Question

2213. 由单个字符重复的最长子字符串

English Version

题目描述

给你一个下标从 0 开始的字符串 s 。另给你一个下标从 0 开始、长度为 k 的字符串 queryCharacters ，一个下标从 0 开始、长度也是 k 的整数 下标 数组 queryIndices ，这两个都用来描述 k 个查询。

第 i 个查询会将 s 中位于下标 queryIndices[i] 的字符更新为 queryCharacters[i] 。

返回一个长度为 k 的数组 lengths ，其中 lengths[i] 是在执行第 i 个查询 之后 s 中仅由 单个字符重复 组成的 最长子字符串 的 长度 _。_

 

示例 1：

输入：s = "babacc", queryCharacters = "bcb", queryIndices = [1,3,3]
输出：[3,3,4]
解释：
- 第 1 次查询更新后 s = "_bbb_acc" 。由单个字符重复组成的最长子字符串是 "bbb" ，长度为 3 。
- 第 2 次查询更新后 s = "bbb_ccc_" 。由单个字符重复组成的最长子字符串是 "bbb" 或 "ccc"，长度为 3 。
- 第 3 次查询更新后 s = "_bbbb_cc" 。由单个字符重复组成的最长子字符串是 "bbbb" ，长度为 4 。
因此，返回 [3,3,4] 。

示例 2：

输入：s = "abyzz", queryCharacters = "aa", queryIndices = [2,1]
输出：[2,3]
解释：
- 第 1 次查询更新后 s = "aba_zz_" 。由单个字符重复组成的最长子字符串是 "zz" ，长度为 2 。
- 第 2 次查询更新后 s = "_aaa_zz" 。由单个字符重复组成的最长子字符串是 "aaa" ，长度为 3 。
因此，返回 [2,3] 。

 

提示：

• 1 <= s.length <= 105
• s 由小写英文字母组成
• k == queryCharacters.length == queryIndices.length
• 1 <= k <= 105
• queryCharacters 由小写英文字母组成
• 0 <= queryIndices[i] < s.length

解法

方法一：线段树

线段树将整个区间分割为多个不连续的子区间，子区间的数量不超过 log(width)。更新某个元素的值，只需要更新 log(width) 个区间，并且这些区间都包含在一个包含该元素的大区间内。区间修改时，需要使用懒标记保证效率。

• 线段树的每个节点代表一个区间；
• 线段树具有唯一的根节点，代表的区间是整个统计范围，如 [1, N]；
• 线段树的每个叶子节点代表一个长度为 1 的元区间 [x, x]；
• 对于每个内部节点 [l, r]，它的左儿子是 [l, mid]，右儿子是 [mid + 1, r], 其中 mid = ⌊(l + r) / 2⌋ (即向下取整)。

对于本题，线段树节点维护的信息有：

1. 前缀最长连续字符个数 lmx；
2. 后缀最长连续字符个数 rmx；
3. 区间最长连续字符个数 mx；
4. 区间长度 size；
5. 区间首尾字符 lc, rc。

class Node:
  def __init__(self):
    self.l = 0
    self.r = 0
    self.lmx = 0
    self.rmx = 0
    self.mx = 0
    self.size = 0
    self.lc = None
    self.rc = None


N = 100010
tr = [Node() for _ in range(N << 2)]


class SegmentTree:
  def __init__(self, s):
    n = len(s)
    self.s = s
    self.build(1, 1, n)

  def build(self, u, l, r):
    tr[u].l = l
    tr[u].r = r
    if l == r:
      tr[u].lmx = tr[u].rmx = tr[u].mx = tr[u].size = 1
      tr[u].lc = tr[u].rc = self.s[l - 1]
      return
    mid = (l + r) >> 1
    self.build(u << 1, l, mid)
    self.build(u << 1 | 1, mid + 1, r)
    self.pushup(u)

  def modify(self, u, x, v):
    if tr[u].l == x and tr[u].r == x:
      tr[u].lc = tr[u].rc = v
      return
    mid = (tr[u].l + tr[u].r) >> 1
    if x <= mid:
      self.modify(u << 1, x, v)
    else:
      self.modify(u << 1 | 1, x, v)
    self.pushup(u)

  def query(self, u, l, r):
    if tr[u].l >= l and tr[u].r <= r:
      return tr[u]
    mid = (tr[u].l + tr[u].r) >> 1
    if r <= mid:
      return self.query(u << 1, l, r)
    if l > mid:
      return self.query(u << 1 | 1, l, r)
    left, right = self.query(u << 1, l, r), self.query(u << 1 | 1, l, r)
    ans = Node()
    self._pushup(ans, left, right)
    return ans

