707909. Find the String
Find the String
Slug: find-the-string
Difficulty: Hard
Id: 707909
Topic Tags: Strings, Graph, Backtracking, Data Structures, Algorithms
Company Tags: Google
Summary
Given a set of strings and their corresponding frequencies, find a string that can be used to create all other strings by replacing certain characters. The frequency of each character in the target string must match its frequency in the original strings.
The problem involves graph traversal, backtracking, and data structures. We will use a combination of these concepts to solve this problem efficiently.
Detailed Explanation
We start by creating an adjacency list representation of the graph where nodes represent characters and edges represent possible replacements between strings. For each string, we create an edge from its last character to all other characters that have the same frequency in the original strings.
Next, we perform a depth-first search (DFS) starting from any node to traverse the graph. During the traversal, we keep track of the current string and its frequency. When we encounter a node with the same frequency as the current string, we can potentially replace the last character of the current string with this new character.
We use backtracking to explore different replacement options. For each possible replacement, we recursively traverse the graph to check if there's another string that can be created by replacing characters. If not, we backtrack and try another replacement option until we find a solution or exhaust all possibilities.
Time complexity analysis: The time complexity is O(N * M), where N is the number of strings and M is the maximum length of a string.
Space complexity analysis: The space complexity is O(M), as we need to store the frequency of each character in the target string.
Optimized Solutions
Java
import java.util.*;
public class FindTheString {
public static String findString(String[] strings, int[] frequencies) {
Map<Character, Integer> graph = new HashMap<>();
for (int i = 0; i < strings.length; i++) {
char lastChar = strings[i].charAt(strings[i].length() - 1);
graph.put(lastChar, frequencies[i]);
}
for (int i = 0; i < strings.length; i++) {
for (char c : strings[i].toCharArray()) {
if (!graph.containsKey(c)) {
graph.put(c, frequencies[i]);
}
}
}
Set<Character> visited = new HashSet<>();
String result = "";
findString(graph, visited, "", result);
return result;
}
private static void findString(Map<Character, Integer> graph, Set<Character> visited, String current, String result) {
if (graph.isEmpty()) {
System.out.println(result);
return;
}
for (Map.Entry<Character, Integer> entry : graph.entrySet()) {
char c = entry.getKey();
int frequency = entry.getValue();
if (!visited.contains(c)) {
visited.add(c);
// Try replacing the last character with this new character
String newString = current + c;
findString(graph, visited, newString, result);
visited.remove(c);
}
}
}
}