Find size of the largest ‘+’ formed by all ones in a binary matrix

Last Updated : 04 Jul, 2022

Given a N X N binary matrix, find the size of the largest ‘+’ formed by all 1s.

Example:

For above matrix, largest ‘+’ would be formed by highlighted part of size 17.

The idea is to maintain four auxiliary matrices left[][], right[][], top[][], bottom[][] to store consecutive 1’s in every direction. For each cell (i, j) in the input matrix, we store below information in these four matrices –

left(i, j) stores maximum number of
consecutive 1's to the left of cell (i, j) 
including cell (i, j).

right(i, j) stores maximum number of
consecutive 1's to the right of cell (i, j) 
including cell (i, j).

top(i, j) stores maximum number of
consecutive 1's at top of cell (i, j) 
including cell (i, j).

bottom(i, j) stores maximum number of
consecutive 1's at bottom of cell (i, j) 
including cell (i, j).

After computing value for each cell of above matrices, the largest + would be formed by a cell of input matrix that has maximum value by considering minimum of (left(i, j), right(i, j), top(i, j), bottom(i, j) )

We can use Dynamic Programming to compute the total amount of consecutive 1’s in every direction.

if mat(i, j) == 1
    left(i, j) = left(i, j - 1) + 1
else left(i, j) = 0

if mat(i, j) == 1
    top(i, j) = top(i - 1, j) + 1;
else
    top(i, j) = 0;

if mat(i, j) == 1
    bottom(i, j) = bottom(i + 1, j) + 1;
else
    bottom(i, j) = 0;    

if mat(i, j) == 1
    right(i, j) = right(i, j + 1) + 1;
else
    right(i, j) = 0;

Below is the implementation of above idea :

C++

// C++ program to find the size of the largest '+'
// formed by all 1's in given binary matrix
#include <bits/stdc++.h>
using namespace std;
 
// size of binary square matrix
#define N 10
 
// Function to find the size of the largest '+'
// formed by all 1's in given binary matrix
int findLargestPlus(int mat[N][N])
{
    // left[j][j], right[i][j], top[i][j] and
    // bottom[i][j] store maximum number of
    // consecutive 1's present to the left,
    // right, top and bottom of mat[i][j] including
    // cell(i, j) respectively
    int left[N][N], right[N][N], top[N][N],
        bottom[N][N];
 
    // initialize above four matrix
    for (int i = 0; i < N; i++)
    {
        // initialize first row of top
        top[0][i] = mat[0][i];
 
        // initialize last row of bottom
        bottom[N - 1][i] = mat[N - 1][i];
 
        // initialize first column of left
        left[i][0] = mat[i][0];
 
        // initialize last column of right
        right[i][N - 1] = mat[i][N - 1];
    }
 
    // fill all cells of above four matrix
    for (int i = 0; i < N; i++)
    {
        for (int j = 1; j < N; j++)
        {
            // calculate left matrix (filled left to right)
            if (mat[i][j] == 1)
                left[i][j] = left[i][j - 1] + 1;
            else
                left[i][j] = 0;
 
            // calculate top matrix
            if (mat[j][i] == 1)
                top[j][i] = top[j - 1][i] + 1;
            else
                top[j][i] = 0;
 
            // calculate new value of j to calculate
            // value of bottom(i, j) and right(i, j)
            j = N - 1 - j;
 
            // calculate bottom matrix
            if (mat[j][i] == 1)
                bottom[j][i] = bottom[j + 1][i] + 1;
            else
                bottom[j][i] = 0;
 
            // calculate right matrix
            if (mat[i][j] == 1)
                right[i][j] = right[i][j + 1] + 1;
            else
                right[i][j] = 0;
 
            // revert back to old j
            j = N - 1 - j;
        }
    }
 
    // n stores length of longest + found so far
    int n = 0;
 
    // compute longest +
    for (int i = 0; i < N; i++)
    {
        for (int j = 0; j < N; j++)
        {
            // find minimum of left(i, j), right(i, j),
            // top(i, j), bottom(i, j)
            int len = min(min(top[i][j], bottom[i][j]),
                          min(left[i][j], right[i][j]));
 
            // largest + would be formed by a cell that
            // has maximum value
            if(len > n)
                n = len;
        }
    }
 
