How do you trade time complexity for space complexity?

Direct answer

What a strong answer covers

How it works

The trade-off is a choice about which resource is scarcer for this problem at this scale. Spending memory buys speed when a lookup table, a hash index, a precomputed prefix sum, or a memoisation cache replaces repeated work, turning a quadratic scan into linear time. Spending time buys memory when the input does not fit, when streaming one pass is enough, or when allocation itself is the bottleneck. The decision needs the real numbers: input size, memory ceiling, whether the operation runs once or a million times, and the cost of a cache miss.

Example

Finding pairs that sum to a target in an array of a million integers takes minutes with a nested loop and milliseconds with a hash set of seen values that uses a few megabytes. The opposite direction: deduplicating a 200 GB log on a machine with 8 GB of RAM cannot hold a set of everything, so it sorts externally in chunks and merges, trading extra passes for bounded memory.

Common mistakes

• Quoting Big-O without stating the input size that makes the difference matter.
• Adding a cache that grows without bound and turns a speed problem into an outage.
• Optimising a function that runs once at startup while the hot loop is untouched.

A useful answer structure

Answer from your own work.

Do not memorise a script: use only examples you can explain honestly in follow-up questions.