Problem Statement
Given a string s and an integer k, the task is to rearrange the characters in s such that identical characters are separated by at least a distance of k. If rearranging the string to meet this condition is not feasible, the function should return an empty string "". This problem involves analyzing the frequency of each character in the string and strategically placing them apart based on the specified distance k.
Examples
Example 1
Input:
s = "aabbcc", k = 3
Output:
"abcabc"
Explanation:
The same letters are at least a distance of 3 from each other.
Example 2
Input:
s = "aaabc", k = 3
Output:
""
Explanation:
It is not possible to rearrange the string.
Example 3
Input:
s = "aaadbbcc", k = 2
Output:
"abacabcd"
Explanation:
The same letters are at least a distance of 2 from each other.
Constraints
1 <= s.length <= 3 * 105sconsists of only lowercase English letters.0 <= k <= s.length
Approach and Intuition
Understanding how to tackle this problem can be grasped better when we break down the examples provided:
Example Analysis
Example 1:
- Input:
s = "aabbcc", k = 3 - Output:
"abcabc" - Explanation: Each character (
a,b,c) appears twice. Given the conditionk = 3, after placing the first occurrence of each character, the second occurrence can be placed right after completing one full cycle of unique characters, satisfying the minimum distance condition between identical characters.
- Input:
Example 2:
- Input:
s = "aaabc", k = 3 - Output:
"" - Explanation: Here, the frequency of
ais 3. Withk = 3, and the total number of characters only 5, it is impossible to position threeas such that each is separated by at least 2 other characters.
- Input:
Example 3:
- Input:
s = "aaadbbcc", k = 2 - Output:
"abacabcd" - Explanation: In this scenario,
aappears three times, but withk = 2, it is feasible to position theas such that each is separated by at least one other different character, which is achieved in the output.
- Input:
From these examples, the key steps to approach this problem include:
- Count the frequency of each character in the string.
- Use a priority queue to manage the order of insertion based on the highest frequency characters.
- Plan the placement of characters to maintain at least
kdistance by filling temporary blocked positions which get unblocked as we place more characters. - If at any point it's not possible to place an element due to reaching an impossible state (i.e., needing to place a character but not having enough distance from its last placement), conclude that it's not possible to rearrange the string as desired.
This type of problem is akin to task scheduling with cooling periods, where you must ensure certain tasks (or characters, in this context) are not executed (or placed) consecutively without a defined break or gap according to the condition provided (k in this problem).
Solutions
- C++
- Java
class Solution {
public:
string shuffleString(string s, int k) {
unordered_map<char, int> charCount;
int highestFreq = 0;
// Calculate character frequencies and determine the highest frequency
for (char ch : s) {
charCount[ch]++;
highestFreq = max(highestFreq, charCount[ch]);
}
unordered_set<char> mostFrequentChars;
unordered_set<char> nextMostFrequentChars;
// Identify characters with the highest and next highest frequencies
for (auto charFreq : charCount) {
if (charFreq.second == highestFreq) {
mostFrequentChars.insert(charFreq.first);
} else if (charFreq.second == highestFreq - 1) {
nextMostFrequentChars.insert(charFreq.first);
}
}
// Create as many segments as the highest frequency
string parts[highestFreq];
// Populate each segment with characters with highest and second highest frequency
for (int i = 0; i < highestFreq; i++) {
for (char ch: mostFrequentChars) {
parts[i] += ch;
}
if (i < highestFreq - 1) {
for (char ch: nextMostFrequentChars) {
parts[i] += ch;
}
}
}
int partIndex = 0;
// Distribute remaining characters across available parts
for (auto charFreq : charCount) {
char currentChar = charFreq.first;
if (mostFrequentChars.count(currentChar) || nextMostFrequentChars.count(currentChar)) {
continue;
}
for (int count = charFreq.second; count > 0; count--) {
parts[partIndex] += currentChar;
partIndex = (partIndex + 1) % (highestFreq - 1);
}
}
// Ensure every part has at least k characters
for (int i = 0; i < highestFreq - 1; i++) {
if (parts[i].length() < k) {
return "";
}
}
string result;
// Concatenate all the parts
for (string part : parts) {
result += part;
}
return result;
}
};
This C++ code provides a solution for rearranging a string so that the same characters are at least k distance apart. Follow this detailed explanation to understand how this code tackles the problem:
Character Frequency Calculation: The code starts by counting the frequency of each character using an
unordered_map. It also tracks the highest frequency found among characters.Identifying Frequent Characters: Next, sets are used to identify characters that have the highest frequency and the second highest frequency.
