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Problem solution · Go

Codeforces 220E — Little Elephant and Inversions

Codeforces 220E — Little Elephant and Inversions: 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
72 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 220E — Little Elephant and Inversions, 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

  • 72 lines of Go from the credited upstream file 220E.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 220E — Little Elephant and Inversions · GoGo
Use this to learn the idea, then write your own version.
package main import (	. "fmt"	"io"	"slices"	"sort") // https://github.com/EndlessChengtype fenwick20 []int func (f fenwick20) update(i, val int) {	for ; i < len(f); i += i & -i {		f[i] += val	}} // [1,i] 的和func (f fenwick20) sum(i int) (res int) {	for ; i > 0; i &= i - 1 {		res += f[i]	}	return} func cf220E(in io.Reader, out io.Writer) {	var n, k, ans int	Fscan(in, &n, &k)	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) 	// 计算不删除时的逆序对（直接从 k 中减掉）	suf := make(fenwick20, m+1)	for i := n - 1; i >= 0; i-- {		a[i] = sort.SearchInts(b, a[i]) + 1 // 离散化		k -= suf.sum(a[i] - 1)		suf.update(a[i], 1)	} 	pre := make(fenwick20, m+1)	l := 0	for r := 1; r < n; r++ {		// 从后缀中删除 a[r-1]，撤销逆序对（a[r-1] 与 pre 和 suf 的逆序对）		suf.update(a[r-1], -1)		k += l - pre.sum(a[r-1]) + suf.sum(a[r-1]-1)		for l < r {			// 尝试往前缀添加 a[l]			inv := l - pre.sum(a[l]) + suf.sum(a[l]-1)			if inv > k { // 逆序对太多了，无法添加				break			}			// 添加后，总逆序对个数 <= k，说明 (l,r) 满足要求			k -= inv			pre.update(a[l], 1)			l++		}		// 右端点为 r 时，左端点可以是 0,1,...,l-1，一共 l 个		ans += l	}	Fprint(out, ans)} //func main() { cf220E(bufio.NewReader(os.Stdin), os.Stdout) }

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