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

We strongly recommend you to minimize your browser and try this yourself first.



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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Python

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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, -11],
      [-2234],
      [ 4578]
    ]
print(countNegative(M, 3, 4))

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP implementation of Naive method
// to count of negative numbers in 
// M[n][m]
  
function countNegative($M, $n, $m)
{
    $count = 0;
  
    // Follow the path shown using 
    // arrows above
    for( $i = 0; $i < $n; $i++)
    {
        for( $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
    $M = array(array(-3, -2, -1, 1),
               array(-2, 2, 3, 4),
               array(4, 5, 7, 8));
      
    echo countNegative($M, 3, 4);
      
// This code is contributed by anuj_67.
?>

chevron_right


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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
  
    // 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][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

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Python

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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, -11],
      [-2234],
      [4,   578]
    ]
print(countNegative(M, 3, 4))

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP implementation of Efficient 
// method to count of negative numbers
// in M[n][m]
  
function countNegative( $M, $n, $m)
{
      
    // initialize result
    $count = 0; 
  
    // 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;
        }
              
        // 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.
?>

chevron_right


Output :



4

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

More Optimal Solution

Here’s a more efficient solution using binary search instead of linear search:

  1. We start from the first row and find the position of the last negative number in the first row using binary search.
  2. Using this information, we find the position of the last negative number in the second row by running binary search only until the position of the last negative number in the row above.
  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:
1. Count is initialized to 0
2. Binary search on full 1st row returns 2 as the index of last negative integer, and we increase count to 0+(2+1) = 3.
3. For 2nd row, we run binary search from index 0 to index 2 and it returns 0 as the index of last negative integer. We increase the count to 3+(0+1) = 4;
4. For 3rd row, first element is > 0, so we end the loop here.

Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of More efficient
// method to count number of negative numbers
// in row-column sorted matrix M[n][m]
import java.util.*;
import java.lang.*;
import java.io.*;
  
class GFG {
  
    // Recursive binary search to get last negative
    // value in a row between a start and an end
    static int getLastNegativeIndex(int array[], int start, int end)
    {
        // Base case
        if (start == end) {
            return start;
        }
  
        // Get the mid for binary search
        int mid = start + (end - start) / 2;
  
        // If current element is negative
        if (array[mid] < 0) {
  
            // If it is the rightmost negative
            // element in the current row
            if (mid + 1 < array.length && array[mid + 1] >= 0) {
                return mid;
            }
  
            // Check in the right half of the array
            return getLastNegativeIndex(array, mid + 1, end);
        }
        else {
  
            // Check in the left half of the array
            return getLastNegativeIndex(array, start, mid - 1);
        }
    }
  
    // Function to return the count of
    // negative numbers in the given matrix
    static int countNegative(int M[][], int n, int m)
    {
  
        // Initialize result
        int count = 0;
  
        // To store the index of the rightmost negative
        // element in the row under consideration
        int nextEnd = m - 1;
  
        // Iterate over all rows of the matrix
        for (int i = 0; i < n; i++) {
  
            // If the first element of the current row
            // is positive then there will be no negatives
            // in the matrix below or after it
            if (M[i][0] >= 0) {
                break;
            }
  
            // Run binary search only until the index of last
            // negative Integer in the above row
            nextEnd = getLastNegativeIndex(M[i], 0, nextEnd);
            count += nextEnd + 1;
        }
        return count;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int M[][] = { { -3, -2, -1, 1 },
                      { -2, 2, 3, 4 },
                      { 4, 5, 7, 8 } };
        int r = M.length;
        int c = M[0].length;
        System.out.println(countNegative(M, r, c));
    }
}
// This code is contributed by Rahul Jain

chevron_right


Output :

4

Here we have replaced the linear search for last negative number with binary search. This should improve the worst case scenario keeping the Worst case to O(n + log(m))

This article is contributed by YK Sugishita and Rahul Jain. 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.

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



My Personal Notes arrow_drop_up

Improved By : Sam007, vt_m, xRahul



Article Tags :
Practice Tags :


2


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.