Segmenting Characters: The characters are then grouped into segments based on their frequency. The number of segments created equals the highest frequency of any character. These segments are filled first with the most frequent characters and then with the next most frequent characters.
Distributing Remaining Characters: After populating the segments with the most and second-most frequent characters, the remaining less frequent characters are distributed evenly across the available segments.
Validation of Segment Size: The code checks whether each of the segments contains at least
kcharacters. If any segment has fewer thankcharacters, an empty string is returned because it’s impossible to rearrange the string as per the given condition.Result Compilation: Finally, it concatenates all the parts together to give a rearranged string.
This implementation ensures that no adjacent characters in the result are the same and are separated by at least k distance, thereby solving the problem effectively. If rearrangement is not possible under the given constraints, the function rightly returns an empty string.
class Solution {
public String shuffleString(String input, int spacing) {
Map<Character, Integer> characterFrequencies = new HashMap<>();
int highestFreq = 0;
for (char ch : input.toCharArray()) {
characterFrequencies.put(ch, characterFrequencies.getOrDefault(ch, 0) + 1);
highestFreq = Math.max(highestFreq, characterFrequencies.get(ch));
}
Set<Character> mostFrequentChars = new HashSet<>();
Set<Character> nextFrequentChars = new HashSet<>();
for (char ch : characterFrequencies.keySet()) {
if (characterFrequencies.get(ch) == highestFreq) {
mostFrequentChars.add(ch);
} else if (characterFrequencies.get(ch) == highestFreq - 1) {
nextFrequentChars.add(ch);
}
}
StringBuilder[] blocks = new StringBuilder[highestFreq];
for (int i = 0; i < highestFreq; i++) {
blocks[i] = new StringBuilder();
for (char ch : mostFrequentChars) {
blocks[i].append(ch);
}
if (i < highestFreq - 1) {
for (char ch : nextFrequentChars) {
blocks[i].append(ch);
}
}
}
int blockIndex = 0;
for (char ch : characterFrequencies.keySet()) {
if (mostFrequentChars.contains(ch) || nextFrequentChars.contains(ch)) {
continue;
}
for (int count = characterFrequencies.get(ch); count > 0; count--) {
blocks[blockIndex].append(ch);
blockIndex = (blockIndex + 1) % (highestFreq - 1);
}
}
for (int i = 0; i < highestFreq - 1; i++) {
if (blocks[i].length() < spacing) {
return "";
}
}
return String.join("", blocks);
}
}
The provided Java program solves the problem of rearranging a string such that the same characters are at least k distance apart using the following strategy:
- Count the frequency of each character in the input string and identify the highest frequency. This helps in understanding how to distribute the characters in the output.
- Create two sets,
mostFrequentCharsfor characters with the highest frequency andnextFrequentCharsfor those just one less than the highest frequency. This categorization aids in handling characters that might be problematic to place within the spacing constraints. - Initialize an array of StringBuilder,
blocks, where each StringBuilder represents a repeated sequential appearance sequence in the eventual string output. Essentially, each block is capable of accumulating characters to be spaced by the specifiedspacingdistance. - Populate these blocks starting with the most frequent characters, ensuring they repeat through the sequence as required by their frequency. Then, add next most frequent characters, adjusting their addition based on their frequency relative to the maximum frequency.
- Cycle through remaining characters, placing them in the sequence blocks in a way that respects the distance constraints until all characters are allocated or until it's determined that arrangement isn't possible (in case one of the blocks is too short to maintain the required spacing).
- Before constructing the final string, check if any of the blocks are shorter than the specified
spacing, indicating that it's impossible to rearrange the string as required. If feasible, concatenate the blocks into a single string to produce the desired output.
Each approach within the function addresses a particular challenge associated with the problem, ensuring that constraints are met, and the string is rearranged or deemed impossible to rearrange correctly. The careful management of character frequencies and their distribution across the blocks array facilitates the achievement of the k-distance spacing requirement.
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