    // 4 directions of length n - 1 and 1 for middle cell
    if (n)
       return 4 * (n - 1) + 1;
 
    // matrix contains all 0's
    return 0;
}
 
/* Driver function to test above functions */
int main()
{
    // Binary Matrix of size N
    int mat[N][N] =
    {
        { 1, 0, 1, 1, 1, 1, 0, 1, 1, 1 },
        { 1, 0, 1, 0, 1, 1, 1, 0, 1, 1 },
        { 1, 1, 1, 0, 1, 1, 0, 1, 0, 1 },
        { 0, 0, 0, 0, 1, 0, 0, 1, 0, 0 },
        { 1, 1, 1, 0, 1, 1, 1, 1, 1, 1 },
        { 1, 1, 1, 1, 1, 1, 1, 1, 1, 0 },
        { 1, 0, 0, 0, 1, 0, 0, 1, 0, 1 },
        { 1, 0, 1, 1, 1, 1, 0, 0, 1, 1 },
        { 1, 1, 0, 0, 1, 0, 1, 0, 0, 1 },
        { 1, 0, 1, 1, 1, 1, 0, 1, 0, 0 }
    };
 
    cout << findLargestPlus(mat);
 
    return 0;
}

C

// C program to find the size of the largest '+'
// formed by all 1's in given binary matrix
#include <stdio.h>
 
// size of binary square matrix
#define N 10
 
int min(int a,int b)
{
   int min = a;
   if(min > b)
     min = b;
   return min;
}
 
// Function to find the size of the largest '+'
// formed by all 1's in given binary matrix
int findLargestPlus(int mat[N][N])
{
    // left[j][j], right[i][j], top[i][j] and
    // bottom[i][j] store maximum number of
    // consecutive 1's present to the left,
    // right, top and bottom of mat[i][j] including
    // cell(i, j) respectively
    int left[N][N], right[N][N], top[N][N], bottom[N][N];
 
    // initialize above four matrix
    for (int i = 0; i < N; i++)
    {
        // initialize first row of top
        top[0][i] = mat[0][i];
 
        // initialize last row of bottom
        bottom[N - 1][i] = mat[N - 1][i];
 
        // initialize first column of left
        left[i][0] = mat[i][0];
 
        // initialize last column of right
        right[i][N - 1] = mat[i][N - 1];
    }
 
    // fill all cells of above four matrix
    for (int i = 0; i < N; i++)
    {
        for (int j = 1; j < N; j++)
        {
            // calculate left matrix (filled left to right)
            if (mat[i][j] == 1)
                left[i][j] = left[i][j - 1] + 1;
            else
                left[i][j] = 0;
 
            // calculate top matrix
            if (mat[j][i] == 1)
                top[j][i] = top[j - 1][i] + 1;
            else
                top[j][i] = 0;
 
            // calculate new value of j to calculate
            // value of bottom(i, j) and right(i, j)
            j = N - 1 - j;
 
            // calculate bottom matrix
            if (mat[j][i] == 1)
                bottom[j][i] = bottom[j + 1][i] + 1;
            else
                bottom[j][i] = 0;
 
            // calculate right matrix
            if (mat[i][j] == 1)
                right[i][j] = right[i][j + 1] + 1;
            else
                right[i][j] = 0;
 
            // revert back to old j
            j = N - 1 - j;
        }
    }
 
    // n stores length of longest + found so far
    int n = 0;
 
    // compute longest +
    for (int i = 0; i < N; i++)
    {
        for (int j = 0; j < N; j++)
        {
            // find minimum of left(i, j), right(i, j),
            // top(i, j), bottom(i, j)
            int len = min(min(top[i][j], bottom[i][j]),
                          min(left[i][j], right[i][j]));
 
            // largest + would be formed by a cell that
            // has maximum value
            if(len > n)
                n = len;
        }
    }
 
    // 4 directions of length n - 1 and 1 for middle cell
    if (n)
       return 4 * (n - 1) + 1;
 
