Count Ways to Build a Target from Increasing Dictionary Columns

You are given equal-length words and a target string. Build the target from left to right. For each target character, choose a matching character from any word at a column strictly greater than the column used previously. Once a column is used, that column and all earlier columns are unavailable across every word. You may choose multiple characters from the same word if their columns increase. Return the number of construction choices modulo 1000000007. ### Constraints and Clarifications The report explicitly identifies [LeetCode 1639](https://leetcode.com/problems/number-of-ways-to-form-a-target-string-given-a-dictionary/description/) but says interview limits were much larger. Exact enlarged bounds are unknown; do not import the reference maxima. Use nonempty equal-length lowercase-English words and a nonempty lowercase target as the referenced domain. Different word indices supplying a matching character count as different choices, including identical word strings. ### Example For words ["aa","ab"] and target "aa", there are two choices at column 0 and one at column 1, so return 2. ### What a Strong Answer Covers Aggregate character frequencies by column, use dynamic programming over target prefixes, and prevent reuse of a column with a descending update order. Explain modulo arithmetic and the input-reading and DP costs separately. ### Follow-up Questions - Why does ascending target-prefix iteration reuse the same column incorrectly? - What if the target is longer than every word?