Row With Maximum Ones
Real-world version: Busiest Row Of The Punch SheetThe statement here is our own: the same inputs, outputs and technique.
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 starts with this. It matches the parent problem's shape, so a solution that works here works on a standard judge unchanged.
def busiest_row(sheet: list[list[int]]) -> list[int]:public int[] busiestRow(int[][] sheet)Checking your plan…
How this problem was checked
- Passed: Reference solution run against 12 test cases: 12 passed
- Passed: 3 worked examples confirmed by running the code
- Passed: 0 eight-word sequences shared with the parent problem's text
- Passed: Agrees with an independent brute-force solution on random inputs
- Passed: Test set shown to catch deliberately broken solutions
- Passed: Reviewed by a separate quality pass
- Passed: Function arity matches the parent problem's