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/*
dp[i][j] implies that if it is possible to obtain substring (i,j) of s3 by using first i characters in s1 and first j characters in s2 */

public class Solution {
    public boolean isInterleave(String s1, String s2, String s3) {
        if (s1 == null || s2 == null || s3 == null) return false;
        int len1 = s1.length(), len2 = s2.length(), len3 = s3.length();
        if (len1 + len2 != len3) return false;
        boolean[][] dp = new boolean[len1 + 1][len2 + 1];
        dp[0][0] = true;
        for (int i = 1; i <= len1; i++) {
            dp[i][0] = dp[i - 1][0] && (s1.charAt(i - 1) == s3.charAt(i - 1));
        }
        for (int j = 1; j <= len2; j++) {
            dp[0][j] = dp[0][j - 1] && (s2.charAt(j - 1) == s3.charAt(j - 1));
        }
        for (int i = 1; i <= len1; i++) {
            for (int j = 1; j <= len2; j++) {
                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[len1][len2];
        
    }
}