Description#
Given a string s, rearrange the characters of s so that any two adjacent characters are not the same.
Return any possible rearrangement of s or return "" if not possible.
Example 1:
Input: s = "aab"
Output: "aba"
Example 2:
Input: s = "aaab"
Output: ""
Constraints:
1 <= s.length <= 500s consists of lowercase English letters.
Solutions#
Solution 1#
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| class Solution:
def reorganizeString(self, s: str) -> str:
n = len(s)
cnt = Counter(s)
mx = max(cnt.values())
if mx > (n + 1) // 2:
return ''
i = 0
ans = [None] * n
for k, v in cnt.most_common():
while v:
ans[i] = k
v -= 1
i += 2
if i >= n:
i = 1
return ''.join(ans)
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| class Solution {
public String reorganizeString(String s) {
int[] cnt = new int[26];
int mx = 0;
for (char c : s.toCharArray()) {
int t = c - 'a';
++cnt[t];
mx = Math.max(mx, cnt[t]);
}
int n = s.length();
if (mx > (n + 1) / 2) {
return "";
}
int k = 0;
for (int v : cnt) {
if (v > 0) {
++k;
}
}
int[][] m = new int[k][2];
k = 0;
for (int i = 0; i < 26; ++i) {
if (cnt[i] > 0) {
m[k++] = new int[] {cnt[i], i};
}
}
Arrays.sort(m, (a, b) -> b[0] - a[0]);
k = 0;
StringBuilder ans = new StringBuilder(s);
for (int[] e : m) {
int v = e[0], i = e[1];
while (v-- > 0) {
ans.setCharAt(k, (char) ('a' + i));
k += 2;
if (k >= n) {
k = 1;
}
}
}
return ans.toString();
}
}
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| class Solution {
public:
string reorganizeString(string s) {
vector<int> cnt(26);
for (char& c : s) ++cnt[c - 'a'];
int mx = *max_element(cnt.begin(), cnt.end());
int n = s.size();
if (mx > (n + 1) / 2) return "";
vector<vector<int>> m;
for (int i = 0; i < 26; ++i) {
if (cnt[i]) m.push_back({cnt[i], i});
}
sort(m.begin(), m.end());
reverse(m.begin(), m.end());
string ans = s;
int k = 0;
for (auto& e : m) {
int v = e[0], i = e[1];
while (v--) {
ans[k] = 'a' + i;
k += 2;
if (k >= n) k = 1;
}
}
return ans;
}
};
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| func reorganizeString(s string) string {
cnt := make([]int, 26)
for _, c := range s {
t := c - 'a'
cnt[t]++
}
mx := slices.Max(cnt)
n := len(s)
if mx > (n+1)/2 {
return ""
}
m := [][]int{}
for i, v := range cnt {
if v > 0 {
m = append(m, []int{v, i})
}
}
sort.Slice(m, func(i, j int) bool {
return m[i][0] > m[j][0]
})
ans := make([]byte, n)
k := 0
for _, e := range m {
v, i := e[0], e[1]
for v > 0 {
ans[k] = byte('a' + i)
k += 2
if k >= n {
k = 1
}
v--
}
}
return string(ans)
}
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| use std::collections::{ HashMap, BinaryHeap, VecDeque };
impl Solution {
#[allow(dead_code)]
pub fn reorganize_string(s: String) -> String {
let mut map = HashMap::new();
let mut pq = BinaryHeap::new();
let mut ret = String::new();
let mut queue = VecDeque::new();
let n = s.len();
// Initialize the HashMap
for c in s.chars() {
map.entry(c)
.and_modify(|e| {
*e += 1;
})
.or_insert(1);
}
// Initialize the binary heap
for (k, v) in map.iter() {
if 2 * *v - 1 > n {
return "".to_string();
} else {
pq.push((*v, *k));
}
}
while !pq.is_empty() {
let (v, k) = pq.pop().unwrap();
ret.push(k);
queue.push_back((v - 1, k));
if queue.len() == 2 {
let (v, k) = queue.pop_front().unwrap();
if v != 0 {
pq.push((v, k));
}
}
}
if ret.len() == n {
ret
} else {
"".to_string()
}
}
}
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Solution 2#
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| class Solution:
def reorganizeString(self, s: str) -> str:
return self.rearrangeString(s, 2)
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 reorganizeString(String s) {
return rearrangeString(s, 2);
}
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 reorganizeString(string s) {
return rearrangeString(s, 2);
}
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 reorganizeString(s string) string {
return rearrangeString(s, 2)
}
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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