  def _pushup(self, root, left, right):
    root.lc, root.rc = left.lc, right.rc
    root.size = left.size + right.size

    root.mx = max(left.mx, right.mx)
    root.lmx, root.rmx = left.lmx, right.rmx

    if left.rc == right.lc:
      if left.lmx == left.size:
        root.lmx += right.lmx
      if right.rmx == right.size:
        root.rmx += left.rmx
      root.mx = max(root.mx, left.rmx + right.lmx)

  def pushup(self, u):
    self._pushup(tr[u], tr[u << 1], tr[u << 1 | 1])


class Solution:
  def longestRepeating(
    self, s: str, queryCharacters: str, queryIndices: List[int]
  ) -> List[int]:
    tree = SegmentTree(s)
    k = len(queryIndices)
    ans = []
    for i, c in enumerate(queryCharacters):
      x = queryIndices[i] + 1
      tree.modify(1, x, c)
      ans.append(tree.query(1, 1, len(s)).mx)
    return ans

class Node {
  int l;
  int r;
  int size;
  int lmx;
  int rmx;
  int mx;
  char lc;
  char rc;
}

class SegmentTree {
  private String s;
  private Node[] tr;

  public SegmentTree(String s) {
    int n = s.length();
    this.s = s;
    tr = new Node[n << 2];
    for (int i = 0; i < tr.length; ++i) {
      tr[i] = new Node();
    }
    build(1, 1, n);
  }

  public void build(int u, int l, int r) {
    tr[u].l = l;
    tr[u].r = r;
    if (l == r) {
      tr[u].lmx = 1;
      tr[u].rmx = 1;
      tr[u].mx = 1;
      tr[u].size = 1;
      tr[u].lc = s.charAt(l - 1);
      tr[u].rc = s.charAt(l - 1);
      return;
    }
    int mid = (l + r) >> 1;
    build(u << 1, l, mid);
    build(u << 1 | 1, mid + 1, r);
    pushup(u);
  }

  void modify(int u, int x, char v) {
    if (tr[u].l == x && tr[u].r == x) {
      tr[u].lc = v;
      tr[u].rc = v;
      return;
    }
    int mid = (tr[u].l + tr[u].r) >> 1;
    if (x <= mid) {
      modify(u << 1, x, v);
    } else {
      modify(u << 1 | 1, x, v);
    }
    pushup(u);
  }

  Node query(int u, int l, int r) {
    if (tr[u].l >= l && tr[u].r <= r) {
      return tr[u];
    }
    int mid = (tr[u].l + tr[u].r) >> 1;
    if (r <= mid) {
      return query(u << 1, l, r);
    }
    if (l > mid) {
      return query(u << 1 | 1, l, r);
    }
    Node ans = new Node();
    Node left = query(u << 1, l, r);
    Node right = query(u << 1 | 1, l, r);
    pushup(ans, left, right);
    return ans;
  }

  void pushup(Node root, Node left, Node right) {
    root.lc = left.lc;
    root.rc = right.rc;
    root.size = left.size + right.size;

    root.mx = Math.max(left.mx, right.mx);
    root.lmx = left.lmx;
    root.rmx = right.rmx;

    if (left.rc == right.lc) {
      if (left.lmx == left.size) {
        root.lmx += right.lmx;
      }
      if (right.rmx == right.size) {
        root.rmx += left.rmx;
      }
      root.mx = Math.max(root.mx, left.rmx + right.lmx);
    }
  }

  void pushup(int u) {
    pushup(tr[u], tr[u << 1], tr[u << 1 | 1]);
  }
}

class Solution {
  public int[] longestRepeating(String s, String queryCharacters, int[] queryIndices) {
    SegmentTree tree = new SegmentTree(s);
    int k = queryCharacters.length();
    int[] ans = new int[k];
    for (int i = 0; i < k; ++i) {
      int x = queryIndices[i] + 1;
      char c = queryCharacters.charAt(i);
      tree.modify(1, x, c);
      ans[i] = tree.query(1, 1, s.length()).mx;
    }
    return ans;
  }
}

class Node {
public:
  int l, r, size, lmx, rmx, mx;
  char lc, rc;
};

class SegmentTree {
private:
  string s;
  vector<Node*> tr;

public:
  SegmentTree(string& s) {
    this->s = s;
    int n = s.size();
    tr.resize(n << 2);
    for (int i = 0; i < tr.size(); ++i) tr[i] = new Node();
    build(1, 1, n);
  }

