Minimum Number of Valid Strings to Form Target II

This approach uses dynamic programming to solve the problem. We define dp[i] as the minimum number of valid strings required to form the prefix of target of length i. The goal is to compute dp[n], where n is the length of target.

The base case is dp[0] = 0, as an empty string requires zero valid strings. For each i from 1 to n, we try to find a split point j (where 0 <= j < i) such that the substring target[j...i-1] is a valid prefix. If it is, we can potentially form target[0...i-1] by taking an optimal solution for target[0...j-1] (which is dp[j]) and adding one more valid string, target[j...i-1].

To check if target[j...i-1] is a valid prefix, we perform a naive search: iterate through every word in the words array and use the startsWith method. We take the minimum over all possible split points j.

1. Let n be the length of the target string.
2. Create a dynamic programming array dp of size n + 1. dp[i] will store the minimum number of valid strings to form the prefix target[0...i-1].
3. Initialize dp[0] = 0 (an empty prefix requires 0 strings) and all other dp[i] to a value representing infinity (e.g., n + 1).
4. Iterate from i = 1 to n: a. For each i, iterate from j = 0 to i - 1. b. Extract the substring sub = target.substring(j, i). c. Check if sub is a valid prefix. This is done by iterating through every word in the words array and checking if word.startsWith(sub). d. If sub is a valid prefix and dp[j] is not infinity, it means we can form target[0...i-1] by appending sub to a valid formation of target[0...j-1]. Update dp[i] = min(dp[i], dp[j] + 1).
5. After the loops complete, if dp[n] is still infinity, it's impossible to form the target. Return -1.
6. Otherwise, dp[n] holds the minimum number of strings, so return dp[n].

The algorithm iterates through all possible ending positions i in the target string and for each i, it considers all possible starting positions j. The substring between j and i is then checked for validity. This validity check is the main performance bottleneck, as it involves iterating through the entire words array for every single substring. This leads to a very high polynomial time complexity.

public int minStrings(String[] words, String target) {
    int n = target.length();
    int[] dp = new int[n + 1];
    java.util.Arrays.fill(dp, n + 1);
    dp[0] = 0;

    for (int i = 1; i <= n; i++) {
        for (int j = 0; j < i; j++) {
            if (dp[j] > n) {
                continue;
            }
            String sub = target.substring(j, i);
            boolean is_valid = false;
            for (String word : words) {
                if (word.startsWith(sub)) {
                    is_valid = true;
                    break;
                }
            }

            if (is_valid) {
                dp[i] = Math.min(dp[i], dp[j] + 1);
            }
        }
    }

    return dp[n] > n ? -1 : dp[n];
}