97.Interleaving String

tags: Medium String DP

97. Interleaving String

題目描述

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.

範例

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?

解答

TypeScript

dp[i][j] 表示 s1 的前 i 個字元和 s2 的前 j 個字元是否可以交錯組合成 s3 的前 i + j 個字元。

從下面兩個方向達到目前狀態 (i,j)：

1. 從 (i - 1, j) 轉移：
   • 如果 dp[i - 1][j] == true，代表著 s1 的前 i - 1 個字元和 s2 的前 j 個字元可以交錯組合成 s3 的前 i + j - 1 個字元。
   • 然後我們只需檢查 s1[i - 1] 是否等於 s3[i + j - 1]。如果相等，則 dp[i][j] 也為 true
2. 從 i, j - 1 轉移：
   • 如果 dp[i, j - 1] == true，代表著 s1 的前 i 個字元和 s2 的前 j - 1 個字元可以交錯組合成 s3 的前 i + j - 1 個字元。
   • 一樣檢查 s2[j - 1] 是否等於 s3[i + j - 1]。如果相等，則 dp[i][j] 也為 true

function isInterleave(s1: string, s2: string, s3: string): boolean {
  if (s1.length + s2.length !== s3.length) return false;

  const dp: boolean[][] = new Array(s1.length + 1)
    .fill(0)
    .map(() => new Array(s2.length + 1).fill(false));
  dp[0][0] = true;

  // s2 = ''
  for (let i = 1; i <= s1.length; i++) {
    dp[i][0] = dp[i - 1][0] && s1[i - 1] === s3[i - 1];
  }

  // s1 = ''
  for (let j = 1; j <= s2.length; j++) {
    dp[0][j] = dp[0][j - 1] && s2[j - 1] === s3[j - 1];
  }

  for (let i = 1; i <= s1.length; i++) {
    for (let j = 1; j <= s2.length; j++) {
      dp[i][j] =
        (dp[i - 1][j] && s1[i - 1] === s3[i + j - 1]) ||
        (dp[i][j - 1] && s2[j - 1] === s3[i + j - 1]);
    }
  }

  return dp[s1.length][s2.length];
}

sheepFri, 25 Aug 2023

Python

class Solution:
    def isInterleave(self, s1: str, s2: str, s3: str) -> bool:
        if len(s1) + len(s2) != len(s3):
            return False

        @cache
        def dfs(i, j):
            if i == 0 and j == 0:
                return True
            if i == 0:
                return s2[j - 1] == s3[j - 1] and dfs(i, j - 1)
            if j == 0:
                return s1[i - 1] == s3[i - 1] and dfs(i - 1, j)
            return (s1[i - 1] == s3[i + j - 1] and dfs(i - 1, j)) \
                or (s2[j - 1] == s3[i + j - 1] and dfs(i, j - 1))

        return dfs(len(s1), len(s2))

Yen-Chi ChenFri, Aug 25, 2023

Reference

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