OpenAI Coding High-Frequency Problem: Count Valid Continuous Ranges Under Conflict Pairs

OpenAI coding interviews also feature problems like this: they look like ordinary interval enumeration at first, but collapse immediately if you try to enumerate all subarrays. This is a classic example.

You are given n elements labeled 1...n, along with conflict pairs (a, b). If a continuous range contains both a and b, that range is invalid. The task is to count how many non-empty continuous ranges are valid.

Single-element ranges are allowed. Empty ranges are not.

The easiest way to go wrong is to treat this as “enumerate every [l, r] and check whether it is valid.” Once n reaches 10^5, O(n²) is already dead.

Why this is not really a brute-force validation problem

Suppose:

n = 5
allergic = [1, 2]
poisonous = [3, 5]

So the conflict pairs are:

(1, 3)
(2, 5)

The valid ranges are:

[1]
[1,2]
[2]
[2,3]
[3]
[2,3,4]
[3,4]
[4]
[3,4,5]
[4,5]
[5]

The most important observation here is not just which ranges are valid. It is this:

For each fixed right endpoint r, what is the leftmost starting point that still makes [l, r] valid?

Once that left boundary is known, the number of valid ranges ending at r becomes immediate.

The real target: the leftmost valid start for each right endpoint

Think of each right endpoint r as one column.

If the leftmost valid starting point for this column is left[r], then the valid ranges ending at r are:

[left[r], r]
[left[r] + 1, r]
...
[r, r]

So the number of valid ranges ending at r is:

r - left[r] + 1

The total answer is the sum of that value over all r.

So the problem has already changed from “count all valid ranges” into:

How do we compute the leftmost valid start for each right endpoint efficiently?

That is the whole reduction.

Step 1: record the largest conflicting position on the left

For every conflict pair (a, b), normalize it so that a < b.

Now treat b as the right endpoint. If a range contains both a and b, it is invalid. That means:

when the right endpoint is b, the left boundary can never be at or before a.

So for each position b, we only care about:

what is the largest index on its left that conflicts with it?

Why only the largest one?

Because it imposes the strongest restriction. Any smaller conflicting index is weaker and does not push the boundary further right.

In the example:

for 1, there is no conflicting point on the left
for 2, there is no conflicting point on the left
for 3, the largest conflicting point on the left is 1
for 4, there is no new conflict
for 5, the largest conflicting point on the left is 2

This part only needs one pass over all conflict pairs.

Step 2: why “largest conflict + 1” is still not enough

At this point, many people think the problem is done:

if the right endpoint is r, the leftmost valid start should just be “largest conflicting index + 1”.

That is only half correct.

The issue is that earlier restrictions do not disappear when the right endpoint moves right.

Using the same example:

when r = 3, because of (1, 3), the left boundary must be at least 2
when r = 4, even though 4 itself introduces no new conflict, [1,4] is still invalid because it still contains both 1 and 3

So when the right endpoint moves to 4, the boundary restriction cannot suddenly move back left.

This means the correct transition is:

current restriction = max(current new restriction, previous restriction)

That is just a prefix maximum propagation.

The correct transition

Let bad[r] mean: when the right endpoint is r, the largest conflicting index on its left.

The real restriction is not raw bad[r], but the prefix maximum of bad.

Why?

Because all earlier restrictions continue to affect later right endpoints.

So while sweeping from left to right:

carry forward the maximum restriction seen so far
the leftmost valid start becomes maxBad[r] + 1
the number of valid ranges ending at r becomes r - maxBad[r]

It is r - maxBad[r] because:

the first valid start is maxBad[r] + 1
the number of choices is r - (maxBad[r] + 1) + 1 = r - maxBad[r]

That is where the linear counting comes from.

Why the whole solution is linear

The flow is compact:

scan all conflict pairs and record the largest conflicting left index for each right endpoint
sweep from 1 to n and propagate prefix maxima
accumulate the valid range contribution for each right endpoint

If the number of conflict pairs is m, then:

Time complexity: O(n + m)
Space complexity: O(n)

That is exactly why the solution survives the 10^5 constraint.

Most common mistakes

1. Brute-force interval enumeration

This is the most direct route to TLE.

If you are still thinking in terms of “fix a left endpoint and expand right,” you are probably already off the intended path. The right endpoint is the side that should be fixed.

2. Only using the current conflict, without carrying previous restrictions

This is the most common logic bug.

Many candidates correctly compute one maximum conflicting left index per endpoint, but forget that earlier restrictions continue to matter. That causes overcounting in later columns.

3. Forgetting to normalize each pair as `a < b`

Conflict pairs are undirected in meaning, but the solution is built around “right endpoint with left-side restriction,” so every pair must first be standardized.

4. Off-by-one mistakes in contribution counting

This kind of problem is often solved conceptually but still fails on indexing:

the leftmost valid start is maxBad + 1
the number of ranges is r - left + 1

If that relationship is not kept clean, the final count becomes subtly wrong.

What OpenAI is really testing here

This looks like a counting problem, but the deeper test is whether you can:

convert “is this interval valid?” into “what restriction does this right endpoint have?”
recognize that the restriction is monotone and inherited
compress local conflicts into a globally sweepable structure

That is also a very typical OpenAI coding-interview pattern:

the brute-force path is obvious but unusable
the accepted solution comes from the right state compression
once the monotone structure is identified, the implementation becomes simple

Final Takeaway

The most important thing to remember from this problem is not one template, but one transformation:

Original question: count all valid continuous ranges
Key reduction: for each right endpoint, find the leftmost valid start
Core structure: record the largest conflicting left index and propagate prefix maxima

Once you see that, the problem drops from O(n²) to O(n + m).

If you are preparing for an OpenAI coding interview, this category of “rewrite interval validity as boundary propagation” is absolutely worth practicing.

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