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Problem solution · Go

Codeforces 1941F — Rudolf and Imbalance

Codeforces 1941F — Rudolf and Imbalance: 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
57 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 1941F — Rudolf and Imbalance, 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

  • 57 lines of Go from the credited upstream file 1941F.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 1941F — Rudolf and Imbalance · GoGo
Use this to learn the idea, then write your own version.
package main import (	. "fmt"	"io"	"slices") // https://github.com/EndlessChengfunc cf1941F(in io.Reader, out io.Writer) {	var T, n, m, k int	for Fscan(in, &T); T > 0; T-- {		Fscan(in, &n, &m, &k)		a := make([]int, n)		for i := range a {			Fscan(in, &a[i])		}		b := make([]int, m)		for i := range b {			Fscan(in, &b[i])		}		c := make([]int, k)		for i := range c {			Fscan(in, &c[i])		} 		ans, se, mxI := a[1]-a[0], 0, 1		for i := 2; i < n; i++ {			d := a[i] - a[i-1]			if d > ans {				ans, se, mxI = d, ans, i			} else if d > se {				se = d			}		}		mid := (a[mxI-1] + a[mxI]) / 2 		slices.Sort(b)		slices.Sort(c)		j := k - 1		for _, v := range b {			for j >= 0 && v+c[j] > mid {				j--			}			if j >= 0 {				ans = min(ans, a[mxI]-v-c[j])			}			if j+1 < k {				ans = min(ans, v+c[j+1]-a[mxI-1])			}		}		Fprintln(out, max(ans, se))	}} //func main() { cf1941F(bufio.NewReader(os.Stdin), os.Stdout) }

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