1129HardArraySegment TreeEnumerationPrefix Sum

Striking Out One Clash to Free the Most Stretches

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Trains the technique from

LeetCode 3480Maximize Subarrays After Removing One Conflicting Pair

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.

Bins stand in a row numbered 1 through n. Each entry of clashes names two different bins that must never both lie inside a chosen stretch, where a stretch is any run of neighbouring bins.

Exactly one entry of clashes must be struck out. After that, count the non-empty stretches that hold no surviving entry's two bins together.

Return the largest count that striking out a single entry can leave.

Examples

Example 1

Input: n = 3, clashes = [[1, 2], [2, 3]]
Output: 4

Whichever entry goes, one clash survives and rules out two of the six stretches, so four are left either way.

Example 2

Input: n = 2, clashes = [[1, 2]]
Output: 3

Striking out the only entry leaves no clash at all, so all three stretches of the two bins count.

Example 3

Input: n = 2, clashes = [[1, 2], [1, 2]]
Output: 2

Both entries name the same two bins, so striking one still leaves the other, and the single stretch covering both bins is ruled out.

Constraints

2 <= n <= 10^5
1 <= clashes.length <= 2 * n
clashes[i].length == 2
1 <= clashes[i][0] <= n
1 <= clashes[i][1] <= n
clashes[i][0] != clashes[i][1]

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_subarrays(n: int, clashes: list[list[int]]) -> int:

Java

public long maxSubarrays(int n, int[][] clashes)

The intended approach

Time O(n + m) over the bins and the entries.Space O(n + m) for the entries grouped by their higher bin.

A stretch is settled by its two ends, so write each entry as its lower bin and its higher bin. An entry spoils a stretch exactly when the stretch starts at or before the lower bin and finishes at or after the higher one. Fix the finishing bin `r`. Only entries whose higher bin is at most `r` can spoil anything, and each of them bars every start at or before its lower bin. So the bar is set by the largest lower bin among them, and the stretches finishing at `r` that survive are those starting past that bar, which is `r` less the bar. Walking `r` upwards and folding in the entries closing at each step keeps that largest lower bin up to date in one pass. Now for the striking out. Removing an entry only helps at those `r` where it alone set the bar, and there the bar drops to the runner-up lower bin, freeing that difference in stretches. So carry the two largest lower bins seen so far and which entry supplied the larger, and credit that entry with the difference at every step. When two entries share the largest lower bin the two are equal and the credit is nothing, which is right, since striking one leaves the other setting the same bar. The answer is the untouched total plus the best credit any single entry earned. One entry must go, so a credit of nothing is still allowed.

Pythonentry point: max_subarrays

def max_subarrays(n, clashes):
    by_close = [[] for _ in range(n + 2)]
    for idx in range(len(clashes)):
        a = clashes[idx][0]
        b = clashes[idx][1]
        lo = a if a < b else b
        hi = b if a < b else a
        by_close[hi].append((lo, idx))
    top = 0
    runner = 0
    owner = -1
    base = 0
    gain = [0] * len(clashes)
    for r in range(1, n + 1):
        for lo, idx in by_close[r]:
            if lo > top:
                runner = top
                top = lo
                owner = idx
            elif lo > runner:
                runner = lo
        base += r - top
        if owner >= 0:
            gain[owner] += top - runner
    extra = 0
    for value in gain:
        if value > extra:
            extra = value
    return base + extra

The obvious version

Try striking out each entry in turn. For each one, walk every pair of ends, test the stretch against every surviving entry outright, and count the survivors, keeping the largest count. Costs the entry count squared times the bin count squared.

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_subarrays(n, clashes):
    m = len(clashes)
    best = 0
    for drop in range(m):
        total = 0
        for lo in range(1, n + 1):
            for hi in range(lo, n + 1):
                ok = True
                for idx in range(m):
                    if idx == drop:
                        continue
                    a = clashes[idx][0]
                    b = clashes[idx][1]
                    if lo <= a and a <= hi and lo <= b and b <= hi:
                        ok = False
                        break
                if ok:
                    total += 1
        if total > best:
            best = total
    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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