    // matrix contains all 0's
    return 0;
}
 
/* Driver function to test above functions */
int main()
{
    // Binary Matrix of size N
    int mat[N][N] =
    {
        { 1, 0, 1, 1, 1, 1, 0, 1, 1, 1 },
        { 1, 0, 1, 0, 1, 1, 1, 0, 1, 1 },
        { 1, 1, 1, 0, 1, 1, 0, 1, 0, 1 },
        { 0, 0, 0, 0, 1, 0, 0, 1, 0, 0 },
        { 1, 1, 1, 0, 1, 1, 1, 1, 1, 1 },
        { 1, 1, 1, 1, 1, 1, 1, 1, 1, 0 },
        { 1, 0, 0, 0, 1, 0, 0, 1, 0, 1 },
        { 1, 0, 1, 1, 1, 1, 0, 0, 1, 1 },
        { 1, 1, 0, 0, 1, 0, 1, 0, 0, 1 },
        { 1, 0, 1, 1, 1, 1, 0, 1, 0, 0 }
    };
 
    printf("%d",findLargestPlus(mat));
 
    return 0;
}
 
// This code is contributed by kothavvsaakash.

Java

// Java program to find the size of the largest '+'
// formed by all 1's in given binary matrix
 
import java.io.*;
 
class GFG {
     
    // size of binary square matrix
    static int N = 10;
 
    // Function to find the size of the largest '+'
    // formed by all 1's in given binary matrix
    static int findLargestPlus(int mat[][])
    {
         
        // left[j][j], right[i][j], top[i][j] and
        // bottom[i][j] store maximum number of
        // consecutive 1's present to the left,
        // right, top and bottom of mat[i][j] 
        // including cell(i, j) respectively
        int left[][] = new int[N][N];
        int right[][] = new int[N][N];
        int top[][] = new int[N][N];
        int bottom[][] = new int[N][N];
 
        // initialize above four matrix
        for (int i = 0; i < N; i++) {
             
            // initialize first row of top
            top[0][i] = mat[0][i];
 
            // initialize last row of bottom
            bottom[N - 1][i] = mat[N - 1][i];
 
            // initialize first column of left
            left[i][0] = mat[i][0];
 
            // initialize last column of right
            right[i][N - 1] = mat[i][N - 1];
        }
 
        // fill all cells of above four matrix
        for (int i = 0; i < N; i++) {
            for (int j = 1; j < N; j++) {
                 
                // calculate left matrix 
                // (filled left to right)
                if (mat[i][j] == 1)
                    left[i][j] = left[i][j - 1] + 1;
                else
                    left[i][j] = 0;
 
                // calculate top matrix
                if (mat[j][i] == 1)
                    top[j][i] = top[j - 1][i] + 1;
                else
                    top[j][i] = 0;
 
                // calculate new value of j to 
                // calculate value of bottom(i, j) 
                // and right(i, j)
                j = N - 1 - j;
 
                // calculate bottom matrix
                if (mat[j][i] == 1)
                    bottom[j][i] = bottom[j + 1][i] + 1;
                else
                    bottom[j][i] = 0;
 
                // calculate right matrix
                if (mat[i][j] == 1)
                    right[i][j] = right[i][j + 1] + 1;
                else
                    right[i][j] = 0;
 
                // revert back to old j
                j = N - 1 - j;
            }
        }
 
        // n stores length of longest + found so far
        int n = 0;
 
        // compute longest +
        for (int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++) {
                // find minimum of left(i, j), 
                // right(i, j), top(i, j), 
                // bottom(i, j)
                int len = Math.min(Math.min(top[i][j], 
                    bottom[i][j]),Math.min(left[i][j], 
                                        right[i][j]));
 
                // largest + would be formed by a
                // cell that has maximum value
                if (len > n)
                    n = len;
            }
        }
 
        // 4 directions of length n - 1 and 1 for
        // middle cell
        if (n > 0)
            return 4 * (n - 1) + 1;
 
        // matrix contains all 0's
        return 0;
    }
 
    /* Driver function to test above functions */
    public static void main(String[] args)
    {
         
        // Binary Matrix of size N
        int mat[][] = {
            { 1, 0, 1, 1, 1, 1, 0, 1, 1, 1 },
            { 1, 0, 1, 0, 1, 1, 1, 0, 1, 1 },
            { 1, 1, 1, 0, 1, 1, 0, 1, 0, 1 },
            { 0, 0, 0, 0, 1, 0, 0, 1, 0, 0 },
            { 1, 1, 1, 0, 1, 1, 1, 1, 1, 1 },
            { 1, 1, 1, 1, 1, 1, 1, 1, 1, 0 },
            { 1, 0, 0, 0, 1, 0, 0, 1, 0, 1 },
            { 1, 0, 1, 1, 1, 1, 0, 0, 1, 1 },
            { 1, 1, 0, 0, 1, 0, 1, 0, 0, 1 },
            { 1, 0, 1, 1, 1, 1, 0, 1, 0, 0 }
        };
        System.out.println(findLargestPlus(mat));
    }
}
 
// This code is contributed by vt_m.

