        y = b;        diameter = length;    }      for(let j = 0; j < 4; j++)    {                 // Move to next cell in x-direction       let cx = a + dx[j];                // Move to next cell in y-direction       let cy = b + dy[j];                // Check if cell is invalid       // then continue       if (cx < 0 || cy < 0 ||           cx >= row || cy >= col ||           lis[cx][cy] == 0 || vis[cx][cy] > 0)       {           continue;       }                // Perform DFS on new cell       dfs(cx, cy, lis);    }          vis[a][b] = 0;      // Decrement the length    length--;}  // Function to find the maximum length of// connected 1s in the given gridfunction findMaximumLength(lis){          // Increment the id    id++;    length = 0;    diameter = 0;      // Traverse the grid[]    for(let i = 0; i < row; i++)    {       for(let j = 0; j < col; j++)       {          if (lis[i][j] != 0)          {                              // Find start point of              // start dfs call              dfs(i, j, lis);              i = row;              break;          }       }    }      id++;    length = 0;    diameter = 0;      // DFS Traversal from cell (x, y)    dfs(x, y, lis);      // Print the maximum length    document.write(diameter);} // Driver code    // Given grid[][]    let grid = [[ 0, 0, 0, 0, 0, 0, 0 ],                     [ 0, 1, 0, 1, 0, 0, 0 ],                     [ 0, 1, 0, 1, 0, 0, 0 ],                     [ 0, 1, 0, 1, 0, 1, 0 ],                     [ 0, 1, 1, 1, 1, 1, 0 ],                     [ 0, 0, 0, 1, 0, 0, 0 ]];      // Function Call    findMaximumLength(grid);   // This code is contributed by sanjoy_62.</script>
9
Time Complexity: O(N*M)
where ‘N’ is the number of rows and ‘M’ is number of columns. 
Amal Kumar Choubey
rutvik_56
sanjoy_62
Kirti_Mangal
arorakashish0911
karandeep1234
BFS
binary-representation
DFS
Competitive Programming
Graph
Greedy
Matrix
Recursion
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