LeetCode-in-Java

2746. Decremental String Concatenation

Medium

You are given a 0-indexed array words containing n strings.

Let’s define a join operation join(x, y) between two strings x and y as concatenating them into xy. However, if the last character of x is equal to the first character of y, one of them is deleted.

For example join("ab", "ba") = "aba" and join("ab", "cde") = "abcde".

You are to perform n - 1 join operations. Let str0 = words[0]. Starting from i = 1 up to i = n - 1, for the ith operation, you can do one of the following:

• Make stri = join(stri - 1, words[i])
• Make stri = join(words[i], stri - 1)

Your task is to minimize the length of strn - 1.

Return an integer denoting the minimum possible length of strn - 1.

Example 1:

Input: words = [“aa”,”ab”,”bc”]

Output: 4

Explanation: In this example, we can perform join operations in the following order to minimize the length of str₂:

str₀ = “aa”

str₁ = join(str₀, “ab”) = “aab”

str₂ = join(str₁, “bc”) = “aabc”

It can be shown that the minimum possible length of str₂ is 4.

Example 2:

Input: words = [“ab”,”b”]

Output: 2

Explanation: In this example, str₀ = “ab”, there are two ways to get str₁: join(str₀, “b”) = “ab” or join(“b”, str₀) = “bab”. The first string, “ab”, has the minimum length. Hence, the answer is 2.

Example 3:

Input: words = [“aaa”,”c”,”aba”]

Output: 6

Explanation: In this example, we can perform join operations in the following order to minimize the length of str₂:

str₀ = “aaa”

str₁ = join(str₀, “c”) = “aaac”

str₂ = join(“aba”, str₁) = “abaaac”

It can be shown that the minimum possible length of str₂ is 6.

Constraints:

• 1 <= words.length <= 1000
• 1 <= words[i].length <= 50
• Each character in words[i] is an English lowercase letter

Solution

public class Solution {
    private Integer[][][] dp;

    public int minimizeConcatenatedLength(String[] words) {
        int n = words.length;
        dp = new Integer[n][26][26];
        String curWord = words[0];
        int curLen = curWord.length();
        char curFirst = curWord.charAt(0);
        char curLast = curWord.charAt(curLen - 1);
        return curLen + solve(1, curFirst, curLast, n, words);
    }

    private int solve(int idx, char prevFirst, char prevLast, int n, String[] words) {
        if (idx == n) {
            return 0;
        }
        if (dp[idx][prevFirst - 'a'][prevLast - 'a'] != null) {
            return dp[idx][prevFirst - 'a'][prevLast - 'a'];
        }
        String curWord = words[idx];
        int curLen = curWord.length();
        char curFirst = curWord.charAt(0);
        char curLast = curWord.charAt(curLen - 1);
        int ans = (int) 1e9;
        if (prevFirst == curLast) {
            ans = Math.min(ans, curLen - 1 + solve(idx + 1, curFirst, prevLast, n, words));
        } else {
            ans = Math.min(ans, curLen + solve(idx + 1, curFirst, prevLast, n, words));
        }
        if (prevLast == curFirst) {
            ans = Math.min(ans, curLen - 1 + solve(idx + 1, prevFirst, curLast, n, words));
        } else {
            ans = Math.min(ans, curLen + solve(idx + 1, prevFirst, curLast, n, words));
        }
        dp[idx][prevFirst - 'a'][prevLast - 'a'] = ans;
        return ans;
    }
}

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