Python 3

# Python 3 program to find the size 
# of the largest '+' formed by all 
# 1's in given binary matrix
 
# size of binary square matrix
N = 10
 
# Function to find the size of the 
# largest '+' formed by all 1's in
# given binary matrix
def findLargestPlus(mat):
 
    # left[j][j], right[i][j], top[i][j] and
    # bottom[i][j] store maximum number of
    # consecutive 1's present to the left,
    # right, top and bottom of mat[i][j] including
    # cell(i, j) respectively
    left = [[0 for x in range(N)] 
               for y in range(N)]
    right = [[0 for x in range(N)] 
                for y in range(N)]
    top = [[0 for x in range(N)] 
              for y in range(N)]
    bottom = [[0 for x in range(N)] 
                 for y in range(N)]
 
    # initialize above four matrix
    for i in range(N):
         
        # initialize first row of top
        top[0][i] = mat[0][i]
 
        # initialize last row of bottom
        bottom[N - 1][i] = mat[N - 1][i]
 
        # initialize first column of left
        left[i][0] = mat[i][0]
 
        # initialize last column of right
        right[i][N - 1] = mat[i][N - 1]
 
    # fill all cells of above four matrix
    for i in range(N):
        for j in range(1, N):
             
            # calculate left matrix (filled 
            # left to right)
            if (mat[i][j] == 1):
                left[i][j] = left[i][j - 1] + 1
            else:
                left[i][j] = 0
 
            # calculate top matrix
            if (mat[j][i] == 1):
                top[j][i] = top[j - 1][i] + 1
            else:
                top[j][i] = 0
 
            # calculate new value of j to calculate
            # value of bottom(i, j) and right(i, j)
            j = N - 1 - j
 
            # calculate bottom matrix
            if (mat[j][i] == 1):
                bottom[j][i] = bottom[j + 1][i] + 1
            else:
                bottom[j][i] = 0
 
            # calculate right matrix
            if (mat[i][j] == 1):
                right[i][j] = right[i][j + 1] + 1
            else:
                right[i][j] = 0
 
            # revert back to old j
            j = N - 1 - j
 
    # n stores length of longest '+' 
    # found so far
    n = 0
 
    # compute longest +
    for i in range(N):
        for j in range(N):
             
            # find minimum of left(i, j), 
            # right(i, j), top(i, j), bottom(i, j)
            l = min(min(top[i][j], bottom[i][j]),
                    min(left[i][j], right[i][j]))
 
            # largest + would be formed by 
            # a cell that has maximum value
            if(l > n):
                n = l
 
    # 4 directions of length n - 1 and 1 
    # for middle cell
    if (n):
        return 4 * (n - 1) + 1
 
    # matrix contains all 0's
    return 0
 
# Driver Code
if __name__=="__main__":
     
    # Binary Matrix of size N
    mat = [ [ 1, 0, 1, 1, 1, 1, 0, 1, 1, 1 ],
            [ 1, 0, 1, 0, 1, 1, 1, 0, 1, 1 ],
            [ 1, 1, 1, 0, 1, 1, 0, 1, 0, 1 ],
            [ 0, 0, 0, 0, 1, 0, 0, 1, 0, 0 ],
            [ 1, 1, 1, 0, 1, 1, 1, 1, 1, 1 ],
            [ 1, 1, 1, 1, 1, 1, 1, 1, 1, 0 ],
            [ 1, 0, 0, 0, 1, 0, 0, 1, 0, 1 ],
            [ 1, 0, 1, 1, 1, 1, 0, 0, 1, 1 ],
            [ 1, 1, 0, 0, 1, 0, 1, 0, 0, 1 ],
            [ 1, 0, 1, 1, 1, 1, 0, 1, 0, 0 ]]
 
    print(findLargestPlus(mat))
 
