DIY Wooden Ladder
Hazrat Ali
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?
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.
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.
4 4 1 3 1 3 3 3 3 2 5 2 3 3 4 2 3 1 1 2
2 1 2 0
Solution
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
vector<int> a(n);
for (int i = 0; i < n; i++) {
cin >> a[i];
}
sort(a.begin(), a.end());
int ans = (n > 2) ? min(n - 2, a[n - 2] - 1) : 0;
cout << ans << endl;
}
return 0;
}