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 = s$_1$ + s$_2$ + … + s$_n$
• t = t$_1$ + t$_2$ + … + t$_m$
• |n - m| <= 1
• The interleaving is s$_1$ + t$_1$ + s$_2$ + t$_2$ + s$_3$ + t$_3$ + … or t$_1$ + s$_1$ + t$_2$ + s$_2$ + t$_3$ + s$_3$ + …

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)

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

        vector<bool> formable(n + 1, false);
        formable.front() = true;
        for (int i = 0; i <= m; ++i) {
            for (int j = 0; j <= n; ++j) {
                if (i == 0 && j == 0) {
                    continue;
                }

                bool from_s1 = i > 0 && formable[j] && s1[i - 1] == s3[i + j - 1];
                bool from_s2 = j > 0 && formable[j - 1] && s2[j - 1] == s3[i + j - 1];
                formable[j] = from_s1 || from_s2;
            }
        }

        return formable.back();
    }
};