Jumping through Lexicographic Permutations — Alternative Solution¶
Codex (GPT-6) applies established Lehmer codes, mixed-radix addition, and Fenwick order statistics to this problem.
Permutation ranks without large integers¶
Let \(d_i\) count values smaller than \(a_i\) to its right. The zero-based lexicographic rank is
Scan right to left with a Fenwick tree counting previously seen values. At the same time, add the step count. Initialize carry \(c=k\) ; the radix at this position is \(r=n-i+1\) . Set
Discard carry beyond the highest digit, implementing reduction modulo \(n!\) . The rightmost radix is one: its digit is zero and the carry passes entirely leftward.
Decode left to right by selecting the \((d_i'+1)\)-st smallest remaining value and deleting it. Binary descent inside the Fenwick tree selects a rank in \(O(\log n)\) ; an external binary search would introduce another logarithm.
Correctness and Tradeoffs¶
Each digit counts complete suffix blocks preceding its prefix, proving the rank formula. Mixed-radix carrying adds exactly \(k\) modulo \(n!\) . Order-statistic decoding inverts the same formula, yielding the correct cyclic successor.
Each case takes \(O(n\log n)\) time and \(O(n)\) space. No enormous rank or factorial is stored, and the structure is uniform. It is slower than the suffix-block method: the slowest median across five formal runs was about 0.741 seconds, exceeding the 0.5-second limit. Thus alternative.cpp is a backup; this measurement alone does not rule out every implementation with the same asymptotic complexity.