# 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 ` `using` `namespace` `std; ` ` `  `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 ` `int` `main ()  ` `{ ` `    ``int` `M = { {-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``)) `

## C#

 `// C# implementation of Naive method ` `// to count of negative numbers in  ` `// M[n][m] ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `    ``// Function to count ` `    ``// negative number ` `    ``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 Code ` `    ``public` `static` `void` `Main ()  ` `    ``{ ` `    ``int` `[,]M = { {-3, -2, -1, 1}, ` `                ``{-2, 2, 3, 4}, ` `                ``{4, 5, 7, 8} }; ` `     `  `    ``Console.WriteLine(countNegative(M, 3, 4)); ` `    ``} ` `} ` ` `  `// This code is contributed by Sam007 `

## PHP

 ` `

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 ` `using` `namespace` `std; ` ` `  `int` `countNegative(``int` `M[], ``int` `n, ``int` `m) ` `{  ` `    ``// initialize result ` `    ``int` `count = 0;  ` ` `  `    ``// Start with top right corner ` `    ``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, -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``;  ` ` `  `        ``// Start with top right corner ` `        ``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 ` ` `  `    ``# Start with top right corner ` `    ``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``)) `

## C#

 `// C# implementation of Efficient  ` `// method to count of negative  ` `// numbers in M[n][m] ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `// Function to count  ` `// negative number ` `static` `int` `countNegative(``int` `[,]M, ``int` `n, ` `                         ``int` `m) ` `    ``{  ` `         `  `        ``// initialize result ` `        ``int` `count = 0;  ` ` `  `        ``// Start with top right corner ` `        ``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 Code ` `    ``public` `static` `void` `Main ()  ` `    ``{ ` `    ``int` `[,]M = { {-3, -2, -1, 1}, ` `                    ``{-2, 2, 3, 4}, ` `                    ``{4, 5, 7, 8} }; ` `     `  `    ``Console.WriteLine(countNegative(M, 3, 4)); ` `    ``} ` `} ` ` `  `// This code is contributed by Sam007 `

## PHP

 `= 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; ` `        ``} ` `             `  `        ``// move to the left and see  ` `        ``// if we can find a negative ` `        ``// number there ` `        ``else` `        ``\$j` `-= 1; ` `    ``} ` ` `  `    ``return` `\$count``; ` `} ` ` `  `    ``// Driver Code ` `    ``\$M` `= ``array``(``array``(-3, -2, -1, 1), ` `               ``array``(-2, 2, 3, 4), ` `               ``array``(4, 5, 7, 8)); ` `     `  `    ``echo` `countNegative(``\$M``, 3, 4); ` `     `  `    ``return` `0;  ` ` `  `// This code is contributed by anuj_67. ` `?> `

Output :

`4`

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

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

This article is attributed to GeeksforGeeks.org

## tags:

Matrix Python Amazon Amazon Matrix

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