Problem

You want to form a target string of lowercase letters.

At the beginning, your sequence is target.length '?' marks.  You also have a stamp of lowercase letters.

On each turn, you may place the stamp over the sequence, and replace every letter in the sequence with the corresponding letter from the stamp.  You can make up to 10 * target.length turns.

For example, if the initial sequence is “?????”, and your stamp is "abc",  then you may make “abc??”, “?abc?”, “??abc” in the first turn.  (Note that the stamp must be fully contained in the boundaries of the sequence in order to stamp.)

If the sequence is possible to stamp, then return an array of the index of the left-most letter being stamped at each turn.  If the sequence is not possible to stamp, return an empty array.

For example, if the sequence is “ababc”, and the stamp is "abc", then we could return the answer [0, 2], corresponding to the moves “?????” -> “abc??” -> “ababc”.

Also, if the sequence is possible to stamp, it is guaranteed it is possible to stamp within 10 * target.length moves.  Any answers specifying more than this number of moves will not be accepted.

Example 1:

Input: stamp = "abc", target = "ababc"
Output: [0,2]
([1,0,2] would also be accepted as an answer, as well as some other answers.)

Input: stamp = "abca", target = "aabcaca"
Output: [3,0,1]

1. 1 <= stamp.length <= target.length <= 1000
2. stamp and target only contain lowercase letters.

Solution: Greedy + Reverse Simulation

Reverse the stamping process. Each time find a full or partial match. Replace the matched char to ‘?’.

Don’t forget the reverse the answer as well.

T = “ababc”, S = “abc”

T = “ab???”, index = 2

T = “?????”, index = 0

ans = [0, 2]

Time complexity: O((T – S)*S)

Space complexity: O(T)

// Author: Huahua, running time: 12 ms
class Solution {
public:
  vector<int> movesToStamp(string stamp, string target) {
    vector<int> ans;
    vector<int> seen(target.length());
    int total = 0;
    while (total < target.length()) {      
      bool found = false;
      for (int i = 0; i <= target.length() - stamp.length(); ++i) {
        if (seen[i]) continue;
        int l = unStamp(stamp, target, i);
        if (l == 0) continue;
        seen[i] = 1;
        total += l;
        ans.push_back(i);
        found = true;
      }
      if (!found) return {};
    }
    reverse(begin(ans), end(ans));
    return ans;
  }
private:
  int unStamp(const string& stamp, string& target, int s) {    
    int l = stamp.size();
    for (int i = 0; i < stamp.length(); ++i) {
      if (target[s + i] == '?')
        --l;
      else if (target[s + i] != stamp[i])
        return 0;
    }
    
    if (l != 0)
      std::fill(begin(target) + s, begin(target) + s + stamp.length(), '?');
    return l;
  }
};

The post 花花酱 LeetCode 936. Stamping The Sequence appeared first on Huahua's Tech Road.

Given an array A of 0s and 1s, divide the array into 3 non-empty parts such that all of these parts represent the same binary value.

If it is possible, return any [i, j] with i+1 < j, such that:

• A[0], A[1], ..., A[i] is the first part;
• A[i+1], A[i+2], ..., A[j-1] is the second part, and
• A[j], A[j+1], ..., A[A.length - 1] is the third part.
• All three parts have equal binary value.

If it is not possible, return [-1, -1].

Note that the entire part is used when considering what binary value it represents.  For example, [1,1,0] represents 6 in decimal, not 3.  Also, leading zeros are allowed, so [0,1,1] and [1,1] represent the same value.

Example 1:

Input: [1,0,1,0,1]
Output: [0,3]

Example 2:

Input: [1,1,0,1,1]
Output: [-1,-1]

Note:

1. 3 <= A.length <= 30000
2. A[i] == 0 or A[i] == 1

Solution:

each part should have the same number of 1 s.

Find the suffix (without leading os) of the last part which should have 1/3 of the total ones.

Time complexity: O(n^2) in theory but close to O(n) in practice

Space complexity: O(n)

// Author: Huahua, 44 ms
class Solution {
public:
  vector<int> threeEqualParts(vector<int>& A) {    
    string s(begin(A), end(A));
    int ones = accumulate(begin(A), end(A), 0);
    if (ones % 3 != 0) return {-1, -1};
    if (ones == 0) return {0, A.size() - 1};
    ones /= 3;
    int right = A.size() - 1;
    while (ones) if (A[right--]) --ones;
    string suffix(begin(s) + right + 1, end(s));
    size_t l = suffix.length();
    size_t left = s.find(suffix);
    if (left == std::string::npos) return {-1, -1};
    size_t mid = s.find(suffix, left + l);
    if (mid == std::string::npos 
       || mid + 2 * l > s.length()) return {-1, -1};
    return {left + l - 1, mid + l};
  }
};

The post 花花酱 LeetCode 927. Three Equal Parts appeared first on Huahua's Tech Road.