Let’s dive into an exciting problem that challenges our understanding of dynamic programming! 🚀 We’re tasked with figuring out the number of ways to form a target string using characters from a list of strings, under some specific constraints. Ready to explore this fascinating challenge? Let’s go! 🎉

📜 Problem Statement

You are given:

• A list of strings words, where each string has the same length.
• A string target.

Your task is to determine how many ways you can form the target string by selecting characters from words. The rules are as follows:

1. Target formation: Form the target string from left to right.
2. Character selection: To form the i-th character of target, you can pick any character from the same column (k-th position) of any string in words if it matches the character in target[i].
3. Index restriction: Once you use a character at index k from any string in words, all characters at indices ≤ k in every string become unavailable.
4. Repetition allowed: You can use multiple characters from the same string.

Return the number of ways to form the target string, modulo \(10^9 + 7\).

🖼️ Examples

Example 1

Input:

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

Output:

6

Explanation:

There are six ways to form "aba":

1. "aba" -> index 0 (“acca”), index 1 (“bbbb”), index 3 (“caca”)
2. "aba" -> index 0 (“acca”), index 2 (“bbbb”), index 3 (“caca”)
3. "aba" -> index 0 (“acca”), index 1 (“bbbb”), index 3 (“acca”)
4. "aba" -> index 0 (“acca”), index 2 (“bbbb”), index 3 (“acca”)
5. "aba" -> index 1 (“caca”), index 2 (“bbbb”), index 3 (“acca”)
6. "aba" -> index 1 (“caca”), index 2 (“bbbb”), index 3 (“caca”)

Example 2

Input:

words = ["abba", "baab"]
target = "bab"

Output:

4

Explanation:

There are four ways to form "bab":

1. "bab" -> index 0 (“baab”), index 1 (“baab”), index 2 (“abba”)
2. "bab" -> index 0 (“baab”), index 1 (“baab”), index 3 (“baab”)
3. "bab" -> index 0 (“baab”), index 2 (“baab”), index 3 (“baab”)
4. "bab" -> index 1 (“abba”), index 2 (“baab”), index 3 (“baab”)

🧠 Thought Process

Intuition and Thought Process 🧠

Before diving into the solution, let’s take a step back and understand the intuition behind solving the problem. The problem asks us to find the number of ways to form a target string from a list of words. This might seem challenging at first due to the constraints, but breaking it into smaller subproblems reveals an efficient approach using dynamic programming.

Step 1: Understand the Problem 🔍

The primary challenge is:

1. Forming the target string: Each character in the target can be chosen from any string in the words list, but the rules for selection are strict:
   • Characters in each word are accessed left-to-right.
   • Once a character is used, all earlier indices in that word are “locked” and cannot be reused.

This means:

• To form a character in the target, we need to count how many times that character appears at a given index across all words.
• We need to combine these counts recursively to determine the total number of ways to form the entire target.

Step 2: Break Down the Problem 🧩

To form the target[i] using the characters in words, we must decide:

1. Skip the current index k in words and move to the next index. This decision ensures we leave the current column untouched for forming later parts of the target.
2. Use the current column to form target[i]. Here, the number of options depends on how many times target[i] appears in column k of words. We then recursively attempt to form the rest of the target (target[i+1:]) from the remaining columns.

Step 3: Dynamic Programming (DP) Insight 💡

Dynamic programming helps us efficiently compute overlapping subproblems:

• Define dp[i][k] as the number of ways to form the substring target[i:] using columns k to the end of words.
• Base cases:
  • If i == len(target), we’ve formed the entire target, so return 1.
  • If k == len(words[0]), we’ve exhausted the columns and cannot form the target, so return 0.
• Recurrence relation:
  • Skip the current column: dp[i][k] += dp[i][k+1].
  • Use the current column if target[i] exists in column k: dp[i][k] += frequency[target[i]][k] * dp[i+1][k+1].

The step

dp[(i, k)] += cnt[(k, c)] * dfs(i + 1, k + 1)

is the heart of the dynamic programming solution. It combines two essential concepts:

1. Counting Possibilities:
   • The term cnt[(k, c)] represents the number of times the character c (which is target[i]) appears in column k of the words matrix. This gives us the number of ways we can choose the current character from this column.
2. Recursing to the Next State:
   • The function dfs(i + 1, k + 1) computes the number of ways to form the remainder of the target string (target[i+1:]) using the columns starting from k + 1.

Why Multiplication? 🤔

The multiplication combines these two quantities because they are independent events. Here’s the reasoning:

• If cnt[(k, c)] is the number of ways to pick the current character target[i] from column k, and
• dfs(i + 1, k + 1) is the number of ways to form the rest of the target string (target[i+1:]) from the subsequent columns (k + 1 onwards),

then the total number of ways to form the target string from this step is the product of these two numbers. This is because:

1. Each choice of target[i] from column k can independently lead to forming the rest of the target in dfs(i + 1, k + 1).
2. For every valid way to pick target[i], there are dfs(i + 1, k + 1) ways to complete the rest of the target string.

