# Count Negative Numbers in a Column-Wise and Row-Wise Sorted Matrix

Find the number of negative numbers in a column-wise / row-wise sorted matrix M[][]. Suppose M has n rows and m columns.

Example:

```Input:  M =  [-3, -2, -1,  1]
[-2,  2,  3,  4]
[4,   5,  7,  8]
Output : 4
We have 4 negative numbers in this matrix
```

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Naive Solution
Here’s a naive, non-optimal solution.

We start from the top left corner and count the number of negative numbers one by one, from left to right and top to bottom.

With the given example:

```[-3, -2, -1,  1]
[-2,  2,  3,  4]
[4,   5,  7,  8]

Evaluation process

[→,  →,  →,  1]
[→,  2,  3,  4]
[4,  5,  7,  8]```

Below is the implementation of above idea.

## C++

```// CPP implementation of Naive method
// to count of negative numbers in
// M[n][m]
#include <bits/stdc++.h>
using namespace std;

int countNegative(int M[][4], int n, int m)
{
int count = 0;

// Follow the path shown using
// arrows above
for(int i = 0; i < n; i++)
{
for(int j = 0; j < m; j++)
{
if( M[i][j] < 0 )
count += 1;

// no more negative numbers
// in this row
else
break;
}
}
return count;
}

// Driver program to test above functions
int main ()
{
int M[3][4] = { {-3, -2, -1, 1},
{-2, 2, 3, 4},
{4, 5, 7, 8} };

cout << countNegative(M, 3, 4);
return 0;
}
// This code is contributed by Niteesh Kumar
```

## Java

```//Java implementation of Naive method
// to count of negative numbers in
// M[n][m]
import java.util.*;
import java.lang.*;
import java.io.*;

class GFG
{
static int countNegative(int M[][], int n,
int m)
{
int count = 0;

// Follow the path shown using
// arrows above
for(int i = 0; i < n; i++)
{
for(int j = 0; j < m; j++)
{
if( M[i][j] < 0 )
count += 1;

// no more negative numbers
// in this row
else
break;
}
}
return count;
}

// Driver program to test above functions
public static void main (String[] args)
{
int M[][] = { {-3, -2, -1, 1},
{-2, 2, 3, 4},
{4, 5, 7, 8} };

System.out.println(countNegative(M, 3, 4));
}
}
// This code is contributed by Chhavi
```

## Python

```# Python implementation of Naive method to count of
# negative numbers in M[n][m]

def countNegative(M, n, m):
count = 0

# Follow the path shown using arrows above
for i in range(n):
for j in range(m):

if M[i][j] < 0:
count += 1

else:
# no more negative numbers in this row
break
return count

# Driver code
M = [
[-3, -2, -1,  1],
[-2,  2,  3,  4],
[ 4,  5,  7,  8]
]
print(countNegative(M, 3, 4))
```

Output :
`4`

In this approach we are traversing through the all the elements and therefore, in the worst case scenario (when all numbers are negative in the matrix), this takes O(n * m) time.

Optimal Solution

Here’s a more efficient solution:

1. We start from the top right corner and find the position of the last negative number in the first row.
2. Using this information, we find the position of the last negative number in the second row.
3. We keep repeating this process until we either run out of negative numbers or we get to the last row.
```With the given example:
[-3, -2, -1,  1]
[-2,  2,  3,  4]
[4,   5,  7,  8]

Here's the idea:
[-3, -2,  ↓,  ←] -> Found 3 negative numbers in this row
[ ↓,  ←,  ←,  4] -> Found 1 negative number in this row
[ ←,  5,  7,  8] -> No negative numbers in this row ```

## C++

```// CPP implementation of Efficient
// method to count of negative numbers
// in M[n][m]
#include <bits/stdc++.h>
using namespace std;

int countNegative(int M[][4], int n, int m)
{
// initialize result
int count = 0;

int i = 0;
int j = m - 1;

// Follow the path shown using
// arrows above
while( j >= 0 && i < n )
{
if( M[i][j] < 0 )
{
// j is the index of the
// last negative number
// in this row. So there
// must be ( j+1 )
count += j + 1;

// negative numbers in
// this row.
i += 1;
}

// move to the left and see
// if we can find a negative
// number there
else
j -= 1;
}

return count;
}

// Driver program to test above functions
int main ()
{
int M[3][4] = { {-3, -2, -1, 1},
{-2, 2, 3, 4},
{4, 5, 7, 8} };

cout << countNegative(M, 3, 4);

return 0;
}
// This code is contributed by Niteesh Kumar
```

## Java

```// Java implementation of Efficient
// method to count of negative numbers
// in M[n][m]
import java.util.*;
import java.lang.*;
import java.io.*;

class GFG
{
static int countNegative(int M[][], int n,
int m)
{
// initialize result
int count = 0;

int i = 0;
int j = m - 1;

// Follow the path shown using
// arrows above
while( j >= 0 && i < n )
{
if( M[i][j] < 0 )
{
// j is the index of the
// last negative number
// in this row. So there
// must be ( j+1 )
count += j + 1;

// negative numbers in
// this row.
i += 1;
}

// move to the left and see
// if we can find a negative
// number there
else
j -= 1;
}
return count;
}

// Driver program to test above functions
public static void main (String[] args)
{
int M[][] = { {-3, -2, -1, 1},
{-2, 2, 3, 4},
{4, 5, 7, 8} };

System.out.println(countNegative(M, 3, 4));
}
}
// This code is contributed by Chhavi
```

## Python

```# Python implementation of Efficient method to count of
# negative numbers in M[n][m]

def countNegative(M, n, m):

count = 0 # initialize result

i = 0
j = m - 1

# Follow the path shown using arrows above
while j >= 0 and i < n:

if M[i][j] < 0:

# j is the index of the last negative number
# in this row.  So there must be ( j+1 )
count += (j + 1)

# negative numbers in this row.
i += 1

else:
# move to the left and see if we can
# find a negative number there
j -= 1
return count

# Driver code
M = [
[-3, -2, -1,  1],
[-2,  2,  3,  4],
[4,   5,  7,  8]
]
print(countNegative(M, 3, 4))
```

Output :
`4`

With this solution, we can now solve this problem in O(n + m) time.

This article is contributed by YK Sugishita. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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