0936 - Stamping The Sequence (Hard)

Problem Link

https://leetcode.com/problems/stamping-the-sequence

Problem Statement

You are given two strings stamp and target. Initially, there is a string s of length target.length with all s[i] == '?'.

In one turn, you can place stamp over s and replace every letter in the s with the corresponding letter from stamp.

• For example, if stamp = "abc" and target = "abcba", then s is "?????" initially. In one turn you can:

  • place stamp at index 0 of s to obtain "abc??",
  • place stamp at index 1 of s to obtain "?abc?", or
  • place stamp at index 2 of s to obtain "??abc".

  Note that stamp must be fully contained in the boundaries of s in order to stamp (i.e., you cannot place stamp at index 3 of s).

We want to convert s to target using at most 10 * target.length turns.

Return an array of the index of the left-most letter being stamped at each turn. If we cannot obtain target from s within 10 * target.length turns, return an empty array.

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 1: Greedy

We first try the first stamp and mark those characters to$*$. Then build some new stamps to check if they exist in target, if so mark them and repeat the process until all characters are stamped.

Written by @wkw

class Solution {
public:
    vector<int> movesToStamp(string stamp, string target) {
        int n = stamp.size(), total = 0, k = -1;
        vector<int> ans;
        // "abca"
        // "aabcaca"
        while (k) {
            k = 0;
            for (int j = n; j > 0; j--) {
                for (int i = 0; i <= n - j; i++) {
                    // build the new stamp
                    string new_stamp = string(i, '*') + stamp.substr(i, j) + string(n - j - i, '*');
                    // abca
                    // abc*
                    // *bca
                    // ab**
                    // *bc*
                    // **ca
                    // a***
                    // *b**
                    // **c*
                    // ***a

                    // check if we can use this new_stamp to cover some characters
                    auto p = target.find(new_stamp);
                    while (p != string::npos) {
                        // if so, mark this position
                        ans.push_back(p);
                        // and replace those characters
                        // e.g.
                        // aabcaca
                        // a****ca
                        // a****ca
                        // a******
                        // a******
                        // *******
                        fill(target.begin() + p, target.begin() + p + n, '*');
                        // try to find if we can cover another set
                        p = target.find(new_stamp);
                        // j is how many characters we've covered
                        k += j;
                    }
                }
            }
            // update the total covered characters
            total += k;
        }
        reverse(ans.begin(), ans.end());
        return total == target.size() ? ans : vector<int>{};
    }
};