880379. Group Shifted String
Group Shifted String
Slug: group-shifted-string Difficulty: Medium Id: 880379 Topic Tags: Strings, Hash, Algorithms Company Tags: Google
Summary
Given a string s and an integer k, the problem is to divide the string into groups of size k+1, such that each group contains all shifted versions of the same substring. For example, if s = "awdabnamowl" and k = 4, the output should be [["a", "w", "d", "ab"], ["n", "am", "ow", "l"]]. The key concepts involved are strings, hashing, and algorithms.
Detailed Explanation
The problem can be solved by using a hashmap to store the frequency of each substring. We first generate all substrings of length k+1 from the input string s. Then, for each substring, we calculate its shifted versions (by shifting each character by 1 position) and increment their frequencies in the hashmap. Finally, we iterate over the hashmap and group the substrings based on their frequencies.
Here is a step-by-step breakdown of the solution:
- Generate all substrings of length
k+1from the input strings. - For each substring, calculate its shifted versions (by shifting each character by 1 position).
- Increment the frequency of each shifted version in a hashmap.
- Iterate over the hashmap and group the substrings based on their frequencies.
The time complexity of this solution is O(n), where n is the length of the input string s. This is because we need to generate all substrings, which takes O(n) time. The space complexity is also O(n), as we need to store the frequency of each substring in the hashmap.
+---------------+
| Input String |
+---------------+
|
v
+---------------+
| Generate Substrings |
+---------------+
|
v
+---------------+
| Calculate Shifted Versions |
+---------------+
|
v
+---------------+
| Increment Frequency in Hashmap |
+---------------+
|
v
+---------------+
| Group Substrings based on Frequencies |
+---------------+
Optimized Solutions
Java
import java.util.HashMap;
import java.util.Map;
public class GroupShiftedString {
public static String[] groupStrings(String s, int k) {
Map<String, Integer> map = new HashMap<>();
for (int i = 0; i <= s.length() - k - 1; i++) {
String substring = s.substring(i, i + k + 1);
for (int j = 0; j < k + 1; j++) {
String shiftedSubstring = substring.substring(1) + substring.charAt(0);
map.put(shiftedSubstring, map.getOrDefault(shiftedSubstring, 0) + 1);
}
}
String[] result = new String[map.size()];
int index = 0;
for (Map.Entry<String, Integer> entry : map.entrySet()) {
while (entry.getValue() > 0) {
for (int j = 0; j < k + 1; j++) {
result[index++] = substring.substring(0, j + 1);
substring = substring.substring(1);
}
entry.setValue(entry.getValue() - 1);
}
}
return result;
}
}