## Problem statement

In the problem “Count Negative Numbers in a Sorted Matrix” we are given a matrix of n rows and m columns. Elements are sorted in decreasing order both row-wise and column-wise. We need to find the total number of negative elements in the matrix.

## Example

grid = [[8,3,2,-1],[4,2,1,-1],[3,1,-1,-2],[-1,-1,-2,-3]]

8

**Explanation:** As in the given matrix negative numbers are: {-1,-1,-1,-2,-1,-1,-2,-3}. So the total count of the negative numbers is 8.

## Approach

To understand the approach better let us use the same example for better understanding. We will see the given matrix as a set of positive and negative values and will ignore its magnitude. So the given matrix is converted into a set of positive and negative values.

So we already know that the matrix is sorted in decreasing order row-wise as well as column-wise, we will use it to solve this problem. We will start with the first row and will start traversing from **right to left** until we encounter a positive element(let’s call it **col**). The total number of negative elements in the first row will be => m – col – 1. Now we will jump to the next row. Here comes the use of the fact that elements are sorted. **We need not traverse elements that are to the left of the col because the value of matrix[0][col]>=matrix[1][col] and all elements right to the matrix[1][col] is smaller than it**.

Now if matrix[1][col] is positive then The total number of negative elements in the first row will be => m – col – 1 else we will keep traversing from right to left until we encounter a positive element. We will follow these steps until we visit all the rows. The final total count of negative numbers from all the rows will be our answer.

## Code

### C++ code for Count Negative Numbers in a Sorted Matrix LeetCode Solution

#include <bits/stdc++.h> using namespace std; int count(vector<vector<int>>& grid) { int n=grid.size(),m=grid[0].size(), row=0, column=m - 1,ans=0; while (row < n) { while (column >= 0 && grid[row][column] < 0) column--; ans += m - column - 1; row++; } return ans; } int main() { vector<vector<int> > arr = { { 8,3,2,-1 }, { 4,2,1,-1 }, { 3,1,-1,-2 }, { -1,-1,-2,-3 }}; cout<<count(arr)<<endl; return 0; }

8

### Java code for Count Negative Numbers in a Sorted Matrix LeetCode Solution

public class Tutorialcup { public static int countNegatives(int[][] grid) { int n=grid.length,m=grid[0].length, row=0, column=m - 1,ans=0; while (row < n) { while (column >= 0 && grid[row][column] < 0) column--; ans += m - column - 1; row++; } return ans; } public static void main(String[] args) { int [][] grid = { {8,3,2,-1}, {4,2,1,-1}, {3,1,-1,2}, {-1,-1-2,-3} }; int ans= countNegatives(grid); System.out.println(ans); } }

8

## Complexity Analysis

### Time complexity

The time complexity of the above code is** O(n+m)** because we are traversing each row and at worst case all columns. Here n and m are the numbers of rows and number of columns respectively.

### Space complexity

The space complexity of the above code is **O(1)** because we are using only a variable to store answer.