97. Interleaving String

Medium (Trung bình) C++ 🔗 Xem trên LeetCode

📋 Đề Bài

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

An interleaving of two strings s and t is a configuration where s and t are divided into n and m substrings respectively, such that:

  • s = s1 + s2 + ... + sn
  • t = t1 + t2 + ... + tm
  • |n - m| <= 1
  • The interleaving is s1 + t1 + s2 + t2 + s3 + t3 + ... or t1 + s1 + t2 + s2 + t3 + s3 + ...

Note: a + b is the concatenation of strings a and b.

 

Example 1:

Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
Output: true
Explanation: One way to obtain s3 is:
Split s1 into s1 = "aa" + "bc" + "c", and s2 into s2 = "dbbc" + "a".
Interleaving the two splits, we get "aa" + "dbbc" + "bc" + "a" + "c" = "aadbbcbcac".
Since s3 can be obtained by interleaving s1 and s2, we return true.

Example 2:

Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
Output: false
Explanation: Notice how it is impossible to interleave s2 with any other string to obtain s3.

Example 3:

Input: s1 = "", s2 = "", s3 = ""
Output: true

 

Constraints:

  • 0 <= s1.length, s2.length <= 100
  • 0 <= s3.length <= 200
  • s1, s2, and s3 consist of lowercase English letters.

 

Follow up: Could you solve it using only O(s2.length) additional memory space?

🧠 Thuật Toán & Kỹ Thuật

Dynamic Programming (Quy hoạch động)Matrix (Ma trận)String (Chuỗi)
⏱️ Thời gian O(n×m)
💾 Không gian O(n×m)

💻 Lời Giải

C++ 0097-interleaving-string.cpp 
int memo[101][101][201];

class Solution {
private:
    string s1, s2, s3;
    int size1, size2, size3;
    
public:
    int dp(int i, int j, int k) {
        if (i == size1 and j == size2 and k == size3) {
            return true;
        }
        if (memo[i][j][k] != -1) {
            return memo[i][j][k];
        }
        
        int ans = 0;
        
        if (i < size1 and s1[i] == s3[k]) {
            ans |= dp(i + 1, j, k + 1);
        }
        if (ans == 1) {
            return memo[i][j][k] = ans;
        }
        if (j < size2 and s2[j] == s3[k]) {
            ans |= dp(i, j + 1, k + 1);
        }
        return memo[i][j][k] = ans;
    }
    
    bool isInterleave(string s1, string s2, string s3) {
        memset(memo, -1, sizeof(memo));
        this->s1 = s1;
        this->s2 = s2;
        this->s3 = s3;
        this->size1 = s1.size();
        this->size2 = s2.size();
        this->size3 = s3.size();
        return dp(0, 0, 0) == 1;
    }
};