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Problem solution · Go

Codeforces 724D — Dense Subsequence

Codeforces 724D — Dense Subsequence: a Go solution using sliding window or two pointers. Learn the idea, check the complexity, and read the full code, with credit to EndlessCheng Codeforces Go.

Technique
Sliding window or two pointers
Source
EndlessCheng Codeforces Go
Length
98 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

Sliding window or two pointers

For Codeforces 724D — Dense Subsequence, the implementation maintains a moving interval and updates only the information that enters or leaves the window.

  1. Choose the invariant that makes a window valid or useful.
  2. Advance the right boundary and add the new element.
  3. Move the left boundary only as needed while maintaining the invariant and updating the answer.

Code notes

  • 98 lines of Go from the credited upstream file 724D.go.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Confirm that neither pointer moves backwards; if so, the scan is usually linear apart from the window’s data-structure operations.

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 724D — Dense Subsequence · GoGo
Use this to learn the idea, then write your own version.
package main import (	"bufio"	"bytes"	. "fmt"	"io") // 反思：这题能不能一次过？// 0. 首先 O(26n) 当然是可以做的，但是我想试试 O(n) 的做法// 想出 O(n) 的做法需要想明白一些细节：// 1. 确认清楚目标是什么：确保所有 L 都被覆盖，即对位置 p，覆盖 [p-m+1,p]，最后检查 [0, n-m+1] 都被覆盖了（这是 L 的范围）// 2. 遍历时，每次先计算出下一个没有被覆盖的位置，这里用 find(0) 来计算// 3. 然后贪心地算出最靠右的覆盖 find(0) 的位置，若没有则当前字母全选，进入下一个字母 // github.com/EndlessCheng/codeforces-gofunc CF724D(_r io.Reader, _w io.Writer) {	in := bufio.NewReader(_r)	out := bufio.NewWriter(_w)	defer out.Flush() 	var m int	var s []byte	Fscan(in, &m, &s)	n := len(s)	pos := [26][]int{}	for i, b := range s {		b -= 'a'		pos[b] = append(pos[b], i)	}	var fa []int	initFa := func(n int) {		fa = make([]int, n+1)		for i := range fa {			fa[i] = i		}	}	var find func(int) int	find = func(x int) int {		if fa[x] != x {			fa[x] = find(fa[x])		}		return fa[x]	}	mergeRange := func(l, r int) (merged bool) {		if l < 0 {			l = 0		}		for i := find(l); i < r; i = find(i + 1) {			fa[i] = r			merged = true		}		return	} 	initFa(n)	ans := make([]byte, 0, n)outer:	for i, ps := range pos {		b := byte(i + 'a')		left := len(ps)		for j := 0; j < len(ps); j++ {			check := find(0)			found := false			for ; j < len(ps); j++ {				if ps[j]-m+1 > check {					break				}				found = true			}			if !found {				for _, p := range ps[j:] {					mergeRange(p-m+1, p+1)				}				break			}			if j > 0 {				j--			}			p := ps[j]			if mergeRange(p-m+1, p+1) {				ans = append(ans, b)				left--				if find(0) >= n-m+1 {					break outer				}			}		}		ans = append(ans, bytes.Repeat([]byte{b}, left)...)	}	Fprint(out, string(ans))} //func main() {//	CF724D(os.Stdin, os.Stdout)//}

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