You have to find a permutation of the string where no letter is followed by another letter and no digit is followed by another digit. That is, no two adjacent characters have the same type.

Return the reformatted string or return an empty string if it is impossible to reformat the string.

Example 1:

Input: s = "a0b1c2"
Output: "0a1b2c"
Explanation: No two adjacent characters have the same type in "0a1b2c". "a0b1c2", "0a1b2c", "0c2a1b" are also valid permutations.

Example 2:

Input: s = "leetcode"
Output: ""
Explanation: "leetcode" has only characters so we cannot separate them by digits.

Example 3:

Input: s = "1229857369"
Output: ""
Explanation: "1229857369" has only digits so we cannot separate them by characters.

Example 4:

Input: s = "covid2019"
Output: "c2o0v1i9d"

Example 5:

Input: s = "ab123"
Output: "1a2b3"

Constraints:

• 1 <= s.length <= 500
• s consists of only lowercase English letters and/or digits.

Solution: Two streams

Create two stacks, one for alphas, another for numbers. If the larger stack has more than one element than the other one then no solution, return “”. Otherwise, interleave two stacks, start with the larger one.

Time complexity: O(n)
Space complexity: O(n)

// Author: Huahua
class Solution {
public:
  string reformat(string s) {
    string q1;
    string q2;
    for (char c : s) {
      if (isalpha(c))
        q1 += c;
      else
        q2 += c;
    }    
    if (q1.length() < q2.length())
      swap(q1, q2);
    if (q1.length() > q2.length() + 1) 
      return "";
    string ans;
    while (!q1.empty()) {
      ans += q1.back();
      q1.pop_back();
      if (q2.empty()) break;
      ans += q2.back();
      q2.pop_back();
    }
    return ans;
  }
};

The post 花花酱 LeetCode 1417. Reformat The String appeared first on Huahua's Tech Road.

Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.

Example 1:

Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
Output: true

Example 2:

Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
Output: false

Solution: DP

Subproblems : whether s3[0:i+j] can be formed by interleaving s1[0:i] and s2[0:j].

Time complexity: O(mn)

Space complexity: O(mn)

// Author: Huhaua
// Running time: 0 ms
class Solution {
public:
  bool isInterleave(string s1, string s2, string s3) {
    int l1 = s1.length();
    int l2 = s2.length();
    int l3 = s3.length();
    if (l1 + l2 != l3) return false;
    // dp[i][j]: whehter s3[0:i+j] is a interleva of s1[0:i] and s2[0:j]
    vector<vector<int>> dp(l1 + 1, vector<int>(l2 + 1));
    dp[0][0] = true;
    for (int i = 0; i <= l1; ++i)
      for (int j = 0; j <= l2; ++j) {        
        if (i > 0) dp[i][j] |= dp[i - 1][j] && s1[i - 1] == s3[i + j - 1];
        if (j > 0) dp[i][j] |= dp[i][j - 1] && s2[j - 1] == s3[i + j - 1];        
      }
    return dp[l1][l2];
  }
};

Recursion + Memorization

// Author: Huahua
// Running time: 0 ms
class Solution {
public:
  bool isInterleave(string s1, string s2, string s3) {
    m_ = vector<vector<int>>(s1.length() + 1, vector<int>(s2.length() + 1, INT_MIN));
    return dp(s1, s1.length(), s2, s2.length(), s3, s3.length());
  }
private:
  // m_[i][j]: whehter s3[0:i+j] is a interleva of s1[0:i] and s2[0:j]
  vector<vector<int>> m_;
  
  bool dp(const string& s1, int l1, const string& s2, int l2, const string& s3, int l3) {
    if (l1 + l2 != l3) return false;
    if (l1 == 0 && l2 == 0) return true;
    if (m_[l1][l2] != INT_MIN) return m_[l1][l2];
    if (l1 > 0 & s3[l3 - 1] == s1[l1 - 1] && dp(s1, l1 - 1, s2, l2, s3, l3 - 1)
      ||l2 > 0 & s3[l3 - 1] == s2[l2 - 1] && dp(s1, l1, s2, l2 - 1, s3, l3 - 1))
      m_[l1][l2] = 1;
    else
      m_[l1][l2] = 0;
    return m_[l1][l2];
  }
};

The post 花花酱 LeetCode 97. Interleaving String appeared first on Huahua's Tech Road.