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:
targetshould be formed from left to right.- To form the
ithcharacter (0-indexed) oftarget, you can choose thekthcharacter of thejthstring inwordsiftarget[i] = words[j][k]. - Once you use the
kthcharacter of thejthstring ofwords, you can no longer use the xth character of any string in words where x <= k. In other words, all characters to the left of or at indexkbecome 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
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")
Solution
Great solution found here.
function solve(i, j, freq, target, t) {
//Define mod, m, k
const MOD = 1000000007;
const m = target.length;
const k = freq[0].length;
if (i == m) return 1;
if (j == k) return 0;
if (t[i][j] != -1) return t[i][j];
// Define not_taken as the number of ways to form target[i] using words[j] and the characters to the left of words[j]
const not_taken = solve(i, j + 1, freq, target, t) % MOD;
//Define taken as the number of ways to form target[i] using words[j] and the characters to the right of words[j]
const taken =
(freq[target[i].charCodeAt() - 97][j] *
solve(i + 1, j + 1, freq, target, t)) %
MOD;
//Return the sum of not_taken and taken
return (t[i][j] = (not_taken + taken) % MOD);
}
//Function that returns the number of ways to form target from words
function numWays(words, target) {
const k = words[0].length;
const m = target.length;
const freq = Array.from(new Array(26), () => Array(k).fill(0));
for (let col = 0; col < k; col++) {
for (let word of words) {
//Increment the frequency of the character at index col in word by 1
freq[word.charCodeAt(col) - 97][col]++;
}
}
const t = Array(m)
.fill(-1)
.map(() => Array(k).fill(-1));
return solve(0, 0, freq, target, t);
}