97. Interleaving String

1. 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.

2. Example

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

3. Constraints

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

4. Solutions

Dynamic Programming

m = s1.size(), n = s2.size()
Time complexity: O(mn)
Space complexity: O(n)

class Solution {
public:
    bool isInterleave(const string &s1, const string &s2, const string &s3) {
        const int m = s1.size(), n = s2.size(), l = s3.size();
        if (m + n != l) {
            return false;
        }

        vector<bool> match(n + 1, false);
        match[0] = true;

        for (int i = 0; i <= m; ++i) {
            if (i > 0) {
                match[0] = match[0] && s1[i - 1] == s3[i - 1];
            }

            bool has_valid_solution = match[0];

            for (int j = 1; j <= n; ++j) {
                bool from_s1 = i > 0 && match[j] && s1[i - 1] == s3[i + j - 1];
                bool from_s2 = match[j - 1] && s2[j - 1] == s3[i + j - 1];

                match[j] = from_s1 || from_s2;

                has_valid_solution = has_valid_solution || match[j];
            }

            if (!has_valid_solution) {
                return false;
            }
        }

        return match.back();
    }
};