package com.fishercoder.solutions;

/**
 * Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.
 * <p>
 * For example,
 * Given:
 * s1 = "aabcc",
 * s2 = "dbbca",
 * <p>
 * When s3 = "aadbbcbcac", return true.
 * When s3 = "aadbbbaccc", return false.
 */
public class _97 {
    public boolean isInterleave(String s1, String s2, String s3) {
        // write your code here
        int m = s1.length();
        int n = s2.length();
        if (m + n != s3.length()) {
            return false;
        }

        boolean[][] dp = new boolean[m + 1][n + 1];

        dp[0][0] = true;

        for (int i = 0; i < m; i++) {
            if (s1.charAt(i) == s3.charAt(i)) {
                dp[i + 1][0] = true;
            } else {
                //if one char fails, that means it breaks, the rest of the chars won't matter any more.
                //Mian and I found one missing test case on Lintcode: ["b", "aabccc", "aabbbcb"]
                //if we don't break, here, Lintcode could still accept this code, but Leetcode fails it.
                break;
            }
        }

        for (int j = 0; j < n; j++) {
            if (s2.charAt(j) == s3.charAt(j)) {
                dp[0][j + 1] = true;
            } else {
                break;
            }
        }

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

        return dp[m][n];
    }
}

package com.fishercoder.solutions;

/**
 * 97. Interleaving String
 *
 * Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.
 * For example,
 * Given:
 * s1 = "aabcc",
 * s2 = "dbbca",
 * When s3 = "aadbbcbcac", return true.
 * When s3 = "aadbbbaccc", return false.
 */
public class _97 {
  public static class Solution1 {
    public boolean isInterleave(String s1, String s2, String s3) {
      int m = s1.length();
      int n = s2.length();
      if (m + n != s3.length()) {
        return false;
      }

      boolean[][] dp = new boolean[m + 1][n + 1];

      dp[0][0] = true;

      for (int i = 0; i < m; i++) {
        if (s1.charAt(i) == s3.charAt(i)) {
          dp[i + 1][0] = true;
        } else {
          //if one char fails, that means it breaks, the rest of the chars won't matter any more.
          //Mian and I found one missing test case on Lintcode: ["b", "aabccc", "aabbbcb"]
          //if we don't break, here, Lintcode could still accept this code, but Leetcode fails it.
          break;
        }
      }

      for (int j = 0; j < n; j++) {
        if (s2.charAt(j) == s3.charAt(j)) {
          dp[0][j + 1] = true;
        } else {
          break;
        }
      }

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

      return dp[m][n];
    }
  }
}

refactor 97