1049HardDynamic ProgrammingTreeDepth-First SearchBinary Search TreeBinary TreeDP on Trees

The Heaviest Sorted Section of a Rig

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Trains the technique from

LeetCode 1373Maximum Sum BST in Binary Tree

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 cargo rig is given as the flat list rig. It hangs from a single top joint, every joint carries at most two joints below it in a first and a second slot, and either slot may be empty.

A flat list is read level by level: its first entry is the top joint's load, and reading left to right, every entry that is not null claims the next two unused positions as its first and second slot in that order, while null marks an empty slot and claims no positions of its own.

A section is one joint together with everything hanging below it. A section is sorted when, for every joint in it, every load in that joint's first slot and below is strictly less than the joint's own load, and every load in its second slot and below is strictly greater.

Return the largest total load any sorted section carries. A section holding no joints at all counts as sorted and carries nothing, so the answer never falls below 0.

Examples

Example 1

Input: rig = [5, 3, 8, 2, 4, 7, 9]
Output: 38

The whole rig is sorted: everything under 5's first slot is below it and everything under its second slot is above, and the same holds at 3 and at 8. So the answer is every load added together.

Example 2

Input: rig = [10, 5, 15, null, null, 6, 20]
Output: 41

The whole rig is not sorted, since the 6 hangs under 10's second slot yet is smaller than 10. The section headed by 15 is sorted, carrying 6, 15 and 20, which is heavier than the section headed by 5 on its own.

Example 3

Input: rig = [-1]
Output: 0

The only section carrying anything carries a negative load, so the empty section, carrying nothing at all, is the heavier one.

Constraints

1 <= rig.length <= 120000
The rig holds between 1 and 4 * 10^4 joints.
-40000 <= rig[i] <= 40000
The first entry of the list is not null.

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_sum_b_s_t(rig: list) -> int:

Java

public int maxSumBST(Integer[] rig)

The intended approach

Time O(n) over the list's length.Space O(n) for the slots and the per-joint facts.

Whether a section is sorted rests on its two slots' sections being sorted and on their loads staying on the right side of the joint. So settle the joints from the bottom up, carrying four facts for each: whether its section is sorted, the smallest and largest load in it, and its total load. A joint's section is sorted when both slots' sections are sorted, the largest load under the first slot is strictly below the joint, and the smallest load under the second slot is strictly above it. Comparing against those extremes is what makes one comparison per slot enough: a load deeper down that broke the rule would already have shown up in the extreme. The section's smallest load is the first slot's smallest, or the joint itself when that slot is empty, and its largest comes the same way from the second slot. Its total is the joint plus both slots' totals. Keep the best total over the sorted sections, starting from nothing so an all-negative rig answers nothing. Reading the flat list into slots is one pass, and walking the joints backwards through that reading order settles every joint after everything below it.

Pythonentry point: max_sum_b_s_t

def max_sum_b_s_t(rig):
    n = len(rig)
    if n == 0:
        return 0
    left = {}
    right = {}
    order = [0]
    level = [0]
    at = 1
    while level and at < n:
        nxt = []
        for owner in level:
            if at < n:
                if rig[at] is not None:
                    left[owner] = at
                    nxt.append(at)
                at += 1
            if at < n:
                if rig[at] is not None:
                    right[owner] = at
                    nxt.append(at)
                at += 1
        order.extend(nxt)
        level = nxt
    fact = {}
    best = 0
    for pos in reversed(order):
        load = rig[pos]
        ok = True
        lowest = load
        highest = load
        total = load
        lo = left.get(pos)
        if lo is not None:
            sub = fact[lo]
            if not sub[0] or sub[2] >= load:
                ok = False
            else:
                lowest = sub[1]
                total += sub[3]
        ro = right.get(pos)
        if ro is not None:
            sub = fact[ro]
            if not sub[0] or sub[1] <= load:
                ok = False
            else:
                highest = sub[2]
                total += sub[3]
        if ok and total > best:
            best = total
        fact[pos] = (ok, lowest, highest, total)
    return best

The obvious version

Test each section on its own. Gather the section's loads by walking it first slot, then joint, then second slot, which for a sorted section has to come out strictly increasing, and check that outright. Add the loads up for the sections that pass and keep the largest total. Costs a walk per joint, so the rig's size squared, rather than four facts carried upward.

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_sum_b_s_t(rig):
    n = len(rig)
    if n == 0:
        return 0
    left = {}
    right = {}
    level = [0]
    at = 1
    while level and at < n:
        nxt = []
        for owner in level:
            if at < n:
                if rig[at] is not None:
                    left[owner] = at
                    nxt.append(at)
                at += 1
            if at < n:
                if rig[at] is not None:
                    right[owner] = at
                    nxt.append(at)
                at += 1
        level = nxt
    joints = [0]
    stack = [0]
    while stack:
        pos = stack.pop()
        for child in (left.get(pos), right.get(pos)):
            if child is not None:
                joints.append(child)
                stack.append(child)
    def reading(pos, out):
        lo = left.get(pos)
        if lo is not None:
            reading(lo, out)
        out.append(rig[pos])
        ro = right.get(pos)
        if ro is not None:
            reading(ro, out)
    best = 0
    for pos in joints:
        loads = []
        reading(pos, loads)
        ok = True
        for k in range(1, len(loads)):
            if loads[k] <= loads[k - 1]:
                ok = False
                break
        if ok:
            total = 0
            for value in loads:
                total += value
            if total > best:
                best = total
    return best

How this problem was checked

Reference solution executed against 17 test cases — 17 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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