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. Follow Up

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

5. Solutions

Dynamic Programming(Follow Up)

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) {
        if (s1.size() + s2.size() != s3.size()) {
            return false;
        }

        // interleave[i][j] means if s1[0..i] and s2[0..j] can form s3[0..i+j]
        vector<bool> interleave(s2.size() + 1);
        interleave[0] = true;
        for (int i = 0; i <= s1.size(); ++i) {
            for (int j = 0; j <= s2.size(); ++j) {
                if (i > 0) {
                    interleave[j] = interleave[j] & interleave[j] && s1[i - 1] == s3[i + j - 1];
                }
                if (j > 0) {
                    interleave[j] =
                        interleave[j] || interleave[j - 1] && s2[j - 1] == s3[i + j - 1];
                }
            }
        }

        return interleave.back();
    }
};