Stamping The Sequence

Problem Statement

The challenge is to transform a string s of the same length as the given target string but initialized with '?' characters into the target string by repeatedly overlaying a given string stamp onto s. Each overlay replaces a section of s with stamp. The task is to transform s to exactly match target within a specific number of turns, where each turn counts as one overlay of the stamp onto s.

In particular:

• You can stamp starting at any position within the string s as long as the stamp does not cross the boundaries. That is, for a stamp of length n, starting from position i in s, you must have i + n <= s.length.
• Your goal is to match s to target using at most 10 * target.length stamplings.
• The output should be the sequence of starting indices for each stamp that successfully transform s into target. If it's not possible to transform within the allowed number of turns, the output should be an empty array.

Examples

Example 1

Input:

stamp = "abc", target = "ababc"

Output:

[0,2]

Explanation:

Initially s = "?????".
- Place stamp at index 0 to get "abc??".
- Place stamp at index 2 to get "ababc".
[1,0,2] would also be accepted as an answer, as well as some other answers.

Example 2

Input:

stamp = "abca", target = "aabcaca"

Output:

[3,0,1]

Explanation:

Initially s = "???????".
- Place stamp at index 3 to get "???abca".
- Place stamp at index 0 to get "abcabca".
- Place stamp at index 1 to get "aabcaca".

Constraints

• 1 <= stamp.length <= target.length <= 1000
• stamp and target consist of lowercase English letters.

Approach and Intuition

To solve this problem, the objective is to systematically choose where to place the stamp on s to gradually transform it into target. Given the constraints, the approach requires carefully planning each stamp overlay to maximize the progress towards the final target. Here's a step-by-step intuition and approach:

1. Reverse Simulation:

   • Instead of starting with a blank s and making it like target, think in reverse by converting target back to s. Start with target and aim to revert it to all '?' by sequentially replacing stamp occurrences with '?'.
   • This approach is akin to unbuilding the target to see where stamps could have been placed, which then provides the stamplings in reverse order.

2. Tracking Progress:

   • Maintain a list or an array to track which characters in s have already been transformed in accordance with target.
   • Use a queue to handle the indices where matches of stamp in target (that can be transformed) are found, simulating the ability to stamp there in the reverse process.

3. Valid Placement:

   • Identify a valid placement of the stamp. This means a segment of s that either:
     • Directly corresponds to the stamp.
     • Is a mix of stamp characters and '?' but can be turned all into '?' when overlaid by the stamp.

4. Sequential Stamping:

   • Implement the placement by replacing the segment with '?' in the reverse simulation or with corresponding stamp characters in the direct approach.
   • Record the stamp index in every placement.

5. Optimization and Termination:

   • Continue the process until s is fully transformed into the target or identify if the transformation is impossible within the limited turns.
   • An optimal solution should be reached well before the maximum allowed moves, courtesy of strategic stamp placements.

By positioning each overlay carefully and leveraging the reverse transformation idea, we can effectively crack the transformation sequence required in efficiently matching s to target.

Solutions

class Solver {
    public int[] stampingSequence(String stamp, String target) {
        int stampLen = stamp.length(), targetLen = target.length();
        Queue<Integer> processQueue = new LinkedList<>();
        boolean[] visited = new boolean[targetLen];
        Stack<Integer> output = new Stack<>();
        List<StampWindow> windowsList = new ArrayList<>();
    
        for (int i = 0; i <= targetLen - stampLen; ++i) {
            Set<Integer> achieved = new HashSet();
            Set<Integer> required = new HashSet();
            for (int j = 0; j < stampLen; ++j) {
                if (target.charAt(i+j) == stamp.charAt(j))
                    achieved.add(i+j);
                else
                    required.add(i+j);
            }
    
            windowsList.add(new StampWindow(achieved, required));
                
            if (required.isEmpty()) {
                output.push(i);
                for (int j = i; j < i + stampLen; ++j) if (!visited[j]) {
                    processQueue.add(j);
                    visited[j] = true;
                }
            }
        }
    
        while (!processQueue.isEmpty()) {
            int current = processQueue.poll();
            for (int j = Math.max(0, current-stampLen+1); j <= Math.min(targetLen-stampLen, current); ++j) {
                if (windowsList.get(j).required.contains(current)) {
                    windowsList.get(j).required.remove(current);
                    if (windowsList.get(j).required.isEmpty()) {
                        output.push(j);
                        for (int m : windowsList.get(j).achieved) if (!visited[m]) {
                            processQueue.add(m);
                            visited[m] = true;
                        }
                    }
                }
            }
        }
    
        for (boolean b : visited)
            if (!b) return new int[0];
    
        int[] result = new int[output.size()];
        int idx = 0;
        while (!output.isEmpty())
            result[idx++] = output.pop();
    
        return result;
    }
}
    
class StampWindow {
    Set<Integer> achieved, required;
    StampWindow(Set<Integer> a, Set<Integer> r) {
        achieved = a;
        required = r;
    }
}