String interleaving is a fascinating concept that often appears in algorithmic coding interviews, particularly those focused on string manipulation. At its core, string interleaving checks whether a string can be formed by interleaving two other strings while maintaining the original order of characters from the two source strings. Let’s break this down step by step.
Suppose we have two strings, A
and B
, along with a third string C
. The task is to determine if C
can be formed by interleaving characters from A
and B
. Interleaving means that you can take characters from A
and B
and combine them in a way that preserves the sequence of characters in both strings.
For example:
C
from A
and B
)However, not all combinations will work. Consider:
B
comes after 'e', which violates the sequence defined in B
)When facing a problem of string interleaving, you can approach it in several ways. Below, we will cover a more structured approach using dynamic programming, which is often preferred in coding interviews due to its efficiency.
The dynamic programming (DP) solution involves creating a 2D table (or matrix) to store boolean values that indicate whether a substring of C
can be formed by interleaving substrings of A
and B
. The basic idea is to iterate through the lengths of A
and B
, checking character by character.
Here’s a step-by-step guide to constructing the algorithm:
Initialize a DP Table: Create a 2D array dp[i][j]
where dp[i][j]
will be true
if the first i
characters of A
and the first j
characters of B
can form the first i + j
characters of C
.
Base Case: Set dp[0][0]
to true
since two empty strings can form an empty string.
Filling the DP Table:
A
and B
.dp[i][j]
, check:
A[i-1]
equals C[i+j-1]
, carry over the truth value from dp[i-1][j]
.B[j-1]
equals C[i+j-1]
, carry over the truth value from dp[i][j-1]
.Here’s what the code looks like in Python:
def isInterleave(A, B, C): if len(A) + len(B) != len(C): return False dp = [[False] * (len(B) + 1) for _ in range(len(A) + 1)] dp[0][0] = True for i in range(len(A) + 1): for j in range(len(B) + 1): if i > 0 and A[i - 1] == C[i + j - 1]: dp[i][j] = dp[i][j] or dp[i - 1][j] if j > 0 and B[j - 1] == C[i + j - 1]: dp[i][j] = dp[i][j] or dp[i][j - 1] return dp[len(A)][len(B)]
Let’s examine how the algorithm works using the example where:
dp[0][0]
= True (both strings empty)Eventually, if we complete this table, we can inspect dp[len(A)][len(B)]
to see whether we can interleave A
and B
to create C
.
The time complexity of this approach is O(n * m), where n is the length of string A
and m is the length of string B
. The space complexity can also be optimized to O(min(n, m)) if needed, as only the previous row is required for further calculations.
With this understanding of string interleaving, you're better equipped to tackle related problems that may come up in interviews. Practicing variations of these problems will enhance your string manipulation skills and overall problem-solving abilities in coding contests and assessments.
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