0678EasyArrayBinary SearchMatrix

Freezing Cells on the Weather Board

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Trains the technique from

LeetCode 1351Count Negative Numbers in a Sorted Matrix

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A weather board shows temperatures in whole degrees, laid out in rows and columns. chart[i][j] is the temperature in row i, column j.

The board is ordered in both directions: reading a row from left to right, the temperature never goes up, and reading a column from top to bottom, the temperature never goes up either. Every row holds the same number of cells.

Return how many cells of the board show a temperature below zero.

Examples

Example 1

Input: chart = [[9, 6, 2, -3], [7, 4, 0, -5], [2, 0, -1, -8], [0, -2, -6, -9]]
Output: 7

The freezing cells are -3 in the first row, -5 in the second, -1 and -8 in the third, and -2, -6 and -9 in the fourth.

Example 2

Input: chart = [[6, 0, 0, -1], [0, 0, -2, -3]]
Output: 3

The cells at zero are not below zero, so only -1, -2 and -3 are counted.

Example 3

Input: chart = [[3, 3], [3, 3]]
Output: 0

No cell on this board is below zero.

Constraints

1 <= chart.length <= 100
1 <= chart[i].length <= 100
-100 <= chart[i][j] <= 100
Every row holds the same number of cells.

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def count_freezing(chart: list[list[int]]) -> int:

Java

public int countFreezing(int[][] chart)

The intended approach

Time O(m + n): the row walk is m steps and the column marker moves right at most n times.Space O(1).

Start at the bottom-left cell and keep a marker for the first freezing column of the row being examined. Walk the rows upwards; in each row push the marker right while the cell it points at is zero or warmer, then every cell from the marker to the right-hand edge of that row is freezing. The marker never has to move back left, because a cell one row up is at least as warm as the cell below it, so the whole walk moves at most the row count plus the column count steps.

Pythonentry point: count_freezing

def count_freezing(chart):
    # staircase walk from the bottom-left corner: O(rows + cols) cells are examined
    rows = len(chart)
    cols = len(chart[0])
    col = 0
    total = 0
    for row in range(rows - 1, -1, -1):
        while col < cols and chart[row][col] >= 0:
            col += 1
        total += cols - col
        if col == cols:
            break
    return total

The obvious version

Look at every cell of the board in turn and count the ones below zero.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_count_freezing(chart):
    total = 0
    for row in chart:
        for value in row:
            if value < 0:
                total += 1
    return total

How this problem was checked

Reference solution executed against 10 test cases — 10 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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