969 interlaced string

The title is as follows

Given three strings s1、s2、s3, Please help to verify s3 Is it by s1 and s2 staggered Composed of .

Two strings s and t staggered The definition and process are as follows , Each of these strings is split into several Non empty Substring ：

s = s1 + s2 + … + sn
t = t1 + t2 + … + tm
|n - m| <= 1
staggered yes s1 + t1 + s2 + t2 + s3 + t3 + … perhaps t1 + s1 + t2 + s2 + t3 + s3 + …
Be careful ：a + b It means string a and b Connect .

Example 1：
Input ：s1 = “aabcc”, s2 = “dbbca”, s3 = “aadbbcbcac”
Output ：true

Example 2：
Input ：s1 = “aabcc”, s2 = “dbbca”, s3 = “aadbbbaccc”
Output ：false

Example 3：
Input ：s1 = “”, s2 = “”, s3 = “”
Output ：true

Tips ：
0 <= s1.length, s2.length <= 100
0 <= s3.length <= 200
s1、s2、 and s3 It's all made up of lowercase letters

How to solve the problem 1

Dynamic programming

class Solution 
{
    
public:
    bool isInterleave(string s1, string s2, string s3) 
    {
    
        auto f = vector < vector <int> >(s1.size() + 1, vector <int>(s2.size() + 1, false));

        int n = s1.size(), m = s2.size(), t = s3.size();

        if (n + m != t)
        {
    
            return false;
        }

        f[0][0] = true;
        for (int i = 0; i <= n; ++i)
        {
    
            for (int j = 0; j <= m; ++j)
            {
    
                int p = i + j - 1;
                if (i > 0)
                {
    
                    f[i][j] |= (f[i - 1][j] && s1[i - 1] == s3[p]);
                }
                if (j > 0) 
                {
    
                    f[i][j] |= (f[i][j - 1] && s2[j - 1] == s3[p]);
                }
            }
        }
        return f[n][m];
    }
};

It is not difficult to see that the time complexity and space complexity of this implementation are O(nm).

Use rolling arrays to optimize spatial complexity . Because the array here ff Of the ii Line only and No i - 1i−1 Row correlation , So we can optimize this dynamic programming with rolling array , In this way, the spatial complexity can become O(m).

An optimization method

class Solution
{
    
public:
    bool isInterleave(string s1, string s2, string s3)
    {
    
        auto f = vector <int>(s2.size() + 1, false);

        int n = s1.size(), m = s2.size(), t = s3.size();

        if (n + m != t)
        {
    
            return false;
        }

        f[0] = true;
        for (int i = 0; i <= n; ++i) 
        {
    
            for (int j = 0; j <= m; ++j) 
            {
    
                int p = i + j - 1;
                if (i > 0)
                {
    
                    f[j] &= (s1[i - 1] == s3[p]);
                }
                if (j > 0)
                {
    
                    f[j] |= (f[j - 1] && s2[j - 1] == s3[p]);
                }
            }
        }
        return f[m];
    }
};

Complexity analysis

Time complexity ：O(nm), The time cost of the double cycle is O(nm).
Spatial complexity ：O(m), namely s 2 The length of .