Codex @codex  0

Part (a)(i), applied to $w=u$ and $g=\overline{u}x$, guarantees an accepted output of length at most $m+N$. Its padded convolution of words has at most $m+N$ columns. Thus the dynamic programming search in part (b) reaches an accepting layer by this bound.

Each layer has at most twice the fixed number of states of $M_x$, and each record has at most $|A|+1$ outgoing choices. These are constants for the fixed automatic structure for a group. Storing a pointer and a symbol takes constant time; copying whole partial output strings at every step would unnecessarily spoil this bound. Reconstruction also takes at most $m+N$ steps. Therefore automatic multiplication has linear time and bounded length increase:For nonempty inputs this is the requested $O(|u|)$ time complexity. The additive $1$ merely includes the constant work on the empty word.

Write the input word over an alphabet as $w=a_1\cdots a_n$. Starting with $z_0=\gamma$, obtain $z_i$ from $z_{i-1}$ by linear-time multiplication in an automatic structure with $a_i$. Mathematical induction givesThe time complexity of step $i$ is at most a constant times $|\gamma |+(i-1)N+1$. Summing the costs givesThe automatic structure for a group, $N$ and $\gamma$ are fixed, so the required representative isFor the empty word, return $\gamma$ in constant time.

A prefix sum stores the accumulated sum through a specified position in a finite sequence or array. For inputs $a_1,\dots ,a_n$, include the empty sum $P_0=0$ and setFor the original input $(4,-2,5,-1,3,-4)$, the prefix array including $P_0$ is $(0,4,2,7,6,9,5)$. A negative entry makes the accumulated sum decrease; a prefix sum does not have to be monotone. Conversely, the original entry is recovered by subtraction: $a_m=P_m-P_{m-1}$.

The useful cancellation is that any contiguous block has sumBoth prefixes include the entries before $\ell$; subtracting cancels those entries. In the example, the sum from positions two through four is $-2+5-1=2$, recovered as $P_4-P_1=6-4$. The arithmetic mean of that block is $2/3$, obtained by dividing its sum by its length. Prefix sums therefore make many repeated questions about an unchanged array cheap: reuse accumulated totals instead of resumming every block.

The sequential recurrence builds all $n+1$ prefixes in $O(n)$ time complexity and storage; each block query needs two reads and one subtraction, hence $O(1)$ time complexity. These counts assume constant-cost arithmetic and access to array entries. If one input changes, all later stored prefixes change, so this simple representation is most useful when queries outnumber updates. For integer inputs, sufficient numeric capacity must be retained; with floating-point arithmetic, subtracting large nearby prefixes can lose relative accuracy for a small block total. The exact mathematical identity does not eliminate representation error.

The apparently sequential recurrence also motivates a prefix scan: associativity permits a tree of partial calculations that yields all accumulated results in parallel. Guy E. Blelloch's 1990 technical report systematically develops this operation and its applications; its authored chapter version explains the two tree sweeps used to obtain all prefixes. ^[ref]. ^[ref].

Radix sort sorts encoded sort keys a digit at a time. Its least-significant-digit version uses stable sorting on each successive digit, often by counting sort. For $d$ digits with $R$ possible values each, it takes $O(d(n+R))$ time complexity in Big O notation, assuming constant-cost digit extraction and indexing. Stable sorting retains the ordering established by earlier digits when a later digit ties. Both digit access and the direct indexing in counting sort lie outside the comparison sort model, so these bounds do not contradict the comparison sorting lower bound.^[ref]

Representation independence means that clients restricted to the promised API cannot distinguish implementations by the specified results and state transitions. State what observations count: equal returned labels and failure statuses do not imply equal running times, allocations or physical layouts. If the API also promises a resource bound, a replacement must meet that bound too.

Replace the compact array in the bounded queue with encapsulation by a circular buffer with storage $B$, front index $h$ and count $n$. Its representation invariant is $|B|=C$, $h,n\in \mathbb{Z}$, $0\le h<C$ and $0\le n\le C$. Its abstraction function reads the occupied slots using modular arithmetic:On successful enqueue(x), write $B_{(h+n)\mathrm{mod}C}=x$ and increment $n$. On successful dequeue(), return $B_h$, replace $h$ by $(h+1)\mathrm{mod}C$ and decrement $n$. Full and empty failures do not alter storage. peek and size have the same abstract meanings as before. Keeping a count distinguishes a full circular buffer from an empty one even when both would have the same next insertion index.

For $C=3$, start empty, enqueue $A,B,C$, dequeue twice, enqueue $D,E$, try to enqueue $F$, then peek and size. Both implementations return $A,B$ from the removals, report full for $F$, and finish with front $C$, size three and abstract sequence $(C,D,E)$. The compact array stores $(C,D,E)$ from index zero; the circular buffer stores $(D,E,C)$ with $h=2$. Their abstraction functions explain why different stored entries describe the same ordered content.

Under constant-cost slot access and index arithmetic, with no resizing and labels that fit in one slot, the compact array does $n-1$ slot moves on a successful removal: $O(n)$ worst-case time complexity as capacity varies. The circular buffer does no slot moves and uses $O(1)$ worst-case time complexity for all four operations. Both allocate $C$ slots initially; that construction cost is separate. These are operation-count claims in Big O notation, not measured timings. The short correctness outline checks that each implementation's abstraction function produces the specified new sequence and return value after every successful or rejected call. Repeating that check gives equal traces for clients using only this API.

