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  1. Radix sort for integers of arbitrary length has $\mathcal{O}(mn)$ time complexity, where m is the length of number.

  2. Count digits takes $\mathcal{O}(log(n))$$\mathcal{O}(\log(n))$ time, so, your cycle will take at least (in worst case though) $\mathcal{O}(nlog(n))$$\mathcal{O}(n \log(n))$ time which is the best possible for general case of sorting problem. Because to find complexity of 2-level cycle, you need to find complexity of subcycle and multiply them.

Of course, there exists an algorithm that can sort any array of numbers in $\mathcal{O}(nlog(n)|log(n))$ time|space$\mathcal{O}(n \log(n))$ time and $O(\log(n))$ space.

  1. Radix sort for integers of arbitrary length has $\mathcal{O}(mn)$ time complexity, where m is the length of number.

  2. Count digits takes $\mathcal{O}(log(n))$ time, so, your cycle will take at least (in worst case though) $\mathcal{O}(nlog(n))$ time which is the best possible for general case of sorting problem. Because to find complexity of 2-level cycle, you need to find complexity of subcycle and multiply them.

Of course, there exists an algorithm that can sort any array of numbers in $\mathcal{O}(nlog(n)|log(n))$ time|space.

  1. Radix sort for integers of arbitrary length has $\mathcal{O}(mn)$ time complexity, where m is the length of number.

  2. Count digits takes $\mathcal{O}(\log(n))$ time, so, your cycle will take at least (in worst case though) $\mathcal{O}(n \log(n))$ time which is the best possible for general case of sorting problem. Because to find complexity of 2-level cycle, you need to find complexity of subcycle and multiply them.

Of course, there exists an algorithm that can sort any array of numbers in $\mathcal{O}(n \log(n))$ time and $O(\log(n))$ space.

2 added 23 characters in body
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  1. Radix sort for integers of arbitrary length has $\mathcal{O}(mn)$ time complexity, where m is the length of number.

  2. Count digits takes $\mathcal{O}(log(n))$ time, so, your cycle will take at least (in worst case though) $\mathcal{O}(nlog(n))$ time which is the best possible for general case of sorting problem. Because to find complexity of 2-level cycle, you need to find complexity of subcycle and multiply them.

Of course, there exists an algorithm that can sort any array of numbers in $\mathcal{O}(nlog(n)|log(n))$ time|space.

  1. Radix sort for integers of arbitrary length has $\mathcal{O}(mn)$ time complexity, where m is the length of number.

  2. Count digits takes $\mathcal{O}(log(n))$ time, so, your cycle will take at least $\mathcal{O}(nlog(n))$ time which is the best possible for general case of sorting problem. Because to find complexity of 2-level cycle, you need to find complexity of subcycle and multiply them.

Of course, there exists an algorithm that can sort any array of numbers in $\mathcal{O}(nlog(n)|log(n))$ time|space.

  1. Radix sort for integers of arbitrary length has $\mathcal{O}(mn)$ time complexity, where m is the length of number.

  2. Count digits takes $\mathcal{O}(log(n))$ time, so, your cycle will take at least (in worst case though) $\mathcal{O}(nlog(n))$ time which is the best possible for general case of sorting problem. Because to find complexity of 2-level cycle, you need to find complexity of subcycle and multiply them.

Of course, there exists an algorithm that can sort any array of numbers in $\mathcal{O}(nlog(n)|log(n))$ time|space.

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  1. Radix sort for integers of arbitrary length has $\mathcal{O}(mn)$ time complexity, where m is the length of number.

  2. Count digits takes $\mathcal{O}(log(n))$ time, so, your cycle will take at least $\mathcal{O}(nlog(n))$ time which is the best possible for general case of sorting problem. Because to find complexity of 2-level cycle, you need to find complexity of subcycle and multiply them.

Of course, there exists an algorithm that can sort any array of numbers in $\mathcal{O}(nlog(n)|log(n))$ time|space.