How do you solve Interleaving String on LeetCode?

Interleaving String (LeetCode #97) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m) time complexity and O(n * m) space complexity. Key insight: Understanding the interleaving concept is crucial; it requires matching characters from both strings in a specific order.

What is the brute force solution for Interleaving String?

The brute force approach for Interleaving String is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks every possible way to interleave the two strings and see if we can form the third string. This is done by recursively trying to match characters from s1 and s2 to s3. The optimal Optimal Solution approach improves this to O(n * m).

What algorithm or data structure is used to solve Interleaving String?

Interleaving String uses the following concepts: String, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Recursion.

What are the interview tips for Interleaving String?

Clearly explain your thought process before jumping into coding; this shows your understanding. Consider edge cases, such as empty strings, and how they affect your solution. Practice writing both brute-force and optimal solutions to strengthen your problem-solving skills.

What are common mistakes when solving Interleaving String?

Forgetting to check the combined length of s1 and s2 against s3 before proceeding. Not properly initializing the DP table or mismanaging indices while filling it.

#97

Interleaving String

Medium

StringDynamic ProgrammingDynamic ProgrammingRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * m)
Space	O(1)	O(n * m)

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Intuition

Time O(n * m)Space O(n * m)

The optimal solution uses dynamic programming to build a table that keeps track of whether substrings of s1 and s2 can form substrings of s3. This avoids redundant calculations and improves efficiency.

⚙️

Algorithm

3 steps

1Step 1: Create a 2D DP array where dp[i][j] indicates if s3 up to index i + j can be formed by interleaving s1 up to index i and s2 up to index j.
2Step 2: Initialize the DP array with base cases: dp[0][0] = true, and fill in the first row and column based on matches with s3.
3Step 3: Fill the DP table by checking if the current character from s1 or s2 matches the current character in s3.

solution.py12 lines

1def isInterleave(s1, s2, s3):
2    if len(s1) + len(s2) != len(s3): return False
3    dp = [[False] * (len(s2) + 1) for _ in range(len(s1) + 1)]
4    dp[0][0] = True
5    for j in range(1, len(s2) + 1):
6        dp[0][j] = dp[0][j - 1] and s2[j - 1] == s3[j - 1]
7    for i in range(1, len(s1) + 1):
8        dp[i][0] = dp[i - 1][0] and s1[i - 1] == s3[i - 1]
9    for i in range(1, len(s1) + 1):
10        for j in range(1, len(s2) + 1):
11            dp[i][j] = (dp[i - 1][j] and s1[i - 1] == s3[i + j - 1]) or (dp[i][j - 1] and s2[j - 1] == s3[i + j - 1])
12    return dp[len(s1)][len(s2)]

ℹ

Complexity note: The time complexity is O(n * m) because we fill a 2D DP table where n and m are the lengths of s1 and s2. The space complexity is also O(n * m) due to the storage of the DP table.

1Understanding the interleaving concept is crucial; it requires matching characters from both strings in a specific order.
2Dynamic programming is a powerful technique for problems involving combinations and sequences.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.