LeetCode - 97 Interleaving String

Medium

97. Interleaving String

題目

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.

解題思路

經典字串 DP 題
判斷 s3 能不能由 s1 和 s2 交錯組成

• 先檢查長度：

  • 如果 s1.size() + s2.size() != s3.size()
  • 代表不可能組成，直接回傳 false

• DP 定義：

  • dp[i][j]：代表 s1 的前 i 個字和 s2 的前 j 個字，能不能組成 s3 的前 i + j 個字

• 初始化：

  • dp[0][0] = true
  • 代表兩個空字串可以組成空字串

• 狀態轉移：

  • 如果目前要接的是 s1[i-1]

    • 當 i > 0 且 s1[i-1] == s3[i+j-1]
    • 代表可以從 dp[i-1][j] 轉移過來
      dp[i][j] = dp[i][j] || dp[i-1][j]

  • 如果目前要接的是 s2[j-1]

    • 當 j > 0 且 s2[j-1] == s3[i+j-1]
    • 代表可以從 dp[i][j-1] 轉移過來
      dp[i][j] = dp[i][j] || dp[i][j-1]

• 這裡用 || 是因為：

  • 只要 dp[i-1][j] 或 dp[i][j-1] 其中一個可以成立
  • dp[i][j] 就可以成立

• 最後答案：

  • dp[n][m]

Solution Code

Time Complexity
O(n × m)

Space Complexity
O(n × m)

class Solution {
public:
    bool isInterleave(string s1, string s2, string s3) {
        int n = s1.size(), m = s2.size();
        if(n + m != s3.size()) return false;
        
        vector<vector<bool>> dp(n+1, vector<bool>(m+1, false)); // dp[i][j] 代表 s1 的 i 個和 s2 的 j 個

        dp[0][0] = true;

        for(int i = 0; i <= n; ++i){
            for(int j = 0; j <= m; ++j){
                if(i > 0 && s1[i-1] == s3[i+j-1]){
                    dp[i][j] = dp[i][j] || dp[i-1][j]; // || 因為是 true or false 只要 dp[i-1][j] 和 dp[i][j-1] 有人是 true 就好
                }
                if(j > 0 && s2[j-1] == s3[i+j-1]){
                    dp[i][j] = dp[i][j] || dp[i][j-1]; // || 因為是 true or false 只要 dp[i-1][j] 和 dp[i][j-1] 有人是 true 就好
                }
            }
        }
        return dp[n][m];
    }
};