Problem Description

You have a list of strings words where all strings have the same length, and you need to form a target string target by selecting characters from these strings following specific rules.

Rules for forming the target:

1. Build from left to right: You must construct target character by character from index 0 to the end.

2. Character selection: To form the i-th character of target (0-indexed), you can pick the k-th character from any string j in words, as long as words[j][k] equals target[i].

3. Position restriction: Once you use a character at position k from any string in words, all characters at positions 0 through k become permanently unavailable across all strings in words. This means you can only move forward in the character positions.

4. Multiple uses allowed: You can use multiple characters from the same string in words, as long as you follow the position restriction rule.

Example visualization:

• If words = ["acca", "bbbb", "caca"] and you want to form target = "aba"
• You could pick 'a' from position 0 of words[0] for target[0]
• Then pick 'b' from position 2 of words[1] for target[1] (positions 0-2 are now unusable)
• Finally pick 'a' from position 3 of words[2] for target[2]

The task is to count how many different ways you can form the complete target string following these rules. Since the answer can be very large, return it modulo 10^9 + 7.

Intuition

When we look at this problem, we need to make a series of choices: for each character in target, we need to decide which position across all strings in words to pick it from. The key constraint is that once we use position k, we can't use any position ≤ k again.

This naturally leads us to think about dynamic programming or recursion with memoization. At each step, we have two indices to track:

• i: which character of target we're trying to match
• j: the current position in the strings of words we're considering

For any state (i, j), we have two choices:

1. Skip position j: We don't use any character from position j of any string, and move to position j+1 while still trying to match target[i]
2. Use position j: If there are characters at position j that match target[i], we can use them and move to match target[i+1] from position j+1 onwards

The second choice brings up an important observation: since all strings in words have the same length, at any position j, multiple strings might have the character we need. If 3 strings have 'a' at position j and we need 'a' for target[i], we have 3 different ways to pick it.

This suggests we should preprocess the data: for each position j and each character c, count how many strings have character c at position j. This gives us cnt[j][c] - the frequency of character c at position j across all strings.

With this preprocessing, our recursive formula becomes clearer:

• dfs(i, j) = number of ways to form target[i:] starting from position j in words
• dfs(i, j) = dfs(i, j+1) + cnt[j][target[i]] * dfs(i+1, j+1)

The first term represents skipping position j, and the second term represents using position j (with cnt[j][target[i]] different ways to do so, then continuing from position j+1 for the next character).

Learn more about Dynamic Programming patterns.

Solution Approach

The solution implements a top-down dynamic programming approach with memoization using the insights from our intuition.

Step 1: Preprocessing

First, we build a frequency table cnt where cnt[j][c] stores how many strings in words have character c at position j:

cnt = [[0] * 26 for _ in range(n)]  # n is the length of each string
for w in words:
    for j, c in enumerate(w):
        cnt[j][ord(c) - ord('a')] += 1

This preprocessing takes O(len(words) × n) time and allows us to quickly look up how many ways we can pick a specific character from a specific position.

Step 2: Recursive Function with Memoization

We define dfs(i, j) which returns the number of ways to form target[i:] when we can only use characters from positions j onwards in words:

@cache
def dfs(i: int, j: int) -> int:
    if i >= m:  # All characters of target are matched
        return 1
    if j >= n:  # No more positions available in words
        return 0
  
    # Option 1: Use position j to match target[i]
    ans = dfs(i + 1, j + 1) * cnt[j][ord(target[i]) - ord('a')]
  
    # Option 2: Skip position j
    ans = (ans + dfs(i, j + 1)) % mod
  
    return ans

Base Cases:

• i >= m: We've successfully matched all characters in target, so return 1
• j >= n: We've exhausted all positions in words but haven't matched all of target, so return 0

Recursive Cases:

• Use position j: If we use position j to match target[i], we have cnt[j][target[i]] ways to do so. After using position j, we move to target[i+1] and can only use positions from j+1 onwards. This gives us cnt[j][target[i]] × dfs(i+1, j+1) ways.
• Skip position j: We don't use position j at all and try to match target[i] from position j+1 onwards. This gives us dfs(i, j+1) ways.

