Efficiency of growing a dynamic array with a fixed constant every time?

Question

So, when a dynamic array doubles in size each time an element is added, I understand how the complexity of the time to expand is O (n) n, which is the elements. How about whether the array will be copied and moved to a new array, the size of which will be only 1 when it is full? (instead of doubling) When we resize using some constant C, is it always time complexity O (n)?

Answer 1

If you grow on some fixed constant C, then no, the runtime will not be O (n). Instead, it will be & Theta; (n ² ).

To see this, think about what happens if you perform a sequence of sequential operations C. Of these operations, C-1 of them will take O (1) time, because space already exists. The last operation will take O (n) time, because it needs to redistribute the array, add a space and copy everything. Therefore, any sequence of operations C takes time O (n + c).

Now consider what happens if you complete a sequence of n operations. Break these operations into blocks of size C; there will be n / C. The overall work required to complete these operations will be

(c + c) + (2c + c) + (3c + c) + ... + (n + c)

= cn/c + (c + 2c + 3c +... + nc/c)

= n + c (1 + 2 + 3 +... + n/c)

= n + c (n/c) (n/c + 1)/2

= n + n (n/c + 1)/2

= n + n ²/c + n/2

= & Theta; (n ²)

, , :

1 + 2 + 4 + 8 + 16 + 32 +... + n

= 1 + 2 + 4 + 8 +... + 2 ^{log n}

= 2 ^{log n + 1} - 1

= 2n - 1

= & Theta; (n)

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^SO.

2 - 1 + 2 + 4 + 8 + 16 +...

2 ⁰ + 2 ¹ + 2 ² +... + 2 ^n-1

2 ⁿ - 1. , , 32- , 2 ³² - 1.