Question

2569. Handling Sum Queries After Update

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Description

You are given two 0-indexed arrays nums1 and nums2 and a 2D array queries of queries. There are three types of queries:

1. For a query of type 1, queries[i] = [1, l, r]. Flip the values from 0 to 1 and from 1 to 0 in nums1 from index l to index r. Both l and r are 0-indexed.
2. For a query of type 2, queries[i] = [2, p, 0]. For every index 0 <= i < n, set nums2[i] = nums2[i] + nums1[i] * p.
3. For a query of type 3, queries[i] = [3, 0, 0]. Find the sum of the elements in nums2.

Return _an array containing all the answers to the third type queries._

 

Example 1:

Input: nums1 = [1,0,1], nums2 = [0,0,0], queries = [[1,1,1],[2,1,0],[3,0,0]]
Output: [3]
Explanation: After the first query nums1 becomes [1,1,1]. After the second query, nums2 becomes [1,1,1], so the answer to the third query is 3. Thus, [3] is returned.

Example 2:

Input: nums1 = [1], nums2 = [5], queries = [[2,0,0],[3,0,0]]
Output: [5]
Explanation: After the first query, nums2 remains [5], so the answer to the second query is 5. Thus, [5] is returned.

 

Constraints:

• 1 <= nums1.length,nums2.length <= 105
• nums1.length = nums2.length
• 1 <= queries.length <= 105
• queries[i].length = 3
• 0 <= l <= r <= nums1.length - 1
• 0 <= p <= 106
• 0 <= nums1[i] <= 1
• 0 <= nums2[i] <= 109

Solutions

Solution 1: Segment Tree

According to the problem description:

• Operation $1$ is to reverse all numbers in the index range $[l,..r]$ of array nums1, that is, change $0$ to $1$ and $1$ to $0$.
• Operation $3$ is to sum all numbers in array nums2.
• Operation $2$ is to add the sum of all numbers in array nums2 with $p$ times the sum of all numbers in array nums1, that is, $sum(nums2) = sum(nums2) + p * sum(nums1)$.

Therefore, we actually only need to maintain the segment sum of array nums1, which can be implemented through a segment tree.

We define each node of the segment tree as Node, each node contains the following attributes:

• l: The left endpoint of the node, the index starts from $1$.
• r: The right endpoint of the node, the index starts from $1$.
• s: The segment sum of the node.
• lazy: The lazy tag of the node.

The segment tree mainly has the following operations:

• build(u, l, r): Build the segment tree.
• pushdown(u): Propagate the lazy tag.
• pushup(u): Update the information of the parent node with the information of the child nodes.
• modify(u, l, r): Modify the segment sum. In this problem, it is to reverse each number in the segment, so the segment sum $s = r - l + 1 - s$.
• query(u, l, r): Query the segment sum.

First, calculate the sum of all numbers in array nums2, denoted as $s$.

When executing operation $1$, we only need to call modify(1, l + 1, r + 1).

When executing operation $2$, we update $s = s + p \times query(1, 1, n)$.

When executing operation $3$, we just need to add $s$ to the answer array.

The time complexity is $O(n + m \times \log n)$, and the space complexity is $O(n)$. Where $n$ and $m$ are the lengths of arrays nums1 and queries respectively.

class Node:
  def __init__(self):
    self.l = self.r = 0
    self.s = self.lazy = 0


class SegmentTree:
  def __init__(self, nums):
    self.nums = nums
    n = len(nums)
    self.tr = [Node() for _ in range(n << 2)]
    self.build(1, 1, n)

  def build(self, u, l, r):
    self.tr[u].l, self.tr[u].r = l, r
    if l == r:
      self.tr[u].s = self.nums[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, l, r):
    if self.tr[u].l >= l and self.tr[u].r <= r:
      self.tr[u].lazy ^= 1
      self.tr[u].s = self.tr[u].r - self.tr[u].l + 1 - self.tr[u].s
      return
    self.pushdown(u)
    mid = (self.tr[u].l + self.tr[u].r) >> 1
    if l <= mid:
      self.modify(u << 1, l, r)
    if r > mid:
      self.modify(u << 1 | 1, l, r)
    self.pushup(u)

