Today I have done leetcode's August LeetCoding Challenge with cpp.

August LeetCoding Challenge 25

Description

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

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

// Accepted
// 106/106 cases passed (0 ms)
// Your runtime beats 100 % of cpp submissions
// Your memory usage beats 96.97 % of cpp submissions (6.2 MB)