A hash table uses amortized analysis with the banker's method for resizing. If inserting *n* elements requires 3*n* credits, which consequence follows regarding the cost?

💡 Explanation

Using the banker's method, 3*n* credits mean that we can 'charge' each insert a cost of 3, therefore the amortized cost per operation is at most 3, because credits pay for resizing, rather than having individual operations cost 3*n*.

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