Problem: Codeforces 1638A — Reverse · 한국어 · 日本語
The array is a permutation of 1..n. We may choose one segment and reverse it; the goal is to make the permutation lexicographically smallest.
Scan from left to right to find the first index i where p[i] != i + 1. Every earlier position already contains its smallest possible value, so it must stay fixed. The value i + 1 occurs somewhere to the right; let its position be j. Reversing [i, j] moves i + 1 to the first mismatching position while leaving the already-correct prefix unchanged. This is the lexicographically smallest possible result.
If no mismatch exists, the permutation is already sorted, so reversing a length-one segment leaves it unchanged. Finding the mismatch and the target value takes O(n) time, and the input permutation uses O(n) space per test case.
import java.io.BufferedInputStream;
import java.io.IOException;
public class Main {
private static final class FastScanner {
private final BufferedInputStream in = new BufferedInputStream(System.in);
private final byte[] buffer = new byte[1 << 16];
private int ptr;
private int len;
private int read() throws IOException {
if (ptr == len) {
len = in.read(buffer);
ptr = 0;
if (len == -1) return -1;
}
return buffer[ptr++];
}
int nextInt() throws IOException {
int c;
do {
c = read();
} while (c <= ' ' && c != -1);
int value = 0;
while (c > ' ') {
value = value * 10 + c - '0';
c = read();
}
return value;
}
}
public static void main(String[] args) throws IOException {
FastScanner fs = new FastScanner();
int tests = fs.nextInt();
StringBuilder answer = new StringBuilder();
while (tests-- > 0) {
int n = fs.nextInt();
int[] permutation = new int[n];
for (int i = 0; i < n; i++) {
permutation[i] = fs.nextInt();
}
int left = 0;
while (left < n && permutation[left] == left + 1) {
left++;
}
if (left < n) {
int right = left;
while (permutation[right] != left + 1) {
right++;
}
while (left < right) {
int value = permutation[left];
permutation[left++] = permutation[right];
permutation[right--] = value;
}
}
for (int i = 0; i < n; i++) {
if (i > 0) answer.append(' ');
answer.append(permutation[i]);
}
answer.append('\n');
}
System.out.print(answer);
}
}