  def query(self, u, l, r):
    if self.tr[u].l >= l and self.tr[u].r <= r:
      return self.tr[u].s
    self.pushdown(u)
    mid = (self.tr[u].l + self.tr[u].r) >> 1
    res = 0
    if l <= mid:
      res += self.query(u << 1, l, r)
    if r > mid:
      res += self.query(u << 1 | 1, l, r)
    return res

  def pushup(self, u):
    self.tr[u].s = self.tr[u << 1].s + self.tr[u << 1 | 1].s

  def pushdown(self, u):
    if self.tr[u].lazy:
      mid = (self.tr[u].l + self.tr[u].r) >> 1
      self.tr[u << 1].s = mid - self.tr[u].l + 1 - self.tr[u << 1].s
      self.tr[u << 1].lazy ^= 1
      self.tr[u << 1 | 1].s = self.tr[u].r - mid - self.tr[u << 1 | 1].s
      self.tr[u << 1 | 1].lazy ^= 1
      self.tr[u].lazy ^= 1


class Solution:
  def handleQuery(
    self, nums1: List[int], nums2: List[int], queries: List[List[int]]
  ) -> List[int]:
    tree = SegmentTree(nums1)
    s = sum(nums2)
    ans = []
    for op, a, b in queries:
      if op == 1:
        tree.modify(1, a + 1, b + 1)
      elif op == 2:
        s += a * tree.query(1, 1, len(nums1))
      else:
        ans.append(s)
    return ans

class Node {
  int l, r;
  int s, lazy;
}

class SegmentTree {
  private Node[] tr;
  private int[] nums;

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

  private void build(int u, int l, int r) {
    tr[u].l = l;
    tr[u].r = r;
    if (l == r) {
      tr[u].s = nums[l - 1];
      return;
    }
    int mid = (l + r) >> 1;
    build(u << 1, l, mid);
    build(u << 1 | 1, mid + 1, r);
    pushup(u);
  }

  public void modify(int u, int l, int r) {
    if (tr[u].l >= l && tr[u].r <= r) {
      tr[u].lazy ^= 1;
      tr[u].s = tr[u].r - tr[u].l + 1 - tr[u].s;
      return;
    }
    pushdown(u);
    int mid = (tr[u].l + tr[u].r) >> 1;
    if (l <= mid) {
      modify(u << 1, l, r);
    }
    if (r > mid) {
      modify(u << 1 | 1, l, r);
    }
    pushup(u);
  }

  public int query(int u, int l, int r) {
    if (tr[u].l >= l && tr[u].r <= r) {
      return tr[u].s;
    }
    pushdown(u);
    int mid = (tr[u].l + tr[u].r) >> 1;
    int res = 0;
    if (l <= mid) {
      res += query(u << 1, l, r);
    }
    if (r > mid) {
      res += query(u << 1 | 1, l, r);
    }
    return res;
  }

  private void pushup(int u) {
    tr[u].s = tr[u << 1].s + tr[u << 1 | 1].s;
  }

  private void pushdown(int u) {
    if (tr[u].lazy == 1) {
      int mid = (tr[u].l + tr[u].r) >> 1;
      tr[u << 1].s = mid - tr[u].l + 1 - tr[u << 1].s;
      tr[u << 1].lazy ^= 1;
      tr[u << 1 | 1].s = tr[u].r - mid - tr[u << 1 | 1].s;
      tr[u << 1 | 1].lazy ^= 1;
      tr[u].lazy ^= 1;
    }
  }
}

class Solution {
  public long[] handleQuery(int[] nums1, int[] nums2, int[][] queries) {
    SegmentTree tree = new SegmentTree(nums1);
    long s = 0;
    for (int x : nums2) {
      s += x;
    }
    int m = 0;
    for (var q : queries) {
      if (q[0] == 3) {
        ++m;
      }
    }
    long[] ans = new long[m];
    int i = 0;
    for (var q : queries) {
      if (q[0] == 1) {
        tree.modify(1, q[1] + 1, q[2] + 1);
      } else if (q[0] == 2) {
        s += 1L * q[1] * tree.query(1, 1, nums2.length);
      } else {
        ans[i++] = s;
      }
    }
    return ans;
  }
}

class Node {
public:
  int l = 0, r = 0;
  int s = 0, lazy = 0;
};

