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

July LeetCoding Challenge 7

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 they are divided into non-empty substrings 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

Example 2:

Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
Output: false

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 (8 ms)
// Your runtime beats 51.77 % of cpp submissions
// Your memory usage beats 89.89 % of cpp submissions (6.2 MB)