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

Description

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 i^th character (0-indexed) of target, you can choose the k^th character of the j^th string in words if target[i] = words[j][k].
Once you use the k^th character of the j^th string of words, you can no longer use the x^th character of any string in words where x <= k. In other words, all characters to the left of or at index k 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] and target contain only lowercase English letters.

Solutions

Solution 1: Preprocessing + Memory Search

We noticed that the length of each string in the string array $words$ is the same, so let’s remember $n$, then we can preprocess a two-dimensional array $cnt$, where $cnt[j][c]$ represents the string array $words$ The number of characters $c$ in the $j$-th position of.

Next, we design a function $dfs(i, j)$, which represents the number of schemes that construct $target[i,..]$ and the currently selected character position from $words$ is $j$. Then the answer is $dfs(0, 0)$.

The calculation logic of function $dfs(i, j)$ is as follows:

If $i \geq m$, it means that all characters in $target$ have been selected, then the number of schemes is $1$.
If $j \geq n$, it means that all characters in $words$ have been selected, then the number of schemes is $0$.
Otherwise, we can choose not to select the character in the $j$-th position of $words$, then the number of schemes is $dfs(i, j + 1)$; or we choose the character in the $j$-th position of $words$, then the number of schemes is $dfs(i + 1, j + 1) \times cnt[j][target[i] - ‘a’]$.

Finally, we return $dfs(0, 0)$. Note that the answer is taken in modulo operation.

The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Where $m$ is the length of the string $target$, and $n$ is the length of each string in the string array $words$.

Solution 2: Preprocessing + Dynamic Programming

Similar to Solution 1, we can first preprocess a two-dimensional array $cnt$, where $cnt[j][c]$ represents the number of characters $c$ in the $j$-th position of the string array $words$.

Next, we define $f[i][j]$ which represents the number of ways to construct the first $i$ characters of $target$, and currently select characters from the first $j$ characters of each word in $words$. Then the answer is $f[m][n]$. Initially $f[0][j] = 1$, where $0 \leq j \leq n$.

Consider $f[i][j]$, where $i \gt 0$, $j \gt 0$. We can choose not to select the character in the $j$-th position of $words$, in which case the number of ways is $f[i][j - 1]$; or we choose the character in the $j$-th position of $words$, in which case the number of ways is $f[i - 1][j - 1] \times cnt[j - 1][target[i - 1] - ‘a’]$. Finally, we add the number of ways in these two cases, which is the value of $f[i][j]$.

Finally, we return $f[m][n]$. Note the mod operation of the answer.

The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Where $m$ is the length of the string $target$, and $n$ is the length of each string in the string array $words$.

class Solution {
    private int m;
    private int n;
    private String target;
    private Integer[][] f;
    private int[][] cnt;
    private final int mod = (int) 1e9 + 7;

    public int numWays(String[] words, String target) {
        m = target.length();
        n = words[0].length();
        f = new Integer[m][n];
        this.target = target;
        cnt = new int[n][26];
        for (var w : words) {
            for (int j = 0; j < n; ++j) {
                cnt[j][w.charAt(j) - 'a']++;
            }
        }
        return dfs(0, 0);
    }

    private int dfs(int i, int j) {
        if (i >= m) {
            return 1;
        }
        if (j >= n) {
            return 0;
        }
        if (f[i][j] != null) {
            return f[i][j];
        }
        long ans = dfs(i, j + 1);
        ans += 1L * dfs(i + 1, j + 1) * cnt[j][target.charAt(i) - 'a'];
        ans %= mod;
        return f[i][j] = (int) ans;
    }
}

class Solution {
public:
    int numWays(vector<string>& words, string target) {
        const int mod = 1e9 + 7;
        int m = target.size(), n = words[0].size();
        vector<vector<int>> cnt(n, vector<int>(26));
        for (auto& w : words) {
            for (int j = 0; j < n; ++j) {
                ++cnt[j][w[j] - 'a'];
            }
        }
        int f[m][n];
        memset(f, -1, sizeof(f));
        function<int(int, int)> dfs = [&](int i, int j) -> int {
            if (i >= m) {
                return 1;
            }
            if (j >= n) {
                return 0;
            }
            if (f[i][j] != -1) {
                return f[i][j];
            }
            int ans = dfs(i, j + 1);
            ans = (ans + 1LL * dfs(i + 1, j + 1) * cnt[j][target[i] - 'a']) % mod;
            return f[i][j] = ans;
        };
        return dfs(0, 0);
    }
};

class Solution:
    def numWays(self, words: List[str], target: str) -> int:
        @cache
        def dfs(i: int, j: int) -> int:
            if i >= m:
                return 1
            if j >= n:
                return 0
            ans = dfs(i + 1, j + 1) * cnt[j][ord(target[i]) - ord('a')]
            ans = (ans + dfs(i, j + 1)) % mod
            return ans

        m, n = len(target), len(words[0])
        cnt = [[0] * 26 for _ in range(n)]
        for w in words:
            for j, c in enumerate(w):
                cnt[j][ord(c) - ord('a')] += 1
        mod = 10**9 + 7
        return dfs(0, 0)

func numWays(words []string, target string) int {
	m, n := len(target), len(words[0])
	f := make([][]int, m)
	cnt := make([][26]int, n)
	for _, w := range words {
		for j, c := range w {
			cnt[j][c-'a']++
		}
	}
	for i := range f {
		f[i] = make([]int, n)
		for j := range f[i] {
			f[i][j] = -1
		}
	}
	const mod = 1e9 + 7
	var dfs func(i, j int) int
	dfs = func(i, j int) int {
		if i >= m {
			return 1
		}
		if j >= n {
			return 0
		}
		if f[i][j] != -1 {
			return f[i][j]
		}
		ans := dfs(i, j+1)
		ans = (ans + dfs(i+1, j+1)*cnt[j][target[i]-'a']) % mod
		f[i][j] = ans
		return ans
	}
	return dfs(0, 0)
}

function numWays(words: string[], target: string): number {
    const m = target.length;
    const n = words[0].length;
    const f = new Array(m + 1).fill(0).map(() => new Array(n + 1).fill(0));
    const mod = 1e9 + 7;
    for (let j = 0; j <= n; ++j) {
        f[0][j] = 1;
    }
    const cnt = new Array(n).fill(0).map(() => new Array(26).fill(0));
    for (const w of words) {
        for (let j = 0; j < n; ++j) {
            ++cnt[j][w.charCodeAt(j) - 97];
        }
    }
    for (let i = 1; i <= m; ++i) {
        for (let j = 1; j <= n; ++j) {
            f[i][j] = f[i][j - 1] + f[i - 1][j - 1] * cnt[j - 1][target.charCodeAt(i - 1) - 97];
            f[i][j] %= mod;
        }
    }
    return f[m][n];
}

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

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

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

Description

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