1. Abridged Problem Statement  
Given a decimal number N (up to 101 000 digits) and an integer K (< number of digits in N), remove exactly K digits from N so that the resulting number P (preserving the order of the remaining digits) is as large as possible. Output P (without leading zeros, unless it is zero).

2. Detailed Editorial  

We need to delete K digits from N to maximize the remaining number P of length L = |N|−K. A classic greedy approach works in O(|N|·K) time when K is relatively small (here K ≤ 999).

Key idea: build P one digit at a time, from left to right.  
- For the first digit of P, we can choose any digit from N[0…K], because if we pick index i there, we still have K−i deletions left to apply on the suffix N[i+1…]. We pick the maximum digit in that prefix; if it appears multiple times, we record all their positions.  
- Suppose we have chosen t digits so far, and the last chosen position is at some index i. We have used up i−t+1 deletions to get there. Hence we have K′ = K − (i−t+1) deletions remaining. The next digit of P can be chosen from N[i+1…i+1+K′]. We again pick the maximum digit in that window. If there are multiple occurrences of that digit in the reachable set, we keep all of them since they might each lead to the same best prefix.  

Implementation details:  
- Maintain a vector valid_states of all possible indices in N that currently correspond to the last chosen digit of P.  
- At each step, from these states, mark reachable next positions (within the window for each state).  
- Scan all reachable positions to find the maximum digit.  
- Gather the new set of valid_states having that digit.  
- Append that digit to the answer string and repeat until P has length L.  

Time Complexity:  
Each of the L ≃ |N|−K steps examines up to O(|valid_states|·K) positions. Since |valid_states| is at most K+1 and K ≤ 999, this runs comfortably under the time limit for |N| up to 100 000.

3. Provided C++ Solution with Detailed Comments  

```cpp
#include <bits/stdc++.h>
using namespace std;

// Read and write helpers for pairs and vectors
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 k;
string n;

// Read input: N as a string, then integer k
void read() {
    cin >> n >> k;
}

// Main solve function
void solve() {
    string ans;                    // Will hold the result P
    vector<int> valid_states;      // Indices in n where the last chosen digit could be

    // Stage 1: pick the first digit of P from positions [0..k]
    char best = 0;
    for(int i = 0; i <= k; i++) {
        best = max(best, n[i]);    // find maximum digit in that prefix
    }
    for(int i = 0; i <= k; i++) {
        if(n[i] == best) {
            valid_states.push_back(i);  // record all positions with that best digit
        }
    }
    ans.push_back(best);           // append it to the answer

    // We need L = n.size() - k digits total
    int L = n.size() - k;
    // For each subsequent digit position in P:
    for(int l = 1; l < L; l++) {
        int used_states = valid_states.size();
        vector<bool> can_visit(n.size(), false);

        // Mark reachable next positions from each valid_state
        for(int idx : valid_states) {
            // We have used up (idx - (l-1)) deletions so far:
            int used_del = idx - (l - 1);
            // Allowed extra deletions: k - used_del
            int max_skip = k - used_del;
            // Next index candidates: idx+1 through idx+1+max_skip
            for(int step = 1; step <= max_skip; step++) {
                int nxt = idx + step;
                if(nxt >= (int)n.size() || can_visit[nxt]) break;
                can_visit[nxt] = true;
            }
        }

        // Among marked reachable positions, find the max digit
        char next_best = 0;
        for(int i = 0; i < (int)n.size(); i++) {
            if(can_visit[i]) next_best = max(next_best, n[i]);
        }

        // Build the new list of valid states: positions with digit == next_best
        vector<int> new_states;
        for(int i = 0; i < (int)n.size(); i++) {
            if(can_visit[i] && n[i] == next_best) {
                new_states.push_back(i);
            }
        }

        ans.push_back(next_best);    // append chosen digit
        valid_states.swap(new_states); // swap to reuse memory
    }

    // Output the final answer
    cout << ans << "\n";
}

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

    read();
    solve();
    return 0;
}
```

4. Python Solution with Detailed Comments  

```python
import sys

def max_after_k_removals(n: str, k: int) -> str:
    """
    Greedy stack-based approach:
    We iterate through digits of n, maintaining a stack.
    While we can still remove digits (k > 0) and the current digit
    is bigger than the top of the stack, pop the stack to increase
    the overall value.
    Finally, if we still have removals left, trim from the end.
    """
    stack = []
    removals = k

    for digit in n:
        # While we can remove and the top of stack is smaller than current digit
        while removals > 0 and stack and stack[-1] < digit:
            stack.pop()
            removals -= 1
        stack.append(digit)

    # If removals remain, remove from the end
    if removals:
        stack = stack[:-removals]

    # Build result and strip leading zeros (unless result is all zeros)
    result = ''.join(stack).lstrip('0')
    return result if result else "0"

def main():
    data = sys.stdin.read().split()
    n, k = data[0], int(data[1])
    print(max_after_k_removals(n, k))

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

5. Compressed Editorial  

Remove K digits to maximize the remaining number P. Greedily pick each digit of P from the left: at step t choose the maximum digit in N between the last picked position+1 and the furthest index reachable using remaining deletions. Keep all indices that tie for the max at each step to maintain future options in O(|N|·K) time. Alternatively, a stack-based one-pass solution removes smaller preceding digits while deletions remain.