Count zeros in a row wise and column wise sorted matrix
Given a N x N binary matrix (elements in matrix can be either 1 or 0) where each row and column of the matrix is sorted in ascending order, count number of 0s present in it.
Expected time complexity is O(N).
Examples:
Input:
[0, 0, 0, 0, 1]
[0, 0, 0, 1, 1]
[0, 1, 1, 1, 1]
[1, 1, 1, 1, 1]
[1, 1, 1, 1, 1]
Output: 8
Input:
[0, 0]
[0, 0]
Output: 4
Input:
[1, 1, 1, 1]
[1, 1, 1, 1]
[1, 1, 1, 1]
[1, 1, 1, 1]
Output: 0
The idea is very simple. We start from the bottom-left corner of the matrix and repeat below steps until we find the top or right edge of the matrix.
- Decrement row index until we find a 0.
- Add number of 0s in current column i.e. current row index + 1 to the result and move right to next column (Increment col index by 1).
The above logic will work since the matrix is row-wise and column-wise sorted. The logic will also work for any matrix containing non-negative integers.
Below is the implementation of above idea :
C++
#include <iostream>
using namespace std;
#define N 5
int countZeroes( int mat[N][N])
{
int row = N - 1, col = 0;
int count = 0;
while (col < N)
{
while (mat[row][col])
if (--row < 0)
return count;
count += (row + 1);
col++;
}
return count;
}
int main()
{
int mat[N][N] =
{
{ 0, 0, 0, 0, 1 },
{ 0, 0, 0, 1, 1 },
{ 0, 1, 1, 1, 1 },
{ 1, 1, 1, 1, 1 },
{ 1, 1, 1, 1, 1 }
};
cout << countZeroes(mat);
return 0;
}
|
C
#include <stdio.h>
#define N 5
int countZeroes( int mat[N][N])
{
int row = N - 1, col = 0;
int count = 0;
while (col < N)
{
while (mat[row][col])
if (--row < 0)
return count;
count += (row + 1);
col++;
}
return count;
}
int main()
{
int mat[N][N] =
{
{ 0, 0, 0, 0, 1 },
{ 0, 0, 0, 1, 1 },
{ 0, 1, 1, 1, 1 },
{ 1, 1, 1, 1, 1 },
{ 1, 1, 1, 1, 1 }
};
printf ( "%d" ,countZeroes(mat));
return 0;
}
|
Java
import java.io.*;
class GFG
{
public static int N = 5 ;
static int countZeroes( int mat[][])
{
int row = N - 1 , col = 0 ;
int count = 0 ;
while (col < N)
{
while (mat[row][col] > 0 )
if (--row < 0 )
return count;
count += (row + 1 );
col++;
}
return count;
}
public static void main (String[] args)
{
int mat[][] = { { 0 , 0 , 0 , 0 , 1 },
{ 0 , 0 , 0 , 1 , 1 },
{ 0 , 1 , 1 , 1 , 1 },
{ 1 , 1 , 1 , 1 , 1 },
{ 1 , 1 , 1 , 1 , 1 } };
System.out.println(countZeroes(mat));
}
}
|
Python
def countZeroes(mat):
N = 5 ;
row = N - 1 ;
col = 0 ;
count = 0 ;
while (col < N):
while (mat[row][col]):
if (row < 0 ):
return count;
row = row - 1 ;
count = count + (row + 1 );
col = col + 1 ;
return count;
mat = [[ 0 , 0 , 0 , 0 , 1 ],
[ 0 , 0 , 0 , 1 , 1 ],
[ 0 , 1 , 1 , 1 , 1 ],
[ 1 , 1 , 1 , 1 , 1 ],
[ 1 , 1 , 1 , 1 , 1 ]];
print ( countZeroes(mat));
|
C#
using System;
class GFG
{
public static int N = 5;
static int countZeroes( int [,] mat)
{
int row = N - 1, col = 0;
int count = 0;
while (col < N)
{
while (mat[row,col] > 0)
if (--row < 0)
return count;
count += (row + 1);
col++;
}
return count;
}
public static void Main ()
{
int [,] mat = { { 0, 0, 0, 0, 1 },
{ 0, 0, 0, 1, 1 },
{ 0, 1, 1, 1, 1 },
{ 1, 1, 1, 1, 1 },
{ 1, 1, 1, 1, 1 } };
Console.WriteLine(countZeroes(mat));
}
}
|
PHP
<?php
function countZeroes( $mat )
{
$N = 5;
$row = $N - 1;
$col = 0;
$count = 0;
while ( $col < $N )
{
while ( $mat [ $row ][ $col ])
if (-- $row < 0)
return $count ;
$count += ( $row + 1);
$col ++;
}
return $count ;
}
$mat = array ( array (0, 0, 0, 0, 1),
array (0, 0, 0, 1, 1),
array (0, 1, 1, 1, 1),
array (1, 1, 1, 1, 1),
array (1, 1, 1, 1, 1));
echo countZeroes( $mat );
?>
|
Javascript
<script>
let N = 5;
function countZeroes(mat)
{
let row = N - 1, col = 0;
let count = 0;
while (col < N)
{
while (mat[row][col] > 0)
if (--row < 0)
return count;
count += (row + 1);
col++;
}
return count;
}
let mat = [[ 0, 0, 0, 0, 1 ],
[ 0, 0, 0, 1, 1 ],
[ 0, 1, 1, 1, 1 ],
[ 1, 1, 1, 1, 1 ],
[ 1, 1, 1, 1, 1 ]];
document.write(countZeroes(mat));
</script>
|
Time complexity of above solution is O(n) since the solution follows single path from bottom-left corner to top or right edge of the matrix.
Auxiliary space used by the program is O(1). since no extra space has been taken.
Last Updated :
19 Aug, 2022
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