1. Abridged Problem Statement  
You are given two strings: a text s of length n and a message t of length m, where t is guaranteed to be a subsequence of s. You may remove any number of characters (possibly zero) from the start of s and any number (possibly zero) from the end of s. In how many distinct ways (pairs of prefix‐ and suffix‐removal counts) does t remain a subsequence of the resulting substring? Output that count.

2. Detailed Editorial

Goal  
We need to count all pairs \((x,y)\) with \(0 \le x+y \le n\) such that \(t\) is still a subsequence of the substring \(s[x\,..\,n-1-y]\).

Key idea  
For each possible prefix removal \(x\) (i.e.\ starting index \(i=x\)), determine the smallest ending index \(e\) in \(s\) so that \(t\) can be found as a subsequence of \(s[i..e]\). Once you know that minimal \(e\), any suffix removal \(y\) satisfying \(y \le n-1-e\) keeps \(e\) within bounds and thus maintains the subsequence. Hence for that \(i\) there are exactly \((n-e)\) valid suffix removals.

How to find minimal end index for each start  
We preprocess a “next‐occurrence” table \(\text{nxt}[pos][c]\) = the smallest index \(\ge pos\) where character \(c\) appears in \(s\), or \(n\) if \(c\) does not reappear after \(pos\). This table is built in \(O(n\cdot|\Sigma|)\) by scanning \(s\) from right to left and copying the future values, then updating for the character at the current position.

Once built, for each start \(i\), we initialize a pointer \(cur = i\!-\!1\). Then for each character \(c\) in \(t\), we jump to  
  cur = nxt[cur+1][c]  
If at any point cur becomes \(n\), it means \(t\) cannot be completed from that start and contributes 0. Otherwise, at the end cur is the minimal ending position \(e\). We add \((n-e)\) to the global answer. Summing over all starts \(i=0..n-1\) gives the result.

Time and memory  
- Building nxt: \(O(n\cdot 26)\).  
- Matching \(t\) for every start: \(O(n\cdot m)\).  
Total \(O(n\,(m+26))\), which is fine for \(n\) up to about \(10^6\) and \(m\le100\). Memory is \(O(n\cdot26)\) for nxt.

3. C++ Solution with Detailed Comments

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

// nxt[i][c] = the smallest index ≥ i where character 'a'+c appears in s,
// or n if no such position exists.
static const int MAXN = 1 << 20;  // up to about 1e6
int nxt[MAXN][26];

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

    string s, t;
    cin >> s >> t;
    int n = s.size();
    int m = t.size();

    // Base case: beyond the end of s, no characters appear
    for(int c = 0; c < 26; c++){
        nxt[n][c] = n;
    }

    // Build nxt table from right to left
    for(int i = n - 1; i >= 0; i--){
        // First, copy the values from i+1
        for(int c = 0; c < 26; c++){
            nxt[i][c] = nxt[i+1][c];
        }
        // Then update for the character s[i]
        int cidx = s[i] - 'a';
        nxt[i][cidx] = i;
    }

    long long answer = 0;
    // Try every possible prefix removal i
    for(int i = 0; i < n; i++){
        // We'll walk through t, starting just before i
        int cur = i - 1;
        // For each character in t, jump to its next occurrence
        for(char ch : t){
            int cidx = ch - 'a';
            cur = nxt[cur + 1][cidx];
            if(cur == n){
                // Cannot complete t from this start
                break;
            }
        }
        if(cur < n){
            // If we succeeded, any suffix removal up to n-1-cur works
            answer += (n - cur);
        }
        // If cur == n, contribution is zero, so skip
    }

    cout << answer << "\n";
    return 0;
}
```

4. Python Solution with Detailed Comments

```python
def count_ways(s: str, t: str) -> int:
    n, m = len(s), len(t)

    # Build next-occurrence table
    # nxt[i][c] = smallest index >= i where s[index] == chr(c + ord('a')), or n if none
    nxt = [ [n]*26 for _ in range(n+1) ]
    
    # Base: nxt[n] is all n's
    for i in range(n-1, -1, -1):
        # Copy from the row i+1
        nxt[i][:] = nxt[i+1][:]
        # Update the entry for s[i]
        nxt[i][ord(s[i]) - 97] = i

    answer = 0
    # Try every prefix removal i
    for i in range(n):
        cur = i - 1
        # Try to match t as a subsequence
        for ch in t:
            ci = ord(ch) - 97
            # Jump to next occurrence at or after cur+1
            cur = nxt[cur+1][ci]
            if cur == n:
                # cannot match t from this start
                break
        if cur < n:
            # we found t ending at cur, so suffix removal up to n-1-cur allowed
            answer += (n - cur)
    return answer

if __name__ == "__main__":
    import sys
    data = sys.stdin.read().split()
    s, t = data[0], data[1]
    print(count_ways(s, t))
```

Explanation of key steps in Python:  
- We use a 2D list `nxt` of size \((n+1)\times 26\).  
- We fill it backward so that each `nxt[i]` knows where each letter next appears if you start looking at position `i`.  
- For each `i` from `0` to `n-1` (meaning remove the first `i` chars), we try to greedily match `t` in the remainder by jumping via `nxt`.  
- If the match finishes at position `cur < n`, there are `n-cur` choices for how many characters to strip from the end.

5. Compressed Editorial  
1. Precompute `nxt[i][c]`: next index ≥ i where character c occurs in s, or n if none.  
2. For each starting cut `i`, simulate matching `t` by repeatedly doing `cur = nxt[cur+1][c]`.  
3. If the match ends at `cur<n`, then you can remove any of the `n-cur` suffix characters and still contain `t`.  
4. Sum these counts across all `i`. This runs in \(O(n(26+m))\).