cf1197A A. DIY Wooden Ladder

Let’s denote a k-step ladder as the following structure: exactly k+2 wooden planks, of which

two planks of length at least k+1 — the base of the ladder;
k planks of length at least 1 — the steps of the ladder;
Note that neither the base planks, nor the steps planks are required to be equal.

For example, ladders 1 and 3 are correct 2-step ladders and ladder 2 is a correct 1-step ladder. On the first picture the lengths of planks are [3,3] for the base and [1] for the step. On the second picture lengths are [3,3] for the base and [2] for the step. On the third picture lengths are [3,4] for the base and [2,3] for the steps.

You have n planks. The length of the i-th planks is ai. You don’t have a saw, so you can’t cut the planks you have. Though you have a hammer and nails, so you can assemble the improvised “ladder” from the planks.

The question is: what is the maximum number k such that you can choose some subset of the given planks and assemble a k-step ladder using them?

Input
The first line contains a single integer T (1≤T≤100) — the number of queries. The queries are independent.

Each query consists of two lines. The first line contains a single integer n (2≤n≤105) — the number of planks you have.

The second line contains n integers a1,a2,…,an (1≤ai≤105) — the lengths of the corresponding planks.

It’s guaranteed that the total number of planks from all queries doesn’t exceed 105.

Output
Print T integers — one per query. The i-th integer is the maximum number k, such that you can choose some subset of the planks given in the i-th query and assemble a k-step ladder using them.

Print 0 if you can’t make even 1-step ladder from the given set of planks.

Example
input
4
4
1 3 1 3
3
3 3 2
5
2 3 3 4 2
3
1 1 2
output
2
1
2
0
Note
Examples for the queries 1−3 are shown at the image in the legend section.

The Russian meme to express the quality of the ladders:
**题意：**给出测试样例的数目，每个测试样子包含一个n和n个数，从中选出能搭建楼梯的阶数，如果是k阶楼梯，那么楼梯的两边长度必须为k+1，问最多能搭建几个阶梯。
**思路：**将数据排序，选择最大的两个数做楼梯的两边，剩下的做个判断即可，详情看代码。

#include <iostream>
#include <algorithm>
using namespace std;
int main() {
	int t, n, a[100005];
	scanf("%d", &t);
	while (t--) {
		int ans = 0;
		scanf("%d", &n);
		for (int i = 0; i < n; i++) {
			scanf("%d", &a[i]);
		}
		sort(a, a + n);
		int num = a[n-2]; // 第二大的数 
		if (num <= n-2) { // 如果第二大的数小于等于前面数据的个数，则楼梯阶数最多为第二大数减一 
			ans = num - 1;
		} else {
			if (n >= 3) { // 如果数大于三个，则为除两边的数据的个数 
				ans = n - 2;
			} else { // 否则为0 
				ans = 0;
			}
		}
		printf("%d\n", ans);
	}
	return 0;
}