Asked in: Meesho
Image of the Question
All Testcases Passed ✔
Solution
from functools import lru_cache
from sys import setrecursionlimit
setrecursionlimit(10**6)
def numWays(words, target):
// ... rest of solution available after purchaseExplanation
```
To solve this problem, begin by understanding the key constraints and requirements. You are given an array of strings (`words`), all of the same length, and a `target` string that you want to build. Each character of the target must be formed by selecting a character from some string in the `words` array, and more specifically, the character must come from a column — that is, a specific index position across all strings.
The central rule is that the indices used in forming the target string must be in strictly increasing order. This means that if you pick a character for position i in the target from column j in the words, the next character in the target must be picked from a column index greater than j. You are allowed to choose the character from any of the strings, but the position in the string (i.e., the column index) must strictly increase as you proceed along the target string.
This constraint naturally suggests a dynamic programming approach. Before diving into that, let's clarify how to represent the data efficiently. Since all the strings in the `words` array are the same length, you can think of them as forming a table with each row representing a string and each column containing the characters at that particular index across all strings. For example, in the sample input, column 0 contains 'a', 'a', and 'e'; column 1 contains 'd', 'e', and 'f'; and column 2 contains 'c', 'c', and 'g'.
Your first thought should be to pre-process this table to quickly count how many times each character appears in each column. This is important because for any character in the target string, you need to know how many ways you can choose that character from any column.
With that count information prepared, the next logical step is to use dynamic programming to explore how many ways you can build the target string using these columns under the strictly increasing index constraint. Define a DP structure where the state represents how many ways you can build the target up to a certain character using up to a certain column.
To process this, you iterate through the columns from left to right. For each column, you examine each character in the target string. If the current target character appears in this column (i.e., count > 0), then you consider all the ways you could have built the previous part of the target before this column, and add the new combinations by multiplying the previous count with the number of ways the current character appears in the column. This recursive accumulation ensures you're counting all valid combinations that obey the strictly increasing index requirement.
As you process columns from left to right, and update the number of ways you can build the target up to each character, you're effectively building up the answer step-by-step. It’s crucial that the transitions only occur forward — you must not update states in a way that would allow backtracking or reusing the same or earlier columns for subsequent characters in the target.
Think of each position in the target string as needing to be filled using one of the later columns. You maintain an array of how many ways each prefix of the target can be formed, and you update this array as you sweep through each column, only ever moving forward.
Eventually, after processing all columns, the final cell of your DP structure (corresponding to the full target string) will hold the total number of ways to construct the entire target string from the words under the given constraints.
Now consider why this works:
- The pre-processing step allows constant-time access to the frequency of each character in each column, which greatly optimizes the lookups.
- The strictly increasing index requirement is naturally maintained by the left-to-right sweep of columns.
- By treating each character in the target as needing to be filled by some column to its right, you ensure all valid orderings are covered.
Edge cases to be aware of include:
- If any character in the target string does not appear in any column to its right when needed, then that path terminates — this is implicitly handled by the character frequency count being zero.
- Very large input sizes: Since you are instructed to return the result modulo 10^9 + 7, ensure that all additions and multiplications in your DP update steps are done under modular arithmetic to avoid integer overflow.
In conclusion, this problem is a blend of frequency-based preprocessing and forward dynamic programming. By converting the columnar data into a count table and then using a bottom-up DP strategy constrained by strictly increasing indices, you can efficiently compute the total number of ways to form the target string.
```