Find size of the largest ‘+’ formed by all ones in a binary matrix
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 matrixint 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 10int 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 matrixint 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 matriximport 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 matrixN = 10# Function to find the size of the# largest '+' formed by all 1's in# given binary matrixdef 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 Codeif __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 matrixusing 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$N = 10;// Function to find the size of the largest '+'// formed by all 1's in given binary matrixfunction findLargestPlus($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
17
Time complexity of above solution is O(n2).
Auxiliary space used by the program is O(n2).
This article is contributed by Aditya Goel. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Login to comment...