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358. Rearrange String k Distance Apart πŸ”’

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 * 105
  • s 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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