1003EasyArrayMatrix

Readings Lowest in Their Row and Highest in Their Column

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Trains the technique from

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 panel of readings is given as matrix, and no two readings anywhere on it are equal.

A reading is standout when it is the smallest in its own row and, at the same time, the largest in its own column.

Return every standout reading, listed in the order their rows appear.

Examples

Example 1

Input: matrix = [[2, 6, 9], [11, 14, 20], [24, 30, 35]]
Output: [24]

The reading 24 is the smallest of its row, whose other entries are 30 and 35, and the largest of its column, whose other entries are 2 and 11.

Example 2

Input: matrix = [[4, 12, 7], [19, 3, 18], [25, 21, 10]]
Output: []

The row smallests are 4, 3 and 10, while the column largests are 25, 21 and 18. Nothing appears in both lists, so no reading stands out.

Example 3

Input: matrix = [[9, 4], [3, 1]]
Output: [4]

The reading 4 is smaller than the 9 beside it and larger than the 1 below it.

Constraints

1 <= matrix.length <= 50
1 <= matrix[i].length <= 50
1 <= matrix[i][j] <= 10^5
No two readings on the panel are equal.

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 lucky_numbers(matrix: list[list[int]]) -> list[int]:

Java

public List<Integer> luckyNumbers(int[][] matrix)

The intended approach

Time O(m * n) over the panel's cells.Space O(m + n) for the two collections.

Collect the smallest reading of every row, and separately the largest reading of every column. A reading is standout precisely when it turns up in both collections, so walk the row smallests in row order and keep the ones the column largests also hold. Two sweeps of the panel do it, one gathering row smallests and one gathering column largests, and a lookup structure for the second collection makes the final walk cheap. Worth knowing, though the sweeps never need it: the panel can hold at most one standout reading. Say two sat in different rows and different columns. Look at the reading where the first one's row crosses the second one's column. The first, being smallest in that row, is no larger than it, and the second, being largest in that column, is no smaller, so the first is at most the second. Doing the same the other way round makes the second at most the first, so they would have to be equal, which the panel forbids. Two in the same row or the same column fails for the same reason.

Pythonentry point: lucky_numbers

def lucky_numbers(matrix):
    rows = len(matrix)
    cols = len(matrix[0])
    row_low = []
    for r in range(rows):
        low = matrix[r][0]
        for c in range(1, cols):
            if matrix[r][c] < low:
                low = matrix[r][c]
        row_low.append(low)
    col_high = set()
    for c in range(cols):
        high = matrix[0][c]
        for r in range(1, rows):
            if matrix[r][c] > high:
                high = matrix[r][c]
        col_high.add(high)
    return [value for value in row_low if value in col_high]

The obvious version

Test each cell of the panel outright: scan its whole row for anything smaller and its whole column for anything larger, keeping the reading when neither turns up. Costs a row scan and a column scan per cell rather than two sweeps of the panel.

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_lucky_numbers(matrix):
    rows = len(matrix)
    cols = len(matrix[0])
    out = []
    for r in range(rows):
        for c in range(cols):
            value = matrix[r][c]
            ok = True
            for other in range(cols):
                if matrix[r][other] < value:
                    ok = False
            for other in range(rows):
                if matrix[other][c] > value:
                    ok = False
            if ok:
                out.append(value)
    return out

How this problem was checked

Reference solution executed against 15 test cases — 15 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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