Interleaving String

Description

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?

Code

DFS with memo

Time Complexity: $O(m\ast n)$, Space Complexity: $O(m\ast n)$

注意 memo 的大小是 (m + 1) * (n + 1)，因為遞迴可能會有 i == m 或是 j == n 的 case。

class Solution {
public:
    bool isInterleave(string s1, string s2, string s3) {
        if(s1.length() + s2.length() != s3.length()) return false;
        int m = s1.length(), n = s2.length();
        vector<vector<int>> memo(m + 1, vector<int>(n + 1, -1));
        return dfs(0, 0, s1, s2, s3, memo);
    }
 
    bool dfs(int i, int j, string& s1, string& s2, string& s3, vector<vector<int>>& memo) {
        if(i >= s1.length() && j >= s2.length() && (i + j) == s3.length()) return true;
 
        if(memo[i][j] != -1) return memo[i][j];
 
        bool a = (i < s1.length() && s1[i] == s3[i + j]) ? dfs(i + 1, j, s1, s2, s3, memo) : false;
        bool b = (j < s2.length() && s2[j] == s3[i + j]) ? dfs(i, j + 1, s1, s2, s3, memo) : false;
        memo[i][j] = a || b;
        return memo[i][j];
    }
};

Source

• Interleaving String - LeetCode