InterleavingString.java

/**
 * 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?
 */
class Solution {
    public boolean isInterleave(String s1, String s2, String s3) {
        int n = s1.length(), m = s2.length(), l = s3.length();
        if(n + m != l) return false;
        boolean[][] dp = new boolean[n + 1][m + 1];

        for(int i = 0; i <= n; i++){
            for(int j = 0; j <= m; j++){
                if(i == 0 && j == 0) dp[i][j] = true;
                else{
                    if(i == 0){
                        dp[i][j] = dp[i][j - 1] && s2.charAt(j - 1) == s3.charAt(j - 1);
                    }
                    else if(j == 0){
                        dp[i][j] = dp[i - 1][j] && s1.charAt(i - 1) == s3.charAt(i - 1);
                    }else{
                        dp[i][j] = (dp[i - 1][j] && s1.charAt(i - 1) == s3.charAt(i + j - 1)) ||
                                    (dp[i][j - 1] && s2.charAt(j - 1) == s3.charAt(i + j - 1));
                    }
                }
            }
        }
        return dp[n][m];
    }

    
}