1163HardArrayStackGreedySortingMonotonic Stack

Cutting the Row Into the Most Blocks

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Trains the technique from

LeetCode 768Max Chunks To Make Sorted II

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 row holds the weights weights, and weights may repeat. Cut the row into blocks of neighbouring crates so that sorting each block on its own leaves the whole row in non-decreasing order.

Return the largest number of blocks such a cut can use.

Examples

Example 1

Input: weights = [1, 3, 2, 4]
Output: 3

Cut after the first crate and after the third. The middle block of 3 and 2 sorts into 2 and 3, and the whole row comes out in order.

Example 2

Input: weights = [2, 3, 1, 4]
Output: 2

The 1 has to end up before both the 2 and the 3, so those three crates must share a block. Cutting after them gives two blocks.

Example 3

Input: weights = [1, 1]
Output: 2

Both crates weigh the same, so the cut between them is allowed and each is its own block.

Constraints

1 <= weights.length <= 2000
0 <= weights[i] <= 10^8

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 max_chunks_to_sorted(weights: list[int]) -> int:

Java

public int maxChunksToSorted(int[] weights)

The intended approach

Time O(n) over the row.Space O(n) for the running smallest from the back.

A cut between two positions is allowed exactly when every crate before it is no heavier than every crate after it. If some earlier crate outweighed a later one, sorting within blocks could never get them the right way round, since neither crate ever crosses the cut. And if no earlier crate outweighs a later one, sorting each side puts both sides in order and the heavier side already sits behind. So a cut is allowed when the largest weight up to it is at most the smallest weight past it. Both of those are running quantities: sweep once from the back keeping the smallest seen so far at each position, then sweep forwards keeping the largest so far and count the positions where the one is at most the other. Since cuts are judged one at a time and any allowed cut can be taken alongside any other, taking every allowed cut gives the most blocks. Equal weights on either side of a cut are fine, which is why the comparison allows equality.

Pythonentry point: max_chunks_to_sorted

def max_chunks_to_sorted(weights):
    n = len(weights)
    tail = [0] * (n + 1)
    tail[n] = 1 << 60
    for i in range(n - 1, -1, -1):
        tail[i] = weights[i] if weights[i] < tail[i + 1] else tail[i + 1]
    blocks = 0
    run = -1
    for i in range(n):
        if weights[i] > run:
            run = weights[i]
        if run <= tail[i + 1]:
            blocks += 1
    return blocks

The obvious version

Try every set of cuts as a bit pattern over the gaps between crates, sort each block the pattern makes, stitch the sorted blocks back together and see whether the result matches the whole row sorted, keeping the pattern with the most blocks. Costs two to the power of the gaps.

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_max_chunks_to_sorted(weights):
    n = len(weights)
    want = sorted(weights)
    best = 1
    for mask in range(1 << (n - 1)):
        pieces = []
        start = 0
        for cut in range(n - 1):
            if mask >> cut & 1:
                pieces.append(weights[start:cut + 1])
                start = cut + 1
        pieces.append(weights[start:])
        built = []
        for piece in pieces:
            built.extend(sorted(piece))
        if built == want and len(pieces) > best:
            best = len(pieces)
    return best

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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