Description#
Given a string s and an integer k, rearrange s such that the same characters are at least distance k from each other. If it is not possible to rearrange the string, return an empty string "".
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 * 105s consists of only lowercase English letters.0 <= k <= s.length
Solutions#
Solution 1#
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| class Solution:
def rearrangeString(self, s: str, k: int) -> str:
h = [(-v, c) for c, v in Counter(s).items()]
heapify(h)
q = deque()
ans = []
while h:
v, c = heappop(h)
v *= -1
ans.append(c)
q.append((v - 1, c))
if len(q) >= k:
w, c = q.popleft()
if w:
heappush(h, (-w, c))
return "" if len(ans) != len(s) else "".join(ans)
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| class Solution {
public String rearrangeString(String s, int k) {
int n = s.length();
int[] cnt = new int[26];
for (char c : s.toCharArray()) {
++cnt[c - 'a'];
}
PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> b[0] - a[0]);
for (int i = 0; i < 26; ++i) {
if (cnt[i] > 0) {
pq.offer(new int[] {cnt[i], i});
}
}
Deque<int[]> q = new ArrayDeque<>();
StringBuilder ans = new StringBuilder();
while (!pq.isEmpty()) {
var p = pq.poll();
int v = p[0], c = p[1];
ans.append((char) ('a' + c));
q.offer(new int[] {v - 1, c});
if (q.size() >= k) {
p = q.pollFirst();
if (p[0] > 0) {
pq.offer(p);
}
}
}
return ans.length() == n ? ans.toString() : "";
}
}
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| class Solution {
public:
string rearrangeString(string s, int k) {
unordered_map<char, int> cnt;
for (char c : s) ++cnt[c];
priority_queue<pair<int, char>> pq;
for (auto& [c, v] : cnt) pq.push({v, c});
queue<pair<int, char>> q;
string ans;
while (!pq.empty()) {
auto [v, c] = pq.top();
pq.pop();
ans += c;
q.push({v - 1, c});
if (q.size() >= k) {
auto p = q.front();
q.pop();
if (p.first) {
pq.push(p);
}
}
}
return ans.size() == s.size() ? ans : "";
}
};
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| func rearrangeString(s string, k int) string {
cnt := map[byte]int{}
for i := range s {
cnt[s[i]]++
}
pq := hp{}
for c, v := range cnt {
heap.Push(&pq, pair{v, c})
}
ans := []byte{}
q := []pair{}
for len(pq) > 0 {
p := heap.Pop(&pq).(pair)
v, c := p.v, p.c
ans = append(ans, c)
q = append(q, pair{v - 1, c})
if len(q) >= k {
p = q[0]
q = q[1:]
if p.v > 0 {
heap.Push(&pq, p)
}
}
}
if len(ans) == len(s) {
return string(ans)
}
return ""
}
type pair struct {
v int
c byte
}
type hp []pair
func (h hp) Len() int { return len(h) }
func (h hp) Less(i, j int) bool {
a, b := h[i], h[j]
return a.v > b.v
}
func (h hp) Swap(i, j int) { h[i], h[j] = h[j], h[i] }
func (h *hp) Push(v any) { *h = append(*h, v.(pair)) }
func (h *hp) Pop() any { a := *h; v := a[len(a)-1]; *h = a[:len(a)-1]; return v }
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