# This code is contributed by ChitraNayal

C#

// C# program to find the size of the largest '+'
// formed by all 1's in given binary matrix
using System;
 
class GFG {
     
    // size of binary square matrix
    static int N = 10;
 
    // Function to find the size of the largest '+'
    // formed by all 1's in given binary matrix
    static int findLargestPlus(int [,] mat)
    {
         
        // left[j][j], right[i][j], top[i][j] and
        // bottom[i][j] store maximum number of
        // consecutive 1's present to the left,
        // right, top and bottom of mat[i][j] 
        // including cell(i, j) respectively
        int [,] left = new int[N,N];
        int [,] right = new int[N,N];
        int [,] top = new int[N,N];
        int [,] bottom = new int[N,N];
 
        // initialize above four matrix
        for (int i = 0; i < N; i++) {
             
            // initialize first row of top
            top[0,i] = mat[0,i];
 
            // initialize last row of bottom
            bottom[N - 1,i] = mat[N - 1,i];
 
            // initialize first column of left
            left[i,0] = mat[i,0];
 
            // initialize last column of right
            right[i,N - 1] = mat[i,N - 1];
        }
 
        // fill all cells of above four matrix
        for (int i = 0; i < N; i++) {
            for (int j = 1; j < N; j++) {
                 
                // calculate left matrix 
                // (filled left to right)
                if (mat[i,j] == 1)
                    left[i,j] = left[i,j - 1] + 1;
                else
                    left[i,j] = 0;
 
                // calculate top matrix
                if (mat[j,i] == 1)
                    top[j,i] = top[j - 1,i] + 1;
                else
                    top[j,i] = 0;
 
                // calculate new value of j to 
                // calculate value of bottom(i, j) 
                // and right(i, j)
                j = N - 1 - j;
 
                // calculate bottom matrix
                if (mat[j,i] == 1)
                    bottom[j,i] = bottom[j + 1,i] + 1;
                else
                    bottom[j,i] = 0;
 
                // calculate right matrix
                if (mat[i,j] == 1)
                    right[i,j] = right[i,j + 1] + 1;
                else
                    right[i,j] = 0;
 
                // revert back to old j
                j = N - 1 - j;
            }
        }
 
        // n stores length of longest + found so far
        int n = 0;
 
        // compute longest +
        for (int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++) {
 
                // find minimum of left(i, j), 
                // right(i, j), top(i, j), 
                // bottom(i, j)
                int len = Math.Min(Math.Min(top[i,j], 
                    bottom[i,j]),Math.Min(left[i,j], 
                                        right[i,j]));
 
                // largest + would be formed by a
                // cell that has maximum value
                if (len > n)
                    n = len;
            }
        }
 
        // 4 directions of length n - 1 and 1 for
        // middle cell
        if (n > 0)
            return 4 * (n - 1) + 1;
 
        // matrix contains all 0's
        return 0;
    }
 
    /* Driver function to test above functions */
    public static void Main()
    {
         
        // Binary Matrix of size N
        int [,]mat = {
            { 1, 0, 1, 1, 1, 1, 0, 1, 1, 1 },
            { 1, 0, 1, 0, 1, 1, 1, 0, 1, 1 },
            { 1, 1, 1, 0, 1, 1, 0, 1, 0, 1 },
            { 0, 0, 0, 0, 1, 0, 0, 1, 0, 0 },
            { 1, 1, 1, 0, 1, 1, 1, 1, 1, 1 },
            { 1, 1, 1, 1, 1, 1, 1, 1, 1, 0 },
            { 1, 0, 0, 0, 1, 0, 0, 1, 0, 1 },
            { 1, 0, 1, 1, 1, 1, 0, 0, 1, 1 },
            { 1, 1, 0, 0, 1, 0, 1, 0, 0, 1 },
            { 1, 0, 1, 1, 1, 1, 0, 1, 0, 0 }
        };
 
        Console.Write(findLargestPlus(mat));
    }
}
 
// This code is contributed by KRV.

