You are asked if you can choose n disjoint subarrays from the array nums such that the ith subarray is equal to groups[i] (0-indexed), and if i > 0, the (i-1)th subarray appears before the ith subarray in nums (i.e. the subarrays must be in the same order as groups).

Return true if you can do this task, and false otherwise.

Note that the subarrays are disjoint if and only if there is no index k such that nums[k] belongs to more than one subarray. A subarray is a contiguous sequence of elements within an array.

Example 1:

Input: groups = [[1,-1,-1],[3,-2,0]], nums = [1,-1,0,1,-1,-1,3,-2,0]
Output: true
Explanation: You can choose the 0th subarray as [1,-1,0,1,-1,-1,3,-2,0] and the 1st one as [1,-1,0,1,-1,-1,3,-2,0].
These subarrays are disjoint as they share no common nums[k] element.

Example 2:

Input: groups = [[10,-2],[1,2,3,4]], nums = [1,2,3,4,10,-2]
Output: false
Explanation: Note that choosing the subarrays [1,2,3,4,10,-2] and [1,2,3,4,10,-2] is incorrect because they are not in the same order as in groups.
[10,-2] must come before [1,2,3,4].

Example 3:

Input: groups = [[1,2,3],[3,4]], nums = [7,7,1,2,3,4,7,7]
Output: false
Explanation: Note that choosing the subarrays [7,7,1,2,3,4,7,7] and [7,7,1,2,3,4,7,7] is invalid because they are not disjoint.
They share a common elements nums[4] (0-indexed).

Constraints:

• groups.length == n
• 1 <= n <= 103
• 1 <= groups[i].length, sum(groups[i].length) <= 103
• 1 <= nums.length <= 103
• -107 <= groups[i][j], nums[k] <= 107

Solution: Brute Force

Time complexity: O(n^2?)
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  bool canChoose(vector<vector<int>>& groups, vector<int>& nums) {
    const int n = nums.size();
    int s = 0;
    for (const auto& g : groups) {
      bool found = false;
      for (int i = s; i <= n - g.size(); ++i)
        if (equal(begin(g), end(g), begin(nums) + i)) {
          s = i + g.size();
          found = true;
          break;
        }
      if (!found) return false;
    }
    return true;
  }
};

The post 花花酱 LeetCode 1764. Form Array by Concatenating Subarrays of Another Array appeared first on Huahua's Tech Road.

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 input string (s) and a pattern (p), implement regular expression matching with support for '.' and '*'.

'.' Matches any single character.
'*' Matches zero or more of the preceding element.

The matching should cover the entire input string (not partial).

Note:

• s could be empty and contains only lowercase letters a-z.
• p could be empty and contains only lowercase letters a-z, and characters like . or *.

Example 1:

Input:
s = "aa"
p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".

Example 2:

Input:
s = "aa"
p = "a*"
Output: true
Explanation: '*' means zero or more of the precedeng element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".

Example 3:

Input:
s = "ab"
p = ".*"
Output: true
Explanation: ".*" means "zero or more (*) of any character (.)".

Example 4:

Input:
s = "aab"
p = "c*a*b"
Output: true
Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore it matches "aab".

Example 5:

Input:
s = "mississippi"
p = "mis*is*p*."
Output: false

Solution 1: Recursion

Time complexity: O((|s| + |p|) * 2 ^ (|s| + |p|))

Space complexity: O(|s| + |p|)

// Author: Huahua, 14 ms
class Solution {
public:
  bool isMatch(string s, string p) {
    return isMatch(s.c_str(), p.c_str());
  }
private:
  bool isMatch(const char* s, const char* p) {
    if (*p == '\0') return *s == '\0';
        
    // normal case, e.g. 'a.b','aaa', 'a'
    if (p[1] != '*' || p[1] == '\0') {
      // no char to match
      if (*s == '\0') return false;

      if (*s == *p || *p == '.')
        return isMatch(s + 1, p + 1);
      else
        return false;
    }
    else {
      int i = -1;
      while (i == -1 || s[i] == p[0] || p[0] == '.') {
          if (isMatch(s + i + 1, p + 2)) return true;
          if (s[++i] == '\0') break;
      }
      return false;
    }
    
    return false;
  }
};

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