1. Abridged Problem Statement  
Given N chains of lengths L1…LN, you wish to join them into a single line. In one minute you may open exactly one link (taking it out of its chain), use that open link to connect chains, then close it. Compute the minimum minutes needed to connect all chains into one continuous chain.

2. Detailed Editorial  
Goal: starting from N separate chains, we must perform exactly N–1 connections to end up with a single chain. Each connection consumes one original link (we open it, hook on another chain, and close it), so opening a link removes it from the chain it came from. We must choose which links to sacrifice as connectors.

Greedy idea: if you repeatedly break links from the currently shortest chain, you keep larger chains intact and available to be hooked on one by one. Once the shortest chain is exhausted, you move on to the next-shortest chain. Meanwhile you always attach the largest remaining chains, since spending all links of a small chain both supplies connectors and removes that small chain from the list still needing attachment. Concretely:

1. Sort the array of lengths in nondecreasing order.  
2. Let left = 0 (points at shortest chain), right = N−1 (points at the chain we're about to attach next), time = 0.  
3. While left < right:  
   a. If lengths[left] > 0:  
      – "Break" one link from chain[left] (lengths[left]--)  
      – Attach chain[right] by using that link (right--)  
      – time++  
      – If chain[left] is now empty (length 0), advance left++ to pick the next chain for breaking.  
   b. Else (lengths[left] == 0), just advance left++.  
4. When left == right, we have connected all other chains into the one at index right; output time.

Correctness: This uses exactly one link per connection. By always using the currently shortest chain's links first, we ensure that no larger chain is "wasted" as a source of connectors until needed. Sorting plus two pointers gives an O(N log N) solution.

3. C++ Solution

```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

int n;
vector<int> lengths;

void read() {
    cin >> n;
    lengths.resize(n);
    cin >> lengths;
}

void solve() {
    // To join all chains into one line we need N - 1 connections, and each
    // connection costs one minute of opening a link. The cheapest links to
    // sacrifice are those of the shortest chains: opening every link of a small
    // chain both spends those links as connectors and removes that chain from
    // the list to be attached. Sort the chains ascending and use two pointers:
    // left points at the chain we are dismantling for links, right at the
    // current largest chain still needing attachment. Each minute we open one
    // link from the left chain (decrementing its remaining length) and use it
    // to absorb the right chain (right--, time++); when the left chain is
    // exhausted we advance left. We stop once left meets right, i.e. all
    // chains are connected.

    sort(lengths.begin(), lengths.end());

    int left = 0, right = n - 1, time = 0;
    while(left < right) {
        if(lengths[left] > 0) {
            lengths[left]--;
            right--;
            time++;
            if(lengths[left] == 0) {
                left++;
            }
        } else {
            left++;
        }
    }

    cout << time << '\n';
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

4. Python Solution

```python
def minimum_time_to_connect_chains():
    import sys
    data = sys.stdin.read().split()
    n = int(data[0])
    # Read all chain lengths and sort ascending
    lengths = sorted(map(int, data[1:]))

    left, right = 0, n - 1   # two pointers: source chains (left) and chains to attach (right)
    time = 0

    # While there is more than one chain remaining
    while left < right:
        if lengths[left] > 0:
            # Open one link from the shortest available chain
            lengths[left] -= 1
            # Use it to attach the chain at 'right'
            right -= 1
            # Count one minute
            time += 1
            # If we've exhausted this chain as a source, move to next
            if lengths[left] == 0:
                left += 1
        else:
            # If no links remain in this source chain, skip it
            left += 1

    print(time)


if __name__ == "__main__":
    minimum_time_to_connect_chains()
```

5. Compressed Editorial  
Sort the chain lengths. Repeatedly open one link from the current shortest chain (decrement its length) to attach the largest remaining chain, counting one minute per operation. Move to the next shortest chain when it's depleted. This greedy two-pointer approach uses exactly one link per merge and runs in O(N log N).
