Part one was fine, just iterate and pick the min of each adjacent pair, handle odd lengths by appending the last element.
Clarify the rules and edge cases, then simulate each round by iterating through the array in steps of two, comparing adjacent players and advancing the stronger (lower rank). Handle the odd-length case by carrying the last player forward as a bye. Repeat until one player remains, recording the array after each round.
Pro tip: Mention that you can optimize space by reusing the input array or using a queue, but prioritize clarity in the first pass. Also, explicitly state the time complexity: O(n) per round, O(n log n) overall, since the array halves each round.
Ask about input size, whether ranks are unique, and how byes are handled. Confirm output format: list of arrays after each round.
Iterate over rounds until one player remains. For each round, create a new list for winners and process pairs.
For i from 0 to length-2 step 2, compare arr[i] and arr[i+1], append the smaller. If length is odd, append the last element as a bye.
After each round, add the current winners list to the result. Set the winners list as the new array and continue until length is 1.
State time complexity O(n log n) and space O(n). Walk through a small example to verify correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Model the tournament as a binary tree where each internal node represents a match and leaves are players. Assign ranks so that the depth of each player's leaf is non-increasing with rank (stronger players have deeper leaves). For non-powers of two, use byes or a balanced tree with dummy leaves to ensure the property holds.
Pro tip: Mention that the solution is essentially a 'seeding' problem and that byes should be given to the strongest players to maximize their potential rounds. This shows you understand real-world tournament design.
Clarify that 'eliminated in later rounds' means stronger players must survive more rounds than weaker ones. The permutation is the initial ordering of players in the bracket.
Represent the tournament as a full binary tree with n leaves (players). Each internal node is a match; the winner advances. The round of elimination is the depth of the leaf from the root.
For ranks i < j (i stronger), the depth of leaf i must be >= depth of leaf j. So stronger players must be placed deeper in the tree.
For n=2^k, a complete binary tree works. Assign ranks in order of increasing depth: rank 1 at deepest leaf, rank n at root's child (depth 1). This ensures stronger players last longer.
For non-powers of two, add dummy leaves (byes) to make n' = next power of two. Assign byes to strongest players (they get a free pass). Then map ranks to leaves ensuring depth non-increasing with rank.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.