Count Negative Number in Matrix

Given a m x n matrix grid which is sorted in non-increasing order both row-wise and column-wise, return the number of negative numbers in grid.



Example 1:


Input: grid = [[4,3,2,-1],[3,2,1,-1],[1,1,-1,-2],[-1,-1,-2,-3]]

Output: 8

Explanation: There are 8 negatives number in the matrix.

Example 2:


Input: grid = [[3,2],[1,0]]

Output: 0


Brute Force:


class Solution {
    public int countNegatives(int[][] grid) {

     int row=grid.length;
     int col=grid[0].length;
     int negativeCount=0;

     for(int i=0;i<row;i++)
     {
        for(int j=0;j<col;j++)
        {
            if(grid[i][j]<0)
            {
                negativeCount++;
            }
        }
     }
return negativeCount;
    }
}




Optimal Approach: Binary Search (O(m log n))

Since each row is sorted in decreasing order, we can apply binary search on each row to efficiently find the first negative number's index.

Algorithm:

  1. Initialize count = 0.

  2. Iterate through each row in the matrix.

  3. Apply binary search to find the first negative number in the row.

  4. The number of negative numbers in the row is total_columns - index_of_first_negative.

  5. Accumulate this count for the final answer.


class Solution {
    public int countNegatives(int[][] grid) {

        // Get the number of rows and columns in the matrix
        int row = grid.length;
        int col = grid[0].length;
        int negativeCount = 0; // Counter to store the total count of negative numbers

        // Iterate through each row of the matrix
        for (int i = 0; i < row; i++) {
            int start = 0;        // Left pointer for binary search
            int end = col - 1;    // Right pointer for binary search

            // Perform binary search in the current row
            while (start <= end) {
                int mid = start + (end - start) / 2; // Calculate the middle index

                if (grid[i][mid] < 0) {
                    // If the mid element is negative, move left (search for earlier negatives)
                    end = mid - 1;
                } else {
                    // If mid is non-negative, move right to find first negative
                    start = mid + 1;
                }
            }

            // After binary search, `start` points to the first negative element in the row
            // All elements from `start` to `col - 1` are negative
            negativeCount += (col - start);
        }

        return negativeCount; // Return the total count of negative numbers in the matrix
    }
}


 

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