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This approach uses a dynamic programming table to ascertain whether the third string, s3, is an interleaving of s1 and s2. We use a 2D DP array dp[i][j] which signifies if s3[0:i+j] is an interleaving of s1[0:i] and s2[0:j]. A bottom-up approach is taken to fill this table.
Time Complexity: O(n * m), where n and m are the lengths of s1 and s2, respectively.
Space Complexity: O(n * m), required for the dp array.
1class Solution:
2 def isInterleave(self, s1: str, s2: str, s3: str) -> bool:
3 n, m, l = len(s1), len(s2), len(s3)
4 if n + m != l:
5 return False
6 dp = [[False] * (m + 1) for _ in range(n + 1)]
7 dp[0][0] = True
8 for i in range(n + 1):
9 for j in range(m + 1):
10 if i > 0:
11 dp[i][j] = dp[i][j] or (dp[i - 1][j] and s1[i - 1] == s3[i + j - 1])
12 if j > 0:
13 dp[i][j] = dp[i][j] or (dp[i][j - 1] and s2[j - 1] == s3[i + j - 1])
14 return dp[n][m]In the Python implementation, the dynamic programming table dp is filled similarly. The table tracks whether s3[0:i+j] can be formed by s1[0:i] and s2[0:j]. We check each character's addition from either string matches s3.
This recursive approach tries to construct the interleaved string by attempting to use characters from s1 and s2 to match each character of s3. To optimize, memoization is applied to avoid recalculating results for the same subproblems.
Time Complexity: O(n * m), due to memoization allowing only one calculation per subproblem.
Space Complexity: O(n * m) for the memoization table.
1
The JavaScript function employs recursion with memoization, stored in a Map object, to try forming s3 incrementally from s1 and s2 by sequentially validating or skipping characters.