0529MediumArrayMatrixEnumerationPrefix Sum

Splitting the Yield Plot in Two

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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 field is surveyed as plot, a grid of parcels where plot[r][c] is the yield of one parcel. Every yield is at least 1.

The field is to be divided between two tenants by a single straight fence. The fence runs the whole way across the field, either along a line between two rows of parcels or along a line between two columns, and it never cuts through a parcel. Both sides must get at least one parcel.

Return true if some such fence gives the two sides the same total yield, and false otherwise.

Examples

Example 1

Input: plot = [[4, 1], [2, 3]]
Output: true

A fence between the two rows gives the top side 4 + 1 = 5 and the bottom side 2 + 3 = 5.

Example 2

Input: plot = [[3, 1], [2, 4]]
Output: true

A fence between the two columns gives the left side 3 + 2 = 5 and the right side 1 + 4 = 5.

Example 3

Input: plot = [[2, 1], [1, 4]]
Output: false

The whole field yields 8. The fence between the rows gives 3 and 5, and the fence between the columns gives 3 and 5, so no fence splits it evenly.

Constraints

1 <= rows == plot.length <= 10^5
1 <= cols == plot[r].length <= 10^5
2 <= rows * cols <= 10^5
1 <= plot[r][c] <= 10^5

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 can_split_plot(plot: list[list[int]]) -> bool:

Java

public boolean canSplitPlot(int[][] plot)

The intended approach

Time O(rows * cols): the grid is read twice and each candidate fence costs one addition.Space O(rows + cols) for the strip totals.

Total the whole field first. An odd total can never be halved, so that case is settled at once. Otherwise total each row, then walk the row boundaries carrying a running total of everything above the fence and compare it with half the field; do the same with column totals for the vertical fences. Stopping one boundary short of each edge is what keeps both sides non-empty.

Pythonentry point: can_split_plot

def can_split_plot(plot):
    # O(rows*cols) time, O(rows+cols) space: running strip totals mean each
    # candidate cut is checked with one addition.
    rows = len(plot)
    cols = len(plot[0])
    row_totals = [sum(row) for row in plot]
    whole = sum(row_totals)
    if whole % 2 != 0:
        return False
    half = whole // 2
    run = 0
    for r in range(rows - 1):
        run += row_totals[r]
        if run == half:
            return True
    col_totals = [0] * cols
    for row in plot:
        for c in range(cols):
            col_totals[c] += row[c]
    run = 0
    for c in range(cols - 1):
        run += col_totals[c]
        if run == half:
            return True
    return False

The obvious version

Try every fence position and, for each one, add up all the parcels on each side from scratch by scanning the whole grid.

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_can_split_plot(plot):
    rows = len(plot)
    cols = len(plot[0])
    for cut in range(1, rows):
        above = 0
        below = 0
        for r in range(rows):
            for c in range(cols):
                if r < cut:
                    above += plot[r][c]
                else:
                    below += plot[r][c]
        if above == below:
            return True
    for cut in range(1, cols):
        left = 0
        right = 0
        for r in range(rows):
            for c in range(cols):
                if c < cut:
                    left += plot[r][c]
                else:
                    right += plot[r][c]
        if left == right:
            return True
    return False

How this problem was checked

Reference solution executed against 14 test cases — 14 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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