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Problem solution · Go

Codeforces 1144G — Two Merged Sequences

Codeforces 1144G — Two Merged Sequences: a Go solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to EndlessCheng Codeforces Go.

Technique
Direct simulation
Source
EndlessCheng Codeforces Go
Length
45 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

Direct simulation

For Codeforces 1144G — Two Merged Sequences, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 45 lines of Go from the credited upstream file 1144G.go.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

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 1144G — Two Merged Sequences · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	. "fmt"	"io") // https://github.com/EndlessChengfunc cf1144G(in io.Reader, _w io.Writer) {	out := bufio.NewWriter(_w)	defer out.Flush()	var n int	Fscan(in, &n)	a := make([]int, n+1)	for i := range n {		Fscan(in, &a[i])	} 	inc, dec := -1, int(1e9)	for i, v := range a[:n] {		// 如果 a[i+1] > v，把 v 放到递增使得 a[i+1] 仍有机会放入递增（因为 a[i+1] > v = new last_inc），		// 若把 v 放到递减，则 a[i+1] 也可能没法放进递减（取决于 last_dec），因此把 v 放进递增通常更有利		// 如果 a[i+1] <= v，下一个元素无法进入以 v 为 last 的递增序列，		// 但可能可以进入以 v 为 last 的递减序列；于是把 v 放入递减更合适		if v > inc && (v >= dec || v < a[i+1]) {			inc = v			a[i] = 0		} else if v < dec {			dec = v			a[i] = 1		} else {			Fprint(out, "NO")			return		}	} 	Fprintln(out, "YES")	for _, v := range a[:n] {		Fprint(out, v, " ")	}} //func main() { cf1144G(bufio.NewReader(os.Stdin), os.Stdout) }

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