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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:

img

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:

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)