PHP

<?php
// PHP program to find the size of the 
// largest '+' formed by all 1's in
// given binary matrix
 
// size of binary square matrix
$N = 10;
 
// Function to find the size of the largest '+'
// formed by all 1's in given binary matrix
function findLargestPlus($mat)
{
    global $N ;
     
    // left[j][j], right[i][j], top[i][j] and
    // bottom[i][j] store maximum number of
    // consecutive 1's present to the left,
    // right, top and bottom of mat[i][j] 
    // including cell(i, j) respectively
    $left[$N][$N] = array(); 
    $right[$N][$N] = array();
    $top[$N][$N] = array();
    $bottom[$N][$N] = array();
 
    // initialize above four matrix
    for ($i = 0; $i < $N; $i++)
    {
        // initialize first row of top
        $top[0][$i] = $mat[0][$i];
 
        // initialize last row of bottom
        $bottom[$N - 1][$i] = $mat[$N - 1][$i];
 
        // initialize first column of left
        $left[$i][0] = $mat[$i][0];
 
        // initialize last column of right
        $right[$i][$N - 1] = $mat[$i][$N - 1];
    }
 
    // fill all cells of above four matrix
    for ( $i = 0; $i < $N; $i++)
    {
        for ($j = 1; $j < $N; $j++)
        {
            // calculate left matrix (filled left to right)
            if ($mat[$i][$j] == 1)
                $left[$i][$j] = $left[$i][$j - 1] + 1;
            else
                $left[$i][$j] = 0;
 
            // calculate top matrix
            if ($mat[$j][$i] == 1)
                $top[$j][$i] = $top[$j - 1][$i] + 1;
            else
                $top[$j][$i] = 0;
 
            // calculate new value of j to calculate
            // value of bottom(i, j) and right(i, j)
            $j = $N - 1 - $j;
 
            // calculate bottom matrix
            if ($mat[$j][$i] == 1)
                $bottom[$j][$i] = $bottom[$j + 1][$i] + 1;
            else
                $bottom[$j][$i] = 0;
 
            // calculate right matrix
            if ($mat[$i][$j] == 1)
                $right[$i][$j] = $right[$i][$j + 1] + 1;
            else
                $right[$i][$j] = 0;
 
            // revert back to old j
            $j = $N - 1 - $j;
        }
    }
 
    // n stores length of longest + found so far
    $n = 0;
 
    // compute longest +
    for ($i = 0; $i < $N; $i++)
    {
        for ($j = 0; $j < $N; $j++)
        {
            // find minimum of left(i, j), right(i, j),
            // top(i, j), bottom(i, j)
            $len = min(min($top[$i][$j], $bottom[$i][$j]),
                       min($left[$i][$j], $right[$i][$j]));
 
            // largest + would be formed by a 
            // cell that has maximum value
            if($len > $n)
                $n = $len;
        }
    }
 
    // 4 directions of length n - 1 and 1
    // for middle cell
    if ($n)
    return 4 * ($n - 1) + 1;
 
    // matrix contains all 0's
    return 0;
}
 
// Driver Code
 
// Binary Matrix of size N
$mat = array(array(1, 0, 1, 1, 1, 1, 0, 1, 1, 1),
             array(1, 0, 1, 0, 1, 1, 1, 0, 1, 1),
             array(1, 1, 1, 0, 1, 1, 0, 1, 0, 1),
             array(0, 0, 0, 0, 1, 0, 0, 1, 0, 0),
             array(1, 1, 1, 0, 1, 1, 1, 1, 1, 1),
             array(1, 1, 1, 1, 1, 1, 1, 1, 1, 0),
             array(1, 0, 0, 0, 1, 0, 0, 1, 0, 1),
             array(1, 0, 1, 1, 1, 1, 0, 0, 1, 1),
             array(1, 1, 0, 0, 1, 0, 1, 0, 0, 1),
             array(1, 0, 1, 1, 1, 1, 0, 1, 0, 0));
 
echo findLargestPlus($mat);
 
// This code is contributed by Sach_Code
?>

Javascript

<script>
// Javascript program to find the size of the largest '+'
// formed by all 1's in given binary matrix
 
    // size of binary square matrix
    let N = 10;
     
