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Length of longest connected 1’s in a Binary Grid
  • Difficulty Level : Hard
  • Last Updated : 04 May, 2021

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:
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:
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++




// C++ program for the above approach
 
#include <bits/stdc++.h>
#define row 6
#define col 7
using namespace std;
 
int vis[row + 1][col + 1], id;
int diameter = 0, length = 0;
 
// Keeps a track of directions
// that is up, down, left, right
int dx[] = { -1, 1, 0, 0 };
int dy[] = { 0, 0, -1, 1 };
 
// Function to perform the dfs traversal
void 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 grid
void 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 Code
int 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




// Java program for the above approach
import 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, right
static int dx[] = { -1, 1, 0, 0 };
static int dy[] = { 0, 0, -1, 1 };
 
// Function to perform the dfs traversal
static 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 grid
static 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 Code
public 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




# Python3 program for the above approach
row = 6
col = 7
  
vis = [[0 for i in range(col + 1)]
          for j in range(row + 1)]
id = 0
diameter = 0
length = 0
  
# Keeps a track of directions
# that is up, down, left, right
dx = [ -1, 1, 0, 0 ]
dy = [ 0, 0, -1, 1 ]
  
# Function to perform the dfs traversal
def 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 grid
def 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 Code
if __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#




// C# program for the above approach
using 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, right
static int []dx = { -1, 1, 0, 0 };
static int []dy = { 0, 0, -1, 1 };
 
// Function to perform the dfs traversal
static 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 grid
static 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 Code
public 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

Javascript




<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 array
for (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, right
let dx = [ -1, 1, 0, 0 ];
let dy = [ 0, 0, -1, 1 ];
  
// Function to perform the dfs traversal
function 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 grid
function 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 polet 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);
  
    // Prlet 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>
Output: 
9

Time Complexity: O(rows + cols)
 

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