97. Interleaving String

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97. Interleaving String

Description

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 + …ort1 + s1 + t2 + s2 + t3 + s3 + …

Note: a + b is the concatenation of strings a and b.

Example 1:

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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:

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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:

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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?

Hints/Notes

  • 2024/12/26
  • dp
  • No solution from 0x3F

Solution

Language: C++

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class Solution {
public:
    vector<vector<int>> dp;

    bool isInterleave(string s1, string s2, string s3) {
        int m = s1.size(), n = s2.size();
        if (m + n != s3.size()) {
            return false;
        }
        dp.resize(m + 1, vector<int>(n + 1, -1));
        return dfs(0, 0, s1, s2, s3);
    }

    bool dfs(int i, int j, string& s1, string& s2, string& s3) {
        if (i == s1.size() && j == s2.size()) {
            return true;
        }
        if (dp[i][j] != -1) {
            return dp[i][j];
        }
        int res = 0;
        if (i < s1.size() && s1[i] == s3[i + j]) {
            res = dfs(i + 1, j, s1, s2, s3);
        }
        if (res) {
            dp[i][j] = res;
            return res;
        }
        if (j < s2.size() && s2[j] == s3[i + j]) {
            res = dfs(i, j + 1, s1, s2, s3);
        }
        dp[i][j] = res;
        return res;
    }
};