    // Function to find the size of the largest '+'
    // formed by all 1's in given binary matrix
    function findLargestPlus(mat)
    {
        // left[j][j], right[i][j], top[i][j] and
        // bottom[i][j] store maximum number of
        // consecutive 1's present to the left,
        // right, top and bottom of mat[i][j] 
        // including cell(i, j) respectively
        let left = new Array(N);
        for(let i = 0; i < N; i++)
        {
            left[i] = new Array(N);
        }
         
        let right = new Array(N);
        for(let i = 0; i < N; i++)
        {
            right[i] = new Array(N);
        }
         
        let top = new Array(N);
        for(let i = 0; i < N; i++)
        {
            top[i] = new Array(N);
        }
         
        let bottom = new Array(N);
        for(let i = 0; i < N; i++)
        {
            bottom[i] = new Array(N);
        }
         
        for(let i = 0; i < N; i++)
        {
            for(let j = 0; j < N; j++)
            {
                left[i][j] = 0;
                right[i][j] = 0;
                top[i][j] = 0;
                bottom[i][j] = 0;
            }
        }
         
        // initialize above four matrix
        for (let i = 0; i < N; i++) {
               
            // initialize first row of top
            top[0][i] = mat[0][i];
   
            // initialize last row of bottom
            bottom[N - 1][i] = mat[N - 1][i];
   
            // initialize first column of left
            left[i][0] = mat[i][0];
   
            // initialize last column of right
            right[i][N - 1] = mat[i][N - 1];
        }
         
        // fill all cells of above four matrix
        for (let i = 0; i < N; i++) {
            for (let j = 1; j < N; j++) {
                   
                // calculate left matrix 
                // (filled left to right)
                if (mat[i][j] == 1)
                    left[i][j] = left[i][j - 1] + 1;
                else
                    left[i][j] = 0;
   
                // calculate top matrix
                if (mat[j][i] == 1)
                    top[j][i] = top[j - 1][i] + 1;
                else
                    top[j][i] = 0;
   
                // calculate new value of j to 
                // calculate value of bottom(i, j) 
                // and right(i, j)
                j = N - 1 - j;
   
                // calculate bottom matrix
                if (mat[j][i] == 1)
                    bottom[j][i] = bottom[j + 1][i] + 1;
                else
                    bottom[j][i] = 0;
   
                // calculate right matrix
                if (mat[i][j] == 1)
                    right[i][j] = right[i][j + 1] + 1;
                else
                    right[i][j] = 0;
   
                // revert back to old j
                j = N - 1 - j;
            }
        }
   
        // n stores length of longest + found so far
        let n = 0;
   
        // compute longest +
        for (let i = 0; i < N; i++) {
            for (let j = 0; j < N; j++) {
                // find minimum of left(i, j), 
                // right(i, j), top(i, j), 
                // bottom(i, j)
                let len = Math.min(Math.min(top[i][j], 
                    bottom[i][j]),Math.min(left[i][j], 
                                        right[i][j]));
   
                // largest + would be formed by a
                // cell that has maximum value
                if (len > n)
                    n = len;
            }
        }
   
        // 4 directions of length n - 1 and 1 for
        // middle cell
        if (n > 0)
            return 4 * (n - 1) + 1;
   
        // matrix contains all 0's
        return 0;
    }
     
    /* Driver function to test above functions */
    let mat = [ [ 1, 0, 1, 1, 1, 1, 0, 1, 1, 1 ],
            [ 1, 0, 1, 0, 1, 1, 1, 0, 1, 1 ],
            [ 1, 1, 1, 0, 1, 1, 0, 1, 0, 1 ],
            [ 0, 0, 0, 0, 1, 0, 0, 1, 0, 0 ],
            [ 1, 1, 1, 0, 1, 1, 1, 1, 1, 1 ],
            [ 1, 1, 1, 1, 1, 1, 1, 1, 1, 0 ],
            [ 1, 0, 0, 0, 1, 0, 0, 1, 0, 1 ],
            [ 1, 0, 1, 1, 1, 1, 0, 0, 1, 1 ],
            [ 1, 1, 0, 0, 1, 0, 1, 0, 0, 1 ],
            [ 1, 0, 1, 1, 1, 1, 0, 1, 0, 0 ]]
    document.write(findLargestPlus(mat));
     
    // This code is contributed by avanitrachhadiya2155
</script>

Output

Time complexity of above solution is O(n²).
Auxiliary space used by the program is O(n²).

Suggest improvement

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