Sure, here is a Java solution using the concept of Disjoint Set Union (DSU) to solve this problem. The DSU data structure allows us to merge different sets together and perform queries on them in nearly constant time.

```java
import java.util.*;
import java.io.*;

public class Main {
static int parent[];
static StringBuilder str[];

static int find(int a) {
if (parent[a] == a)
return a;
return parent[a] = find(parent[a]);
}

static void union(int a, int b) {
int parent_a = find(a);
int parent_b = find(b);
if (parent_a != parent_b) {
parent[parent_b] = parent_a;
str[parent_a].append(str[parent_b]);
}
}

public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
parent = new int[n + 1];
str = new StringBuilder[n + 1];
for (int i = 1; i

Question

Sure, here is a Java solution using the concept of Disjoint Set Union (DSU) to solve this problem. The DSU data structure allows us to merge different sets together and perform queries on them in nearly constant time.

```java
import java.util.*;
import java.io.*;

public class Main {
    static int parent[];
    static StringBuilder str[];

static int find(int a) {
        if (parent[a] == a)
            return a;
        return parent[a] = find(parent[a]);
    }

static void union(int a, int b) {
        int parent_a = find(a);
        int parent_b = find(b);
        if (parent_a != parent_b) {
            parent[parent_b] = parent_a;
            str[parent_a].append(str[parent_b]);
        }
    }

public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());
        parent = new int[n + 1];
        str = new StringBuilder[n + 1];
        for (int i = 1; i <= n; i++) {
            parent[i] = i;
            str[i] = new StringBuilder(br.readLine());
        }
        for (int i = 1; i < n; i++) {
            StringTokenizer st = new StringTokenizer(br.readLine());
            int a = Integer.parseInt(st.nextToken());
            int b = Integer.parseInt(st.nextToken());
            union(a, b);
        }
        System.out.println(str[find(1)]);
    }
}
```

In this solution, we maintain an array `parent[]` to keep track of the parent of each string, and an array `str[]` of `StringBuilder` objects to store the strings. The `find()` function is used to find the parent of a string, and the `union()` function is used to merge two strings together.

The time complexity of this solution is O(N), where N is the number of strings. This is because we perform a constant amount of work for each string. The space complexity is also O(N), as we need to store the parent and string for each string.

Knowee AI · Accepted Answer

Sure, here is a Java solution using the concept of Disjoint Set Union (DSU) to solve this problem. The DSU data structure allows us to merge different sets together and perform queries on them in nearly constant time.

```java
import java.util.*;
import java.io.*;

public class Main {
    static int parent[];
    static StringBuilder str[];

static int find(int a) {
        if (parent[a] == a)
            return a;
        return parent[a] = find(parent[a]);
    }

static void union(int a, int b) {
        int parent_a = find(a);
        int parent_b = find(b);
        if (parent_a != parent_b) {
            parent[parent_b] = parent_a;
            str[parent_a].append(str[parent_b]);
        }
    }

public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());
        parent = new int[n + 1];
        str = new StringBuilder[n + 1];
        for (int i = 1; i <= n; i++) {
            parent[i] = i;
            str[i] = new StringBuilder(br.readLine());
        }
        for (int i = 1; i < n; i++) {
            StringTokenizer st = new StringTokenizer(br.readLine());
            int a = Integer.parseInt(st.nextToken());
            int b = Integer.parseInt(st.nextToken());
            union(a, b);
        }
        System.out.println(str[find(1)]);
    }
}
```

In this solution, we maintain an array `parent[]` to keep track of the parent of each string, and an array `str[]` of `StringBuilder` objects to store the strings. The `find()` function is used to find the parent of a string, and the `union()` function is used to merge two strings together.

The time complexity of this solution is O(N), where N is the number of strings. This is because we perform a constant amount of work for each string. The space complexity is also O(N), as we need to store the parent and string for each string.

Question

You are given a collection of N non-empty strings, denoted by S 1 , S 2 , … , S n . Then you are given N

S a

1 N | S i | ≤ 10 6 , where | S i | denotes the length of the i t h string. After these N strings, each of the next N

Print the last string which remains at the end of the N

Solution

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