1639. Number of Ways to Form a Target String Given a Dictionary - Detailed Explanation

Problem Statement

You are given a list of words, where each word has the same length m, and a string target. You need to determine how many ways you can form the target string by selecting one letter from each column of the words array. Each column in words represents a position from which you can pick a character (one from any word in that column). Return the number of ways modulo 10^9 + 7 (to handle large results).

Example 1:

Input:

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

Output:

6

Explanation: There are 6 ways to form "aba" by picking letters from each column:

• 'a' from "acca", 'b' from "bbbb", 'a' from "acca"
• 'a' from "acca", 'b' from "bbbb", 'a' from "caca"
• 'a' from "acca", 'b' from "bbbb", 'a' from "caca"
• 'a' from "acca", 'b' from "bbbb", 'a' from "acca"
• 'a' from "caca", 'b' from "bbbb", 'a' from "acca"
• 'a' from "caca", 'b' from "bbbb", 'a' from "caca"

(Each bullet shows one possible selection of letters from the columns of the words list to form "aba".)

Constraints:

• 1 <= words.length <= 1000
• 1 <= words[i].length, target.length <= 100
• words[i] consists of lowercase English letters.
• target consists of lowercase English letters.

Hints to Solve the Problem

1. Brute Force Approach: Consider generating all possible sequences of characters (by choosing one character from each column) and check which ones match the target. This is a direct approach but becomes inefficient as the size of input grows.
2. Optimized Approach: Use dynamic programming (DP). Precompute the frequency of each character in each column of words. Then, use DP to count ways to form the target by iterating through target characters and columns.

Approach 1: Brute Force (Exhaustive Search)

Explanation:

This approach tries all possible ways of picking characters from the columns and counts the ones that form the target. We can use recursion or DFS to explore every combination:

• Recursively go through each column index and try to match the current target character with letters from that column.
• If a word in the current column has the required character for the current position of target, include that choice and move to the next character of target (and next column).
• Also, consider the possibility of skipping the current column (not taking any character from this column for the target) and move to the next column.
• Count all the combinations that successfully build the entire target string.

This brute force approach is straightforward but inefficient for large inputs because it explores every combination of choices (exponential time).

Python Code (Brute Force)