The total number of ways is the sum of both options, taken modulo 10^9 + 7.

Time Complexity: O(m × n) where m = len(target) and n = len(words[0]), since we have at most m × n unique states and each state is computed once due to memoization.

Space Complexity: O(m × n) for the memoization cache plus O(26 × n) for the preprocessing array.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Given:

• words = ["acca", "bbbb", "caca"]
• target = "aba"

Step 1: Preprocessing

First, we build the frequency table cnt[j][c] that counts how many strings have character c at position j:

Position 0: a appears 1 time (in "acca")
           b appears 1 time (in "bbbb")
           c appears 1 time (in "caca")

Position 1: a appears 1 time (in "caca")
           b appears 1 time (in "bbbb")
           c appears 1 time (in "acca")

Position 2: a appears 0 times
           b appears 1 time (in "bbbb")
           c appears 2 times (in "acca", "caca")

Position 3: a appears 2 times (in "acca", "caca")
           b appears 1 time (in "bbbb")
           c appears 0 times

Step 2: Recursive Calculation

Now we calculate dfs(0, 0) - ways to form target[0:] = "aba" starting from position 0.

dfs(0, 0): Form "aba" from position 0

• Option 1: Use position 0 for 'a' → cnt[0]['a'] × dfs(1, 1) = 1 × dfs(1, 1)
• Option 2: Skip position 0 → dfs(0, 1)
• Result: dfs(1, 1) + dfs(0, 1)

dfs(1, 1): Form "ba" from position 1

• Option 1: Use position 1 for 'b' → cnt[1]['b'] × dfs(2, 2) = 1 × dfs(2, 2)
• Option 2: Skip position 1 → dfs(1, 2)
• Result: dfs(2, 2) + dfs(1, 2)

dfs(2, 2): Form "a" from position 2

• Option 1: Use position 2 for 'a' → cnt[2]['a'] × dfs(3, 3) = 0 × 1 = 0
• Option 2: Skip position 2 → dfs(2, 3)
• Result: 0 + dfs(2, 3)

dfs(2, 3): Form "a" from position 3

• Option 1: Use position 3 for 'a' → cnt[3]['a'] × dfs(3, 4) = 2 × 1 = 2
• Option 2: Skip position 3 → dfs(2, 4) = 0 (no more positions)
• Result: 2 + 0 = 2

Working backwards:

• dfs(2, 2) = 0 + 2 = 2
• dfs(1, 2): Form "ba" from position 2
  • Use position 2 for 'b': 1 × dfs(2, 3) = 1 × 2 = 2
  • Skip position 2: dfs(1, 3) = 0 (can't form 'b' from position 3)
  • Result: 2 + 0 = 2
• dfs(1, 1) = 2 + 2 = 4

For dfs(0, 1): Form "aba" from position 1

• Eventually evaluates to 0 (we can't form all of "aba" starting from position 1)

Final Answer: dfs(0, 0) = 4 + 0 = 4

The 4 ways are:

1. words[0][0]='a', words[1][1]='b', words[0][3]='a'
2. words[0][0]='a', words[1][1]='b', words[2][3]='a'
3. words[0][0]='a', words[1][2]='b', words[0][3]='a'
4. words[0][0]='a', words[1][2]='b', words[2][3]='a'

Solution Implementation

Time and Space Complexity

Time Complexity: O(m × n + k × n)

The algorithm consists of two main parts:

1. Preprocessing phase: Building the cnt array takes O(k × n) time, where k is the number of words in the words array and n is the length of each word. We iterate through all characters in all words.