  void build(int u, int l, int r) {
    tr[u]->l = l;
    tr[u]->r = r;
    if (l == r) {
      tr[u]->lmx = tr[u]->rmx = tr[u]->mx = tr[u]->size = 1;
      tr[u]->lc = tr[u]->rc = s[l - 1];
      return;
    }
    int mid = (l + r) >> 1;
    build(u << 1, l, mid);
    build(u << 1 | 1, mid + 1, r);
    pushup(u);
  }

  void modify(int u, int x, char v) {
    if (tr[u]->l == x && tr[u]->r == x) {
      tr[u]->lc = tr[u]->rc = v;
      return;
    }
    int mid = (tr[u]->l + tr[u]->r) >> 1;
    if (x <= mid)
      modify(u << 1, x, v);
    else
      modify(u << 1 | 1, x, v);
    pushup(u);
  }

  Node* query(int u, int l, int r) {
    if (tr[u]->l >= l && tr[u]->r <= r) return tr[u];
    int mid = (tr[u]->l + tr[u]->r) >> 1;
    if (r <= mid) return query(u << 1, l, r);
    if (l > mid) query(u << 1 | 1, l, r);
    Node* ans = new Node();
    Node* left = query(u << 1, l, r);
    Node* right = query(u << 1 | 1, l, r);
    pushup(ans, left, right);
    return ans;
  }

  void pushup(Node* root, Node* left, Node* right) {
    root->lc = left->lc;
    root->rc = right->rc;
    root->size = left->size + right->size;

    root->mx = max(left->mx, right->mx);
    root->lmx = left->lmx;
    root->rmx = right->rmx;

    if (left->rc == right->lc) {
      if (left->lmx == left->size) root->lmx += right->lmx;
      if (right->rmx == right->size) root->rmx += left->rmx;
      root->mx = max(root->mx, left->rmx + right->lmx);
    }
  }

  void pushup(int u) {
    pushup(tr[u], tr[u << 1], tr[u << 1 | 1]);
  }
};

class Solution {
public:
  vector<int> longestRepeating(string s, string queryCharacters, vector<int>& queryIndices) {
    SegmentTree* tree = new SegmentTree(s);
    int k = queryCharacters.size();
    vector<int> ans(k);
    for (int i = 0; i < k; ++i) {
      int x = queryIndices[i] + 1;
      tree->modify(1, x, queryCharacters[i]);
      ans[i] = tree->query(1, 1, s.size())->mx;
    }
    return ans;
  }
};

type segmentTree struct {
  str []byte
  mx  []int
  lmx []int
  rmx []int
}

func newSegmentTree(s string) *segmentTree {
  n := len(s)
  t := &segmentTree{
    str: []byte(s),
    mx:  make([]int, n<<2),
    lmx: make([]int, n<<2),
    rmx: make([]int, n<<2),
  }
  t.build(0, 0, n-1)
  return t
}

func (t *segmentTree) build(x, l, r int) {
  if l == r {
    t.lmx[x] = 1
    t.rmx[x] = 1
    t.mx[x] = 1
    return
  }
  m := int(uint(l+r) >> 1)
  t.build(x*2+1, l, m)
  t.build(x*2+2, m+1, r)
  t.pushup(x, l, m, r)
}

func (t *segmentTree) pushup(x, l, m, r int) {
  lch, rch := x*2+1, x*2+2
  t.lmx[x] = t.lmx[lch]
  t.rmx[x] = t.rmx[rch]
  t.mx[x] = max(t.mx[lch], t.mx[rch])
  // can be merged
  if t.str[m] == t.str[m+1] {
    if t.lmx[lch] == m-l+1 {
      t.lmx[x] += t.lmx[rch]
    }
    if t.rmx[rch] == r-m {
      t.rmx[x] += t.rmx[lch]
    }
    t.mx[x] = max(t.mx[x], t.rmx[lch]+t.lmx[rch])
  }
}

func (t *segmentTree) update(x, l, r, pos int, val byte) {
  if l == r {
    t.str[pos] = val
    return
  }
  m := int(uint(l+r) >> 1)
  if pos <= m {
    t.update(x*2+1, l, m, pos, val)
  } else {
    t.update(x*2+2, m+1, r, pos, val)
  }
  t.pushup(x, l, m, r)
}

func longestRepeating(s string, queryCharacters string, queryIndices []int) []int {
  ans := make([]int, len(queryCharacters))
  t := newSegmentTree(s)
  n := len(s)
  for i, c := range queryCharacters {
    t.update(0, 0, n-1, queryIndices[i], byte(c))
    ans[i] = t.mx[0]
  }
  return ans
}