Example to Illustrate Multiplication

Input:

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

Precomputed Frequency Table:

Case: i = 0, k = 0 (Forming “aba”, starting at column 0)

• We are forming the first character 'a' of the target:
  • cnt[(0, 'a')] = 2 (two ‘a’s in column 0).
  • To compute the number of ways to form the rest of the target ("ba") starting from column 1, we recursively call dfs(1, 1).

Calculation:

The number of ways to form "aba" starting from column 0 is:

Here, 2 accounts for the two ways to choose 'a' in column 0, and for each of those choices, we multiply by the number of ways to form "ba" starting from column 1.

Why Not Addition?

If we were to use addition instead of multiplication, it would imply that the possibilities are mutually exclusive instead of independent. However, in this problem, picking each occurrence of the current character does not exclude other choices for forming the rest of the target.

General Pattern in DP Problems

This step of combining counts of current possibilities (cnt[(k, c)]) with results of future possibilities (dfs(i + 1, k + 1)) through multiplication is a common technique in problems where:

1. Choices at one step are independent of the choices in subsequent steps.
2. Each choice contributes to all possible outcomes of the remaining problem.

This is a hallmark of problems involving combinatorics and recursive enumeration.

Key Observations 🔍

1. Precompute Frequencies: Instead of repeatedly counting occurrences of target[i] in each column, precompute a frequency table freq where freq[c][k] is the count of character c in column k.
2. Recursive Transition: Each state depends on two smaller subproblems: one where the current column is skipped, and one where it contributes to forming the target.
3. Modulo Constraint: Since the answer can be very large, every addition is computed modulo \(10^9 + 7\).

💡 Basic Solution (Brute Force)

Approach:

1. For each character of target, iterate through all columns of words.
2. Count how many characters in the current column match the current target character.
3. Recursively find all possible ways to form the remaining part of target.

Code:

def numWays(words, target):
    MOD = 10**9 + 7

    def countWays(i, k):
        if i == len(target):  # Successfully formed the target
            return 1
        if k == len(words[0]):  # Ran out of columns
            return 0

        # Count the number of matches for target[i] in column k
        matches = sum(1 for word in words if word[k] == target[i])

        # Skip the current column or use it and move to the next character
        return (countWays(i, k + 1) + matches * countWays(i + 1, k + 1)) % MOD

    return countWays(0, 0)

Time Complexity:

The brute force approach is highly inefficient due to repeated work. Each recursive call has two branches, leading to exponential complexity:

• \(O(2^{\text{len(words[0])}} \cdot \text{len(target)})\).

⚡ Optimized Solution (Dynamic Programming)

We can improve the brute force approach by using memoization to store the results of subproblems. This avoids redundant calculations.

Key Ideas:

1. Frequency Table: Precompute the frequency of each character in every column of words.
2. Memoization: Store the results of (i, k) in a dictionary dp to avoid recalculating.
3. Recursion: Use a recursive function to calculate the number of ways to form the target using columns starting from k.

Code:

from collections import defaultdict

class Solution:
    def numWays(self, words, target):
        MOD = 10**9 + 7
        cnt = defaultdict(int)

        # Build frequency table for each column
        for word in words:
            for i, c in enumerate(word):
                cnt[(i, c)] += 1

        dp = {}

        def dfs(i, k):
            if i == len(target):  # Successfully formed the target
                return 1
            if k == len(words[0]):  # Ran out of columns
                return 0
            if (i, k) in dp:  # Return memoized result
                return dp[(i, k)]

            # Option 1: Skip the current column
            dp[(i, k)] = dfs(i, k + 1)

            # Option 2: Use the current column if it matches target[i]
            c = target[i]
            if (k, c) in cnt:
                dp[(i, k)] += cnt[(k, c)] * dfs(i + 1, k + 1)

            dp[(i, k)] %= MOD
            return dp[(i, k)]

        return dfs(0, 0)

Time Complexity:

• Building Frequency Table: \(O(n \cdot m)\), where \(n\) is the number of strings in words and \(m\) is their length.
• Dynamic Programming: \(O(t \cdot m)\), where \(t\) is the length of target.

Total time complexity: \(O(n \cdot m + t \cdot m)\).

Space Complexity:

• Frequency table: \(O(26 \cdot m)\) for storing character counts.
• DP table: \(O(t \cdot m)\).

🔍 Example Walkthrough

Input:

words = ["abba", "baab"]
target = "bab"

1. Frequency Table:

1. Recursive DP Steps:

• Starting at \(i=0, k=0\), match ‘b’ using column 0, 1, or 2.
• Move to \(i=1, k=1\), match ‘a’.
• Move to \(i=2, k=2\), match ‘b’.

1. Final Output: \(4\) ways.

🛠️ Edge Cases

1. No Matching Characters: If target contains characters not in words, return \(0\).
2. Empty Target: If target is an empty string, there’s exactly one way to form it: do nothing.
3. Empty Words: If words is empty, return \(0\).

🌟 Conclusion

This problem highlights the power of dynamic programming in solving complex combinatorial problems efficiently. By using memoization and a precomputed frequency table, we reduced what would have been an exponential time complexity to a manageable polynomial one. 🎉

Ready to tackle more problems like this? 🚀