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A few small steps. No rush.

Follow the connections in this topic

Let’s picture it

Take the biggest snack first?

4
5
4

Values are 4, 5, 4, and neighbors cannot both be taken. The biggest is 5.

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One little change

But the two 4s make 8.

4 ✓
5 ×
4 ✓

The tempting first choice blocked a better total. Looking best now wasn’t enough.

Same idea. One more small step.
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Try to break a greedy rule.

best now?
best overall?

Greedy means committing to a local choice. Before trusting it, ask what that choice might block later.

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Does one failing example disprove “always take the biggest” here?

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Greedy choices

One small step at a time.

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Find 8 in a sorted row.

2
4
6
8
10

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

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Let’s picture it

Fit as many meetings as possible.

ends at 2
ends at 5

Meetings cannot overlap. Choosing the one that ends earliest leaves the most room afterward.

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Can we swap without losing?

later finish
earlier finish ✓

Replace another schedule’s first meeting with the earliest-ending one. Later meetings still fit because we finish no later.

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That swap explains the choice.

safe swap
same count

This is an exchange argument. It shows why earliest finish is safe for maximizing meeting count, not for every meeting objective.

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Would this explanation also prove maximum total reward when meetings pay different amounts?

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Greedy choices

One small step at a time.

This idea builds on these smaller ones. Pick any to revisit it here.

Take the biggest snack first?

4
5
4

Values are 4, 5, 4, and neighbors cannot both be taken. The biggest is 5.

1 / 3
Let’s picture it

Now meetings have rewards.

long: 100
short: 10 + 10

One long meeting pays 100. Two short, compatible meetings pay 10 each.

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More meetings isn’t more reward.

20
100

The two short meetings total 20. The long one wins if we care about payment.

Same idea. One more small step.
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Bring back take or skip.

skip
take + compatible best

Compare skipping this meeting with taking it plus the best compatible earlier reward. This turns into DP.

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Does earliest finish alone still solve the reward version?

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Greedy choices

One small step at a time.

This idea builds on these smaller ones. Pick any to revisit it here.

Fit as many meetings as possible.

ends at 2
ends at 5

Meetings cannot overlap. Choosing the one that ends earliest leaves the most room afterward.

1 / 3
Let’s picture it

Each value extends reach.

index 0
reach through 2

At position 0 with value 2, we can reach positions 1 or 2. We do not need to choose the exact landing yet.

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Keep the farthest reachable boundary.

reachable prefix
farthest

Every index at or before farthest can be processed. Its jump may extend the same reachable prefix.

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A gap ends the proof.

reachable
gap
unreachable

If the scan reaches an index beyond farthest, no earlier reachable position could jump across that gap.

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Must greedy choose one exact jump path?

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Greedy choices

One small step at a time.

This idea builds on these smaller ones. Pick any to revisit it here.

Fit as many meetings as possible.

ends at 2
ends at 5

Meetings cannot overlap. Choosing the one that ends earliest leaves the most room afterward.

1 / 3
Let’s picture it

Deadlines create a prefix constraint.

duration 3
deadline 5

A course taking 3 days with deadline 5 must fit before day 5 alongside all earlier chosen courses.

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One little change

Sort by deadline.

early deadline
later deadline

Processing earliest deadlines first makes each prefix the tightest set that must already fit.

Same idea. One more small step.
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If late, replace the longest course.

too late
remove longest
same count potential

Removing the largest chosen duration frees the most time while losing only one course. A max-heap finds it.

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Why remove the longest chosen duration when a deadline is missed?

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Greedy choices

One small step at a time.

This idea builds on these smaller ones. Pick any to revisit it here.

Fit as many meetings as possible.

ends at 2
ends at 5

Meetings cannot overlap. Choosing the one that ends earliest leaves the most room afterward.

1 / 3
Let’s picture it

Frequent symbols should stay shallow.

freq 10 → shallow
freq 1 → deeper

A symbol used 10 times should usually have a shorter code than one used once. Code cost is frequency times depth.

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Merge the two smallest weights.

small a
small b
parent a+b

Make the least frequent symbols siblings. Their combined weight represents one subtree in the remaining problem.

Same idea. One more small step.
Now it has a name

Repeat with a min-heap.

pop two
add cost
push sum

Each merge adds the two weights to total encoded length. The heap exposes the next two smallest subtrees.

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What does a merged subtree’s weight represent?

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Greedy choices

One small step at a time.

This idea builds on these smaller ones. Pick any to revisit it here.

Only the smallest waiting item matters next.

5
1
4

Tasks cost 5,1,4. We need the cheapest task now, then the next cheapest after adding new tasks.

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