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Problem solution · Go

Codeforces 1838D — Bracket Walk

Codeforces 1838D — Bracket Walk: 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
151 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 1838D — Bracket Walk, 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

  • 151 lines of Go from the credited upstream file 1838D.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 1838D — Bracket Walk · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	. "fmt"	"io"	"time") // https://space.bilibili.com/206214type node38 struct {	lr       [2]*node38	priority uint	key      int	subCnt   int} func (o *node38) size() int {	if o != nil {		return o.subCnt	}	return 0} func (o *node38) maintain() { o.subCnt = 1 + o.lr[0].size() + o.lr[1].size() } func (o *node38) rotate(d int) *node38 {	x := o.lr[d^1]	o.lr[d^1] = x.lr[d]	x.lr[d] = o	o.maintain()	x.maintain()	return x} type treap38 struct {	rd   uint	root *node38} func (t *treap38) fastRand() uint {	t.rd ^= t.rd << 13	t.rd ^= t.rd >> 17	t.rd ^= t.rd << 5	return t.rd} func (t *treap38) size() int { return t.root.size() } func (t *treap38) _put(o *node38, key int) *node38 {	if o == nil {		return &node38{priority: t.fastRand(), key: key, subCnt: 1}	}	if d := o.cmp(key); d >= 0 {		o.lr[d] = t._put(o.lr[d], key)		if o.lr[d].priority > o.priority {			o = o.rotate(d ^ 1)		}	} else {		// 相等	}	o.maintain()	return o} func (t *treap38) put(key int) { t.root = t._put(t.root, key) } func (t *treap38) _delete(o *node38, key int) *node38 {	if o == nil {		return nil	}	if d := o.cmp(key); d >= 0 {		o.lr[d] = t._delete(o.lr[d], key)	} else {		if o.lr[1] == nil {			return o.lr[0]		}		if o.lr[0] == nil {			return o.lr[1]		}		d = 0		if o.lr[0].priority > o.lr[1].priority {			d = 1		}		o = o.rotate(d)		o.lr[d] = t._delete(o.lr[d], key)	}	o.maintain()	return o} func (t *treap38) delete(key int) { t.root = t._delete(t.root, key) } func (o *node38) cmp(a int) int {	b := o.key	if a == b {		return -1	}	if a < b {		return 0	}	return 1} func (t *treap38) min() (min int) {	for o := t.root; o != nil; o = o.lr[0] {		min = o.key	}	return} func (t *treap38) max() (max int) {	for o := t.root; o != nil; o = o.lr[1] {		max = o.key	}	return} func CF1838D(_r io.Reader, _w io.Writer) {	in := bufio.NewReader(_r)	out := bufio.NewWriter(_w)	defer out.Flush() 	var n, q, p int	var s []byte	Fscan(in, &n, &q, &s)	t := &treap38{rd: uint(time.Now().UnixNano())/2 + 1}	for i, b := range s {		if b%2 != byte(i%2) {			t.put(i)		}	}	for ; q > 0; q-- {		Fscan(in, &p)		p--		s[p] ^= 1		if s[p]%2 != byte(p%2) {			t.put(p)		} else {			t.delete(p)		}		if n%2 > 0 || t.size() > 0 && t.min()%2 <= t.max()%2 {			Fprintln(out, "NO")		} else {			Fprintln(out, "YES")		}	}} //func main() { CF1838D(os.Stdin, os.Stdout) }

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