R_sort_2

What is the worst-case time complexity of the Radix Sort algorithm for an array of n elements, where the maximum number of digits in any number is d?

Question

R_sort_2

What is the worst-case time complexity of the Radix Sort algorithm for an array of n elements, where the maximum number of digits in any number is d?

BlackTom AI · Accepted Answer

First, let's consider what Radix Sort does: it processes the input elements digit by digit, performing a stable sort (typically counting sort) for each digit position. Each pass over the n elements costs proportional to n, plus a small constant depending on the digit base.

Option 1: O(n log n). This reflects a comparison-based sorting bound, which Radix Sort avoids by not relying on comparisons for each digit; thus this is not the right characterization for the given radix approach.

Option 2: O(n²). This would imply quadratic behavior in the input size, which would occur in a naïve nested-loop algorithm, not in Radix Sort with per-digit passes.

Option 3: O(n + d). Here the time grows linearly with n and also with d, but it omits the fact that there are multiple passes over the data. Since there is a pass for each digit (up to d passes), the total cost accumulates as n for each pass, making the total effectively proportional to d * n, not n + d.

Option 4: O(d * n). This captures the core idea: there are d digit-positions to process, and each pass touches all n elements, giving a product of the two factors as the overall time. In many analyses this is described alongside a small additional term for the counting-sort range, but the dominant factor in the worst case is indeed d times n.

Together, these evaluations show that the most accurate characterization among the choices is the one where the work scales with both the number of digits and the number of elements, i.e., proportional to d * n.

R_sort_2 What is the worst-case time complexity of the Radix Sort algorithm for an array of n elements, where the maximum number of digits in any number is d?

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