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Comment by kevinventullo | Hacker News Reader

As long as the resizing e.g. doubles the size each time, the total running time will still be O(n). Basically because

  1+2+4+...+2^k = O(2^k)

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I think I get it. With doubling, the sum of all the work done for resizing is roughly (written chonologically backwards): n/2 + n/4 + ... ~= n, and O(n) + O(n) is still O(n). Thanks!

Yep! This is generally referred to as "amortized linear time". It's the difference between considering the cost "per operation" vs. "per algorithm". The former is technically correct (as an upper bound), but too pessimistic when you consider the algorithm as a whole.

See https://en.wikipedia.org/wiki/Amortized_analysis#Dynamic_Arr...

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