- Difficulty: Hard
- Tags: LeetCode, Hard, Array, String, Dynamic Programming, leetcode-1639, O(l * (w + n)), O(n)
Problem
You are given a list of strings of the same length words
and a string target
.
Your task is to form target
using the given words
under the following rules:
target
should be formed from left to right.- To form the
ith
character (0-indexed) oftarget
, you can choose thekth
character of thejth
string inwords
iftarget[i] = words[j][k]
. - Once you use the
kth
character of thejth
string ofwords
, you can no longer use thexth
character of any string inwords
wherex <= k
. In other words, all characters to the left of or at indexk
become unusuable for every string. - Repeat the process until you form the string
target
.
Notice that you can use multiple characters from the same string in words
provided the conditions above are met.
Return the number of ways to form target
from words
. Since the answer may be too large, return it modulo 109 + 7
.
Example 1:
Input: words = ["acca","bbbb","caca"], target = "aba" Output: 6 Explanation: There are 6 ways to form target. "aba" -> index 0 ("acca"), index 1 ("bbbb"), index 3 ("caca") "aba" -> index 0 ("acca"), index 2 ("bbbb"), index 3 ("caca") "aba" -> index 0 ("acca"), index 1 ("bbbb"), index 3 ("acca") "aba" -> index 0 ("acca"), index 2 ("bbbb"), index 3 ("acca") "aba" -> index 1 ("caca"), index 2 ("bbbb"), index 3 ("acca") "aba" -> index 1 ("caca"), index 2 ("bbbb"), index 3 ("caca")
Example 2:
Input: words = ["abba","baab"], target = "bab" Output: 4 Explanation: There are 4 ways to form target. "bab" -> index 0 ("baab"), index 1 ("baab"), index 2 ("abba") "bab" -> index 0 ("baab"), index 1 ("baab"), index 3 ("baab") "bab" -> index 0 ("baab"), index 2 ("baab"), index 3 ("baab") "bab" -> index 1 ("abba"), index 2 ("baab"), index 3 ("baab")
Constraints:
1 <= words.length <= 1000
1 <= words[i].length <= 1000
- All strings in
words
have the same length. 1 <= target.length <= 1000
words[i]
andtarget
contain only lowercase English letters.