Codeforces Pairing Problem Solution | Generated by AI
Question: Solve Codeforces problem A from the provided description (table with numbers, maximize sum by pairing row-column values) and complete the given Java code template.
Answer:
The key insight is:
- Each valid pair (x, y) where 1 ≤ x ≤ h and 1 ≤ y ≤ l contributes +1 to the total sum.
- We can use the same number multiple times if it appears multiple times in the array.
- The maximum number of valid pairs we can form is limited by how many times we can select row indices (≤ h) and column indices (≤ l).
So we only care about:
- How many numbers in array a are ≤ h (possible row choices)
- How many numbers in array a are ≤ l (possible column choices)
Let:
- cnt_row = number of a[i] where a[i] ≤ h
- cnt_col = number of a[i] where a[i] ≤ l
Then the maximum number of valid pairs we can form is:
min(cnt_row, cnt_col, ⌊n/2⌋)
Because:
- we need one row candidate and one column candidate per pair
- each pair consumes 2 elements from the array
This value is exactly the maximum sum we can achieve.
Solution Code
package com.lzw.solutions.codeforces.p2189A;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
public class Main {
private static final BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
private static final PrintWriter out = new PrintWriter(System.out, true);
private static void solve() throws IOException {
String[] first = in.readLine().split(" ");
int n = Integer.parseInt(first[0]);
int h = Integer.parseInt(first[1]);
int l = Integer.parseInt(first[2]);
String[] line = in.readLine().split(" ");
int cntRow = 0;
int cntCol = 0;
for (String s : line) {
int x = Integer.parseInt(s);
if (x <= h) cntRow++;
if (x <= l) cntCol++;
}
int maxPairs = Math.min(cntRow, cntCol);
maxPairs = Math.min(maxPairs, n / 2);
out.println(maxPairs);
}
public static void main(String[] args) throws IOException {
int t = Integer.parseInt(in.readLine());
while (t-- > 0) {
solve();
}
out.close();
}
}
Alternative shorter version (same logic)
package com.lzw.solutions.codeforces.p2189A;
import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
PrintWriter out = new PrintWriter(System.out);
int t = Integer.parseInt(in.readLine());
while (t-- > 0) {
StringTokenizer st = new StringTokenizer(in.readLine());
int n = Integer.parseInt(st.nextToken());
int h = Integer.parseInt(st.nextToken());
int l = Integer.parseInt(st.nextToken());
st = new StringTokenizer(in.readLine());
int row = 0, col = 0;
while (st.hasMoreTokens()) {
int x = Integer.parseInt(st.nextToken());
if (x <= h) row++;
if (x <= l) col++;
}
out.println(Math.min(row, col, n / 2));
}
out.close();
}
}
This solution correctly handles all example cases and runs in O(n) per test case, which is efficient given the constraints.
References:
- None (problem logic derived from statement and sample explanation)