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Same door. One small difference.

You have a key. Your friend doesn’t. The door is the same, but your next move isn’t.

Remember the key.

If we only write down “at the door,” we lose something important.

That little note is a state.

A state remembers what we need to continue. Here: where we are, and whether we have the key.

How many little jobs?

Visit three places. Say hello once at each. That’s three hellos.

Now try every pair.

Three people each greet all three people, including themselves for this toy count. That’s nine greetings.

Count how the work grows.

One pass grows with n. Trying every pair grows with n × n. Code length doesn’t tell us that.

Find 8 in a sorted row.

Start with 2, 4, 6, 8, 10. Look at the middle: 6.

Keep the side that can work.

8 is bigger than 6. Everything on the left is too small, so we can leave it behind.

Keep one promise.

If 8 is present, it stays in the part we keep. That repeating promise is called an invariant.

Split a row into two groups.

Values are 3, 2, 4. Keep their order. We want the smallest possible largest group sum.

Try a cap of 4.

3 and 2 cannot stay together. Neither can 2 and 4. We need three groups, so cap 4 is too small.

Try a cap of 5.

[3, 2] and [4] fit. A cap of 6 would work too. Making the cap larger cannot hurt.

The answers have a boundary.

Too small, too small, then possible forever. Binary search can find that first possible cap.

Search answers, not positions.

Test a cap by greedily filling each group before starting the next. Nonnegative values make this feasibility test safe.

Keep a small part of a row.

Our budget is 5. The values 2 and 3 fit together.

One more makes it too much.

Add 4: the sum is 9. Remove from the left until it fits again.

That moving part is a window.

With nonnegative values, adding cannot lower the sum. That predictable change tells us which boundary to move.

A window meets a negative number.

Find stretches totaling 2 in [2, −1, 1]. Adding −1 shrinks the sum, so the usual window direction is no longer reliable.

Keep a running total instead.

Before anything: 0. After each number: 2, 1, 2. These totals describe the boundaries between items.

Subtract two boundaries.

A current total of 2 minus an earlier total of 0 gives a stretch totaling 2.

Ask for the missing earlier total.

At total s, look for earlier total s − k. Count how often it appeared, then record s.

Prefix sums meet a dictionary.

The two valid stretches are [2] and [2, −1, 1]. We found both without trying every starting point again.

Who is still waiting for warmth?

Temperatures are 73, 71, 74. The first two days have not yet seen a warmer day.

74 answers two questions.

It is warmer than 71, and also warmer than 73. Their waits end together.

Keep only unanswered days.

Store their positions in a stack. Temperatures go downward from the bottom toward the top.

Resolve the top while you can.

When a warmer day arrives, pop each cooler waiting day and record the distance between their positions.

That’s a monotonic stack.

Each day enters once and leaves at most once. The nested while loop still does only linear total popping.

Reuse the waiting stack.

Now the bars have heights 2, 4, 2. We want the biggest rectangle under neighboring bars.

The tallest bar is not enough.

Height 4 across one bar gives area 4. Height 2 across all three gives area 6.

Wait until a shorter bar arrives.

The final 2 stops the height-4 rectangle. We now know its right boundary.

The stack remembers the other boundary.

Pop 4. The previous smaller bar is at position 0; the new smaller bar is at 2. Width: 2 − 0 − 1.

Close every waiting rectangle.

Use a zero-height sentinel at the end. Popping bars computes completed areas; keep the biggest.