0728EasyArrayMatrix

Busiest Row Of The Punch Sheet

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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 punch sheet is recorded as a grid sheet of 0s and 1s, where sheet[r][c] is 1 when the cell in row r and column c has been punched and 0 when it has not. Every row of the sheet has the same number of columns.

Find the row that holds the most punched cells. If several rows hold the same number of punched cells, and that is the most, take the one with the smallest row index.

Return a two-entry array [row, punches], where row is the index of that row and punches is how many punched cells it holds.

Examples

Example 1

Input: sheet = [[0, 1, 1, 0], [1, 0, 0, 0], [0, 1, 1, 1]]
Output: [2, 3]

The rows hold 2, 1 and 3 punched cells, so row 2 with its 3 punches is reported.

Example 2

Input: sheet = [[1, 0, 1], [0, 1, 0], [1, 1, 0]]
Output: [0, 2]

Rows 0 and 2 both hold 2 punched cells and row 1 holds 1. The tie between rows 0 and 2 goes to the smaller index.

Example 3

Input: sheet = [[0, 0, 0], [0, 0, 0], [0, 0, 0]]
Output: [0, 0]

Nothing is punched, so every row holds 0 and the smallest index wins the tie.

Constraints

1 <= sheet.length <= 100
1 <= sheet[i].length <= 100
0 <= sheet[i][j] <= 1
Every row of sheet has the same number of columns.

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 busiest_row(sheet: list[list[int]]) -> list[int]:

Java

public int[] busiestRow(int[][] sheet)

The intended approach

Time O(m * n) for a sheet of m rows and n columns.Space O(1) beyond the two-entry answer.

Walk the rows in order, counting the punched cells of each one, and remember the best row and its count. Because the stored best is only replaced when a later row is strictly busier, a row that merely equals the current best is passed over and the earliest of the tied rows is what comes out.

Pythonentry point: busiest_row

def busiest_row(sheet):
    # One pass over the sheet, keeping the best row seen so far.  The count only
    # replaces the best when it is strictly larger, so an equal count leaves the
    # earlier row in place and the smallest index wins a tie.
    # Time O(m * n).  Space O(1) beyond the two returned numbers.
    best_row = 0
    best_count = 0
    for r in range(len(sheet)):
        punches = 0
        for value in sheet[r]:
            if value == 1:
                punches += 1
        if punches > best_count:
            best_count = punches
            best_row = r
    return [best_row, best_count]

The obvious version

Tally every row into a list of counts, sweep that list for the largest count, then sweep it again from the front and report the first row that reaches it.

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_busiest_row(sheet):
    counts = []
    for row in sheet:
        counts.append(sum(row))
    top = counts[0]
    for value in counts:
        if value > top:
            top = value
    for r in range(len(counts)):
        if counts[r] == top:
            return [r, top]
    return [0, 0]

How this problem was checked

Reference solution executed against 12 test cases — 12 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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