Why do we use the tightest Big O bound?
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MellowPine47
If an algorithm runs in O(n), then O(n²), O(n³), and even O(n!) are technically valid upper bounds too. So why do people expect the smallest or tightest bound when asking for an algorithm's Big O complexity? Does using the tightest bound change what Big O notation means?
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The purpose is comparison, not reporting an exact stopwatch time. If one solution is O(n) and another is O(n²), the difference becomes increasingly important as the input grows. Choosing a deliberately loose bound can make a fast algorithm look worse than it really is and lead to poor design decisions.
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One terminology detail: Big O itself means an upper bound, not necessarily the exact or tightest one. When people say an algorithm is O(n), they often mean that n is the tightest relevant upper bound; formally, Θ(n) expresses a tight asymptotic bound. So an algorithm can be both O(n) and O(n²), but O(n) is the more informative description.
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Big O describes an upper bound, so all of those statements can technically be true. But a loose bound is not very useful. Saying an O(n) algorithm is O(n!) is like answering that a $90 grocery bill was less than $100,000—it’s correct, but it tells you almost nothing. In practice, people give the tightest useful asymptotic bound so algorithms can be compared meaningfully.
QuietLynx22: Exactly. If every algorithm were described as O(n!), the notation would still be technically correct but would stop helping us distinguish efficient algorithms from inefficient ones.
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The goal is to understand how the running time grows as the input grows. An O(1) algorithm stays roughly constant, O(n) grows proportionally with the input, and O(n²) grows much faster. Reporting an unnecessarily large upper bound hides that behavior, which is why the tightest bound is normally preferred.
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MapleViolet64: For example, if the running time is n² + n, it is O(n²), O(n³), and so on. We simplify it to O(n²) because that is the dominant growth rate and the tightest useful bound.