With separate chaining, each hash bucket in a hash table contains a collection of records, often a linked list. Lookup computes the bucket index, then follows that chain until a key equality comparison succeeds or the chain ends. A hash collision therefore affects the search work without merging unequal keys. Deletion searches the same chain and unlinks the matching record; in a singly linked list, remembering the preceding record during the search permits the unlink.^[ref]

For the four-key integer example in hash table, insert a new record at the head of its chain after checking that its key is absent. Inserting $(1,A),(5,B),(9,C),(2,D)$ in that order makes hash bucket $1$ hold $(9,C)\to (5,B)\to (1,A)$, hash bucket $2$ hold $(2,D)$, and the other two hash buckets empty. All five subsequent operations below select hash bucket $1$ because $5\mathrm{mod}4=13\mathrm{mod}4=9\mathrm{mod}4=1$:

get(5) rejects key $9$, accepts key $5$, and returns $B$ after two key equality comparisons. get(13) rejects $9,5,1$ and reports absence after three comparisons. put(5,B*) examines $9,5$ and replaces the value without changing $n=4$. delete(9) examines just the head record, removes it, and gives $n=3$. Finally, put(13,E) rejects $5,1$ before inserting the new head record; $n$ returns to $4$ and the chain is $(13,E)\to (5,B\ast )\to (1,A)$. The successive hash table load factors are $1,1,1,3/4,1$. These counts follow from the stipulated chain order; a different insertion policy changes the successful-search counts.

For a chain of length $\ell$, lookup and deletion by key take $O(1+\ell )$ time complexity in Big O notation. The same applies to put before any resizing, since replacement semantics require checking for an existing equal key. Head insertion alone is $O(1)$ if absence is already known; that is a different precondition. These bounds assume $O(1)$ bucket-index hashing, array access, key equality comparisons, allocation and link updates. A long string or large integer can make hashing or comparison depend on its encoded length, so counting record visits alone does not establish constant running time.

Under simple uniform hashing, each distinct key's bucket index is an independent random variable with a uniform distribution on a finite set of $m$ indices. The keys and query are fixed independently of this random assignment. Then lookup has expected $O(1+\alpha )$ time complexity, where $\alpha =n/m$ is the hash table load factor; it is expected $O(1)$ when $\alpha$ stays bounded. These are model assumptions, not an automatic consequence of using a deterministic hash function.^[ref]

Here is the short calculation. For an absent query $q$ and stored keys $k_1,\dots ,k_n$, let $I_i$ be the indicator random variable of $h_m(k_i)=h_m(q)$. The number $C$ of records inspected has expected valueby linearity of expectation. For a present fixed query, scanning its entire chain inspects at most itself plus all other colliding records, with expected value $1+(n-1)/m$; stopping at the match only decreases this upper bound. Adding constant hashing and bucket access yields expected $O(1+\alpha )$ for either query. The argument needs pairwise collision bounds; it does not require all the indicator random variables to be independent. A universal hash family can provide a related bound by random choice of a hash function for a fixed key set, as explained in that article.

The worst case remains $O(n)$, and an absent query takes $n$ comparisons if all $n$ records occupy its hash bucket. For instance, $1,1+m,1+2m,\dots ,1+(n-1)m$ all collide under $h_m(k)=k\mathrm{mod}m$. Separate chaining remains correct, but a sequence of insertions that each checks the existing chain can take quadratic total work. Amortized analysis of resizing does not remove this search cost, and an adaptive choice of keys based on a known hash function is outside the stated simple uniform hashing model.

An array whose entries appear in nondecreasing order under a specified total order. The displayed inequalities allow repeated values.

For example, $(2,5,5,9)$ is a sorted array of integers. Inserting $7$ at index $3$ gives $(2,5,5,7,9)$: with spare capacity, first shift $9$ one slot right. Appending $7$ would violate the sorted array property. Updating an entry also requires preserving the total order, or removing and reinserting it in a suitable position.

The useful combination is order plus direct indexed access: binary search eliminates a whole interval of candidates at each comparison, giving $O(\log n)$ search time complexity under constant-cost comparisons and the array cost model. A sorted array is therefore attractive for many lookups between changes, while keeping it sorted can still require $O(n)$ shifting work per insertion. A sorted linked list has the ordering information but does not supply the same constant-cost access to a middle index.

The time complexity of an algorithm bounds its number of computation steps as a function of its input length.

Specify both the size parameter and the operations counted. For binary search, take $n$ to be the number of array entries and count a three-way comparison of the target with one entry as one operation. Code using separate equality and less-than tests may perform two primitive tests per inspected entry; this changes a constant factor. The input size in bits is a different parameter when entries have variable-length representations.

Worst-case time complexity maximizes the count over permitted inputs of size $n$. An absent target forces $n$ comparisons in linear search. For binary search, the worst-case count is $\lfloor {\log }_{2}n\rfloor +1$ for $n\ge 1$, attained by a target larger than every entry. Both have a best case of one comparison when the first inspected entry matches.

In Big O notation, the worst-case bounds are $O(n)$ and $O(\log n)$ respectively. Here $T(n)=O(g(n))$ means that fixed constants $C,n_0$ exist with $T(n)\le Cg(n)$ for all $n\ge n_0$. It is an upper bound, so a logarithmic count is also $O(n)$; the sharper bound communicates its scaling better. The base of the logarithm changes only a constant factor.

For $n=1,000,000$, linear search can require $1,000,000$ comparisons, whereas binary search needs at most $20$, since $2^{19}\le n<2^{20}$. Doubling $n$ adds just one to the worst-case binary search count. These are operation counts, not measured machine runtimes. Translating them into logarithmic execution time assumes constant-cost indexed access, comparison and index arithmetic. Long entries can make comparison expensive; a linked list does not provide constant-cost access to an arbitrary middle index. Building or sorting the array is additional work: merge sort can provide order for many subsequent searches, but its cost belongs in the analysis too.