Overview

Given \(N\) cards with distinct integers written on them, this problem asks for the minimum number of groups needed when dividing the cards into “consecutive groups” — groups where, when arranged in ascending order, the values increase by \(1\) at each step.

Analysis

The key insight for solving this problem is that “the final grouping is not affected by the order in which the cards are arranged.” Therefore, it is natural to first sort the cards in ascending order.

For example, if the cards are \(\{3, 5, 4, 8, 7\}\), sorting them gives \(\{3, 4, 5, 7, 8\}\). - \(3\) and \(4\) are consecutive (difference of \(1\)), so they can be placed in the same group. - \(4\) and \(5\) are consecutive (difference of \(1\)), so they can be placed in the same group. - \(5\) and \(7\) are not consecutive (difference of \(2\)), so we must split into a new group here. - \(7\) and \(8\) are consecutive (difference of \(1\)), so they can be placed in the same group.

As a result, we can see that dividing into \(2\) groups, \(\{3, 4, 5\}\) and \(\{7, 8\}\), is optimal (minimum).

In other words, by counting the number of positions where the difference between adjacent elements is not \(1\) in the sorted array, we can determine the required number of groups. A new group is needed whenever a card appears whose value is not consecutive with the previous card.

Algorithm

1. Store the values \(A_i\) of the given \(N\) cards in an array and sort it in ascending order.
2. Initialize the group count (ans) to \(1\) (since if there is at least one card, there is at least one group).
3. Scan the array from the beginning, comparing adjacent elements \(A_i\) and \(A_{i+1}\).
4. If \(A_{i+1} \neq A_i + 1\), the values are not consecutive, so increment ans by \(1\).
5. Output the final value of ans.

Complexity

• Time Complexity: \(O(N \log N)\)
  • Sorting the cards takes \(O(N \log N)\).
  • The subsequent array scan is \(O(N)\).
  • Overall, the sorting complexity is dominant, but for \(N = 2 \times 10^5\), this is well within the time limit.
• Space Complexity: \(O(N)\)
  • \(O(N)\) memory is used to store the \(N\) input cards.

Implementation Notes

• Fast I/O: In Python, using sys.stdin.read().split() allows reading a large amount of input all at once, which helps reduce execution time.

• Comparing adjacent elements: By setting the loop range up to \(N-1\) and comparing A[i] with A[i+1], we can correctly check for consecutiveness while avoiding out-of-bounds array access.

  Source Code

import sys

def main():
    # 標準入力からすべてのデータを取得し、空白（スペースや改行）で分割してリストにする
    # この方法は大量の入力を高速に読み込むのに適しています
    input_data = sys.stdin.read().split()
    
    # 入力が空の場合は何もせず終了
    if not input_data:
        return
    
    # 最初の値がカードの枚数 N
    n = int(input_data[0])
    
    # 残りの値が各カードに書かれた整数 A_i
    # 整数型のリストに変換する
    a = list(map(int, input_data[1:]))
    
    # カードに書かれた整数を昇順にソートする
    # 連番グループを特定しやすくするため
    a.sort()
    
    # カードが0枚の場合はグループ数も0（制約上は 1 <= N）
    if n == 0:
        print(0)
        return
    
    # 最小のグループ数を求める
    # ソートされた状態で、隣り合う要素の差が1でない箇所があるたびに
    # 新しいグループが必要になる
    ans = 1
    for i in range(n - 1):
        # 隣り合う整数の差が1でない場合、そこは連番ではない
        if a[i+1] != a[i] + 1:
            ans += 1
    
    # 結果を出力する
    print(ans)

if __name__ == '__main__':
    main()

This editorial was generated by gemini-3-flash-thinking.