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Problem solution · Go

Codeforces 1367F2 — Flying Sort (Hard Version)

Codeforces 1367F2 — Flying Sort (Hard Version): a Go solution using sorting and greedy selection. Learn the idea, check the complexity, and read the full code, with credit to EndlessCheng Codeforces Go.

Technique
Sorting and greedy selection
Source
EndlessCheng Codeforces Go
Length
58 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Sorting and greedy selection

For Codeforces 1367F2 — Flying Sort (Hard Version), the implementation first exposes a useful order, then scans that order while making locally justified choices.

  1. Choose the key that reveals the greedy or grouping structure.
  2. Sort the relevant records by that key.
  3. Scan in order, maintaining the invariant that makes each local choice safe.

Code notes

  • 58 lines of Go from the credited upstream file 1367F2.go.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Sorting is typically the dominant term unless the subsequent scan uses a more expensive nested operation.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from EndlessCheng Codeforces Go by Σndless (EndlessCheng) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeCodeforces 1367F2 — Flying Sort (Hard Version) · GoGo
Use this to learn the idea, then write your own version.
package main import (	. "fmt"	"io"	"slices"	"sort") // https://github.com/EndlessChengfunc cf1367F2(in io.Reader, out io.Writer) {	var T, n int	for Fscan(in, &T); T > 0; T-- {		Fscan(in, &n)		a := make([]int, n)		for i := range a {			Fscan(in, &a[i])		}		b := slices.Clone(a)		slices.Sort(b)		b = slices.Compact(b) 		m := len(b)		tot := make([]int, m)		for i, v := range a {			a[i] = sort.SearchInts(b, v)			tot[a[i]]++		} 		// f[v] 表示当前以值 v 结尾的最长子序列的长度		f := make([]int, m)		// f2[v] 记录以一整块以值 v 结尾的最长子序列的长度		// 「一整块」指的是子序列中包含了数组中所有的 v		// 只有当所有 v 都遍历到，这个状态才有意义，表示一个「已完成」的、可以被 v+1 安全转移的状态		full := make([]int, m)		cnt := make([]int, m)		for _, v := range a {			if v > 0 {				if cnt[v-1] == tot[v-1] {					f[v] = max(f[v], full[v-1])				} else {					f[v] = max(f[v], cnt[v-1])				}			}			f[v]++ // 虽然在 cnt[v-1] 的基础上 +1 不一定对，但这不是最优的			if cnt[v] > 0 {				full[v]++			} else {				full[v] = f[v]			}			cnt[v]++		}		Fprintln(out, n-slices.Max(f))	}} //func main() { cf1367F2(bufio.NewReader(os.Stdin), os.Stdout) }

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