#97

Medium

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Interleaving String

The Interleaving String problem requires determining if a string can be formed by interleaving two others, utilizing dynamic programming techniques.

string dynamic programming

Problem Statement

Given three strings s1, s2, and s3, the task is to determine if s3 is formed by interleaving s1 and s2. An interleaving of two strings s1 and s2 is a configuration where s1 and s2 are split into substrings that, when interwoven, form the string s3. The split can occur at any valid point within s1 and s2, but the order of characters within s1 and s2 must be preserved.

You need to decide whether it’s possible to interleave s1 and s2 to form s3, considering that both strings can contribute characters in any order, but without rearranging any characters within each individual string.

Examples

Example 1

Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"

Output: true

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

Notice how it is impossible to interleave s2 with any other string to obtain s3.

Example 3

Input: s1 = "", s2 = "", s3 = ""

Output: true

Example details omitted.

Constraints

0 <= s1.length, s2.length <= 100
0 <= s3.length <= 200
s1, s2, and s3 consist of lowercase English letters.

Solution Approach

Dynamic Programming Approach

This problem can be solved using a 2D dynamic programming (DP) table where dp[i][j] represents whether the first i characters of s1 and the first j characters of s2 can form the first i + j characters of s3. You iterate through all possible combinations of i and j, and check whether the character from s1 or s2 matches the corresponding character in s3. If it does, update the DP table accordingly.

State Transition Technique

By using state transitions in the DP table, you ensure that for every valid (i, j) state, the next state either includes the next character from s1 or s2, depending on which string matches the current character of s3. This state transition helps build the solution incrementally and eliminates redundant checks, thus optimizing the time complexity.

Edge Case Handling

Edge cases include empty strings for s1, s2, or s3. When both s1 and s2 are empty, the result is true only if s3 is also empty. Additionally, if s3’s length is not equal to the combined length of s1 and s2, you can immediately return false without further processing.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is O(m * n), where m and n are the lengths of s1 and s2, respectively. This is because we iterate through each combination of characters from both strings. The space complexity is also O(m * n) due to the 2D DP table used to store intermediate results. Optimizations can be applied to reduce the space complexity if needed.

What Interviewers Usually Probe

The candidate should explain how to fill the DP table, especially how the state transitions work for each character comparison.
A correct answer will involve explaining the transition between s1, s2, and s3 and handling edge cases like empty strings.
The interviewer should watch for the candidate's understanding of dynamic programming and how it can optimize the problem-solving process.

Common Pitfalls or Variants

Common pitfalls

Misunderstanding the interleaving concept and attempting to rearrange characters rather than preserving the order from both strings.
Failing to account for edge cases, such as when one or both strings are empty.
Not properly filling the DP table or skipping necessary state transitions during the solution process.

Follow-up variants

The problem could be modified by changing the constraints, such as limiting the maximum lengths of s1 and s2, which would require optimizations in the solution approach.
Another variant could involve allowing characters from s1 and s2 to overlap, requiring more sophisticated DP or backtracking techniques.
A variant could also introduce additional constraints like including special characters or uppercase letters, which would require adaptations in handling input.

FAQ

What is the primary technique for solving the Interleaving String problem?

The primary technique for solving this problem is dynamic programming, specifically using a state transition approach to build up solutions incrementally.

How do we handle empty strings in the Interleaving String problem?

If both s1 and s2 are empty, s3 must also be empty for the result to be true. Any other combination where s3’s length doesn’t match the sum of s1 and s2 lengths will return false.

Why is dynamic programming used in this problem?

Dynamic programming is used because it allows you to solve overlapping subproblems efficiently by breaking down the problem into smaller, manageable states, avoiding redundant calculations.

What are common mistakes when solving the Interleaving String problem?

Common mistakes include not properly managing the state transitions in the DP table, failing to account for edge cases, or misunderstanding the interleaving condition.

How can GhostInterview help me with the Interleaving String problem?

GhostInterview helps by providing a clear breakdown of the problem-solving process, offering interactive examples, and guiding you through the steps needed to apply dynamic programming techniques.

terminal

Solution

Solution 1: Memoization Search

Let's denote the length of string $s_1$ as $m$ and the length of string $s_2$ as $n$. If $m + n \neq |s_3|$, then $s_3$ is definitely not an interleaving string of $s_1$ and $s_2$, so we return `false`.

class Solution:
    def isInterleave(self, s1: str, s2: str, s3: str) -> bool:
        @cache
        def dfs(i: int, j: int) -> bool:
            if i >= m and j >= n:
                return True
            k = i + j
            if i < m and s1[i] == s3[k] and dfs(i + 1, j):
                return True
            if j < n and s2[j] == s3[k] and dfs(i, j + 1):
                return True
            return False

        m, n = len(s1), len(s2)
        if m + n != len(s3):
            return False
        return dfs(0, 0)

Solution 2: Dynamic Programming

We can convert the memoization search in Solution 1 into dynamic programming.

class Solution:
    def isInterleave(self, s1: str, s2: str, s3: str) -> bool:
        @cache
        def dfs(i: int, j: int) -> bool:
            if i >= m and j >= n:
                return True
            k = i + j
            if i < m and s1[i] == s3[k] and dfs(i + 1, j):
                return True
            if j < n and s2[j] == s3[k] and dfs(i, j + 1):
                return True
            return False

        m, n = len(s1), len(s2)
        if m + n != len(s3):
            return False
        return dfs(0, 0)

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