2. Dynamic Programming phase: The dfs function with memoization explores at most m × n unique states, where m is the length of the target string and n is the length of each word. Each state (i, j) is computed only once due to the @cache decorator. For each state, the computation takes O(1) time.

Therefore, the overall time complexity is O(k × n + m × n). If we assume k is relatively small or consider it as part of the input preprocessing, the dominant term becomes O(m × n).

Space Complexity: O(m × n + n)

The space usage includes:

1. Memoization cache: The @cache decorator stores up to m × n states, each taking O(1) space, resulting in O(m × n) space.

2. Count array: The cnt array has dimensions n × 26, which is O(n) space since 26 is a constant.

3. Recursion stack: In the worst case, the recursion depth can be O(m + n).

The dominant term is O(m × n) from the memoization cache, making the overall space complexity O(m × n).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Integer Overflow When Computing Products

The Pitfall: When calculating dp(target_idx + 1, column_idx + 1) * char_count[column_idx][needed_char_index], the multiplication can produce very large numbers that exceed Python's integer limits in other languages or cause issues even before applying the modulo operation. While Python handles arbitrary precision integers, in languages like Java or C++, this would cause overflow. Even in Python, performing modulo operations too late can impact performance.

Example of the Issue:

# Problematic approach - modulo applied after multiplication
use_column = dp(target_idx + 1, column_idx + 1) * char_count[column_idx][needed_char_index]
total_ways = (use_column + skip_column) % MOD  # Modulo applied too late

The Solution: Apply modulo operations immediately after each arithmetic operation to keep numbers manageable:

# Better approach - apply modulo after each operation
use_column = (dp(target_idx + 1, column_idx + 1) * char_count[column_idx][needed_char_index]) % MOD
skip_column = dp(target_idx, column_idx + 1) % MOD
total_ways = (use_column + skip_column) % MOD

2. Incorrect Understanding of the Position Restriction

The Pitfall: A common misunderstanding is thinking that once you use a character from a specific string at position k, you can't use that particular string anymore. The actual rule is that positions 0 through k become unavailable across ALL strings, not just the one you picked from.

Incorrect Mental Model:

# Wrong: Tracking which strings have been used
def dp(target_idx, column_idx, used_strings):
    # This would incorrectly track individual string usage
    pass

Correct Understanding: The restriction is column-based (position-based) across all strings:

# Correct: Only tracking column position
def dp(target_idx, column_idx):
    # When we use column_idx, all columns 0 to column_idx become unavailable
    # We can only proceed with column_idx + 1 onwards
    use_column = dp(target_idx + 1, column_idx + 1) * char_count[column_idx][needed_char_index]

3. Off-by-One Errors in Index Handling

The Pitfall: Mixing up whether indices are inclusive or exclusive, especially in base cases:

Common Mistake:

# Wrong base case checks
if target_idx > target_length:  # Should be >=
    return 1
if column_idx > word_length:    # Should be >=
    return 0

Correct Implementation:

# Correct base case checks
if target_idx >= target_length:  # We've processed all characters (0 to target_length-1)
    return 1
if column_idx >= word_length:    # No more columns available (0 to word_length-1)
    return 0

4. Forgetting to Handle the Case Where Character Doesn't Exist

The Pitfall: Not considering that char_count[column_idx][needed_char_index] might be 0, meaning the needed character doesn't exist at that column position.

Why It's Not Obviously Wrong: The code still works because multiplying by 0 gives 0 ways, which is correct. However, this can lead to unnecessary recursive calls and reduced performance.

Optimization:

# More efficient - skip recursion if character doesn't exist
needed_char_index = ord(target[target_idx]) - ord('a')
if char_count[column_idx][needed_char_index] > 0:
    use_column = (dp(target_idx + 1, column_idx + 1) * 
                  char_count[column_idx][needed_char_index]) % MOD
else:
    use_column = 0  # Skip unnecessary recursion

skip_column = dp(target_idx, column_idx + 1)
return (use_column + skip_column) % MOD