Big-O drops constants and lower-order terms because it tracks growth shape, not exact operation counts. Multiplying by two or adding a fixed setup does not change how a curve bends as the input grows.
Say an algorithm does 3n + 100 steps. For small n the hundred dominates. But as n reaches millions, the 3n term outweighs it, and the factor three does not change that it grows linearly. So the algorithm is O(n).
This makes comparisons hardware-free. A constant might just come from one machine being faster. By ignoring it, Big-O shows which algorithm pulls ahead as the problem gets large, on any computer.
Rewriting in plainer words…
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An algorithm does n^2 + n log n steps. Which term do you keep, and how do you know?Keep n^2. Show that n log n divided by n^2 shrinks toward zero as n grows. Plugging in a single value of n proves nothing.
Big-O drops constants and lower-order terms because it describes how the cost grows with the input, not the exact number of steps.
Doubling the work or adding a fixed setup cost doesn't change the shape of the curve as n gets large.
Take 3n + 100. For small n the 100 dominates, but at a million the 3n term outweighs it, and the 3 doesn't stop it growing linearly, so it's O(n).
Dropping constants also takes hardware out of the comparison, since a constant factor might only mean one machine is faster than another.
What's left tells you which algorithm pulls ahead as the problem gets large, on any computer.
The cost is that Big-O says nothing about small inputs, where a fixed setup or a big constant can decide which one is actually faster.
Weak answers to avoid
Says constants never matterThey matter at small n and in real runtimes. Say Big-O ignores them on purpose to compare growth, then note when a constant can still decide the winner.
Recites the rule with no reasonSaying you just drop them misses the point. Explain that a constant factor or fixed cost doesn't change how the cost grows as n increases.
Drops a term that isn't lower-orderOnly terms that grow strictly slower can go. In O(n + m) the inputs are independent, and n log n is not the same as n.
Treats O(n) as exactly n stepsBig-O is an upper bound on growth, not a step count. 3n + 100 is O(n) even though it never takes exactly n steps.
Interview lens
Likely follow-ups, what you can say, and the weak answers to avoid.
