97. Interleaving String

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?

Solution:

Recursion

class Solution {
    public boolean isInterleave(String s1, String s2, String s3) {
       int n = s1.length();
       int m = s2.length();
       // If the total length doesn't match, it cannot be interleaving
       if (n + m != s3.length()){
        return false;
       } 

       return dfs(n - 1, m -1, s1.toCharArray(), s2.toCharArray(), s3.toCharArray());
    }

    private boolean dfs(int i, int j, char[] s1, char[] s2, char[] s3){
      // If both s1 and s2 are fully used, it's a valid interleaving
        if (i < 0 && j < 0){
            return true;
        }

        boolean res = false;
       // Try using s1's last character
        if (i >= 0 && s1[i] == s3[i + j + 1]){
            if (dfs(i - 1, j, s1, s2, s3)){
                return true;// If successful, return immediately
            }
        }
                // Try using s2's last character
        if (j >= 0 && s2[j] == s3[i + j + 1]){
            if (dfs(i, j - 1, s1, s2, s3)){
                return true;
            }
        }

        return false;

    }
}

// TC: O(2^(M+N))
// SC: O(M+N)

At each step, you may branch into two recursive calls:

• one using s1
• one using s2

Recursion + Memo

class Solution {
    public boolean isInterleave(String s1, String s2, String s3) {
       int n = s1.length();
       int m = s2.length();
       if (n + m != s3.length()){
        return false;
       } 
       // 字符串 注意 空
       int[][] memo = new int[n + 1][m + 1];
       for (int[] row : memo){
        Arrays.fill(row, -1);
       }

       return dfs(n - 1, m -1, s1.toCharArray(), s2.toCharArray(), s3.toCharArray(), memo);
    }

    private boolean dfs(int i, int j, char[] s1, char[] s2, char[] s3, int[][] memo){
        if (i < 0 && j < 0){
            return true;
        }

        if (memo[i + 1][j + 1] != -1){
            if (memo[i + 1][j + 1] == 1){
                return true;
            }else{
                return false;
            }
        }

        boolean res = false;

        if (i >= 0 && s1[i] == s3[i + j + 1]){
            if (dfs(i - 1, j, s1, s2, s3, memo)){
                res = true;
                memo[i + 1][j + 1] = 1;
            }
        }

        if (!res && j >= 0 && s2[j] == s3[i + j + 1]){
            if (dfs(i, j - 1, s1, s2, s3, memo)){
                res = true;
                memo[i + 1][j + 1] = 1;
            }
        }

        if (memo[i + 1][j + 1] == -1){
            memo[i+1][j + 1] = 0;
        }
        return res;

    }
}
// TC: O(m * n)
// SC: O(m * n)

class Solution {
    public boolean isInterleave(String s1, String s2, String s3) {
        if (s1.length() + s2.length() != s3.length()){
            return false;
        }

        Boolean[][] memo = new Boolean[s1.length() + 1][s2.length() + 1];

        return helper(s1, s2, 0, 0, s3, memo);
    }


    private boolean helper(String s1, String s2, int i1, int i2, String s3, Boolean[][] memo){
        if (i1 == s1.length() && i2 == s2.length()){
            return true;
        }

        if (memo[i1][i2] != null){
            return memo[i1][i2];
        }

        Boolean valid = false;

        if (i1 < s1.length() && s1.charAt(i1) == s3.charAt(i1 + i2)){
            valid = helper(s1, s2, i1 + 1, i2, s3, memo);
        }

        if (!valid && i2 < s2.length() && s2.charAt(i2) == s3.charAt(i1+i2)){
            valid = helper(s1, s2, i1, i2+ 1, s3, memo);
        }

        memo[i1][i2] = valid;
        return valid;
    }
}

// TC: O(n*m)
// SC: O(n*m)