Interleaving String

To solve this coding challenge of determining if a string
s3
is formed by interleaving two strings
s1
and
s2
, you need to employ a dynamic programming approach. The idea is to build up a table of boolean values that signify whether a substring of
s3
can be formed by interleaving substrings of
s1
and
s2
. Now we'll go into detailed explanation.

Explanation

First, let's break down the problem step-by-step:
  1. Base Case Handling :
    • If the combined lengths of
      s1
      and
      s2
      do not equal to the length of
      s3
      , return
      False
      immediately. It is impossible to interleave and form
      s3
      if the lengths don't match.
  2. Dynamic Programming Table Setup :
    • Create a 2D table (a list of lists or just a list if optimizing for space) where each entry
      dp[i][j]
      represents if it's possible to form the substring
      s3[:i+j]
      by interleaving
      s1[:i]
      and
      s2[:j]
      .
  3. Initialization :
    • dp[0][0] should be
      True
      since two empty strings can form an empty string.
    • For
      dp[i][0]
      : Each entry should be
      True
      if
      s1[:i]
      equals
      s3[:i]
      .
    • For
      dp[0][j]
      : Each entry should be
      True
      if
      s2[:j]
      equals
      s3[:j]
      .
  4. Filling Up the DP Table :
    • For each cell
      dp[i][j]
      ,
      • If the character from
        s3
        at position
        i+j-1
        matches the character from
        s1
        at position
        i-1
        , and
        dp[i-1][j]
        is
        True
        , then set
        dp[i][j]
        to
        True
        .
      • Or, if the character from
        s3
        at position
        i+j-1
        matches the character from
        s2
        at position
        j-1
        , and
        dp[i][j-1]
        is
        True
        , then set
        dp[i][j]
        to
        True
        .
  5. Result Extraction :
    • The final result will be in
      dp[len(s1)][len(s2)]
      .
    Here's an extremely detailed pseudocode illustrating the above steps: 
                                                
# Base case: check if combined lengths match
if length(s1) + length(s2) != length(s3):
return False

# Create a DP table initialized with 'False'
# dp[i][j] indicates if s3[:i+j] can be formed by interleaving s1[:i] and s2[:j]
dp = 2D array of Size (length(s1) + 1) by (length(s2) + 1) initialized to False

# Initializing the base cases
dp[0][0] = True # Two empty strings form an empty string

# Fill first column: only s1 used
for i from 1 to length(s1):
dp[i][0] = dp[i - 1][0] and s1[i - 1] == s3[i - 1]

# Fill first row: only s2 used
for j from 1 to length(s2):
dp[0][j] = dp[0][j - 1] and s2[j - 1] == s3[j - 1]

# Fill the rest of the dp table
for i from 1 to length(s1):
for j from 1 to length(s2):
# Check if current character of s1 matches with the current character in s3
if dp[i - 1][j] and s1[i - 1] == s3[i + j - 1]:
dp[i][j] = True
# Check if current character of s2 matches with the current character in s3
if dp[i][j - 1] and s2[j - 1] == s3[i + j - 1]:
dp[i][j] = True

# The answer is whether the entire s1 and s2 can form s3
return dp[length(s1)][length(s2)]

    Detailed Steps in Pseudocode

  6. Check Lengths :
    • If the sum of the lengths of
      s1
      and
      s2
      is not equal to the length of
      s3
      , then it is not possible to interleave them to form
      s3
      . This basic check can save time by immediately returning
      false
      if the lengths don’t match.
  7. Initialize DP Array :
    • Create a two-dimensional list
      dp
      with
      len(s1) + 1
      rows and
      len(s2) + 1
      columns. Initialize all entries to
      False
      .
  8. Base Case Setup :
    • dp[0][0]
      is set to
      True
      since empty strings can form an empty string.
  9. First Column Setup :
    • Iterate through each character in
      s1
      to populate the first column. If
      s1
      up to that character matches
      s3
      , mark these entries
      True
      .
  10. First Row Setup :
    • Iterate through each character in
      s2
      to populate the first row. If
      s2
      up to that character matches
      s3
      , mark these entries
      True
      .
  11. Populate DP Table :
    • For each cell in the DP table, use the values from the previous row and column to determine if the current cell should be
      True
      . This represents using characters from either
      s1
      or
      s2
      to build up to the current point in
      s3
      .
  12. Result Extraction :
    • The value at
      dp[len(s1)][len(s2)]
      will give the final answer, indicating if
      s3
      can be formed by interleaving
      s1
      and
      s2
      .
By following these detailed steps, you can solve the problem effectively using dynamic programming, ensuring you get the correct answer efficiently even for longer strings.