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

  void modify(int u, int l, int r) {
    if (tr[u]->l >= l && tr[u]->r <= r) {
      tr[u]->lazy ^= 1;
      tr[u]->s = tr[u]->r - tr[u]->l + 1 - tr[u]->s;
      return;
    }
    pushdown(u);
    int mid = (tr[u]->l + tr[u]->r) >> 1;
    if (l <= mid) {
      modify(u << 1, l, r);
    }
    if (r > mid) {
      modify(u << 1 | 1, l, r);
    }
    pushup(u);
  }

  int query(int u, int l, int r) {
    if (tr[u]->l >= l && tr[u]->r <= r) {
      return tr[u]->s;
    }
    pushdown(u);
    int mid = (tr[u]->l + tr[u]->r) >> 1;
    int res = 0;
    if (l <= mid) {
      res += query(u << 1, l, r);
    }
    if (r > mid) {
      res += query(u << 1 | 1, l, r);
    }
    return res;
  }

private:
  vector<Node*> tr;
  vector<int> nums;

  void build(int u, int l, int r) {
    tr[u]->l = l;
    tr[u]->r = r;
    if (l == r) {
      tr[u]->s = nums[l - 1];
      return;
    }
    int mid = (l + r) >> 1;
    build(u << 1, l, mid);
    build(u << 1 | 1, mid + 1, r);
    pushup(u);
  }

  void pushup(int u) {
    tr[u]->s = tr[u << 1]->s + tr[u << 1 | 1]->s;
  }

  void pushdown(int u) {
    if (tr[u]->lazy) {
      int mid = (tr[u]->l + tr[u]->r) >> 1;
      tr[u << 1]->s = mid - tr[u]->l + 1 - tr[u << 1]->s;
      tr[u << 1]->lazy ^= 1;
      tr[u << 1 | 1]->s = tr[u]->r - mid - tr[u << 1 | 1]->s;
      tr[u << 1 | 1]->lazy ^= 1;
      tr[u]->lazy ^= 1;
    }
  }
};

class Solution {
public:
  vector<long long> handleQuery(vector<int>& nums1, vector<int>& nums2, vector<vector<int>>& queries) {
    SegmentTree* tree = new SegmentTree(nums1);
    long long s = 0;
    for (int& x : nums2) {
      s += x;
    }
    vector<long long> ans;
    for (auto& q : queries) {
      if (q[0] == 1) {
        tree->modify(1, q[1] + 1, q[2] + 1);
      } else if (q[0] == 2) {
        s += 1LL * q[1] * tree->query(1, 1, nums1.size());
      } else {
        ans.push_back(s);
      }
    }
    return ans;
  }
};

type node struct {
  l, r, s, lazy int
}

type segmentTree struct {
  nums []int
  tr   []*node
}

func newSegmentTree(nums []int) *segmentTree {
  n := len(nums)
  tr := make([]*node, n<<2)
  for i := range tr {
    tr[i] = &node{}
  }
  t := &segmentTree{nums, tr}
  t.build(1, 1, n)
  return t
}

func (t *segmentTree) build(u, l, r int) {
  t.tr[u].l, t.tr[u].r = l, r
  if l == r {
    t.tr[u].s = t.nums[l-1]
    return
  }
  mid := (l + r) >> 1
  t.build(u<<1, l, mid)
  t.build(u<<1|1, mid+1, r)
  t.pushup(u)
}

func (t *segmentTree) modify(u, l, r int) {
  if t.tr[u].l >= l && t.tr[u].r <= r {
    t.tr[u].lazy ^= 1
    t.tr[u].s = t.tr[u].r - t.tr[u].l + 1 - t.tr[u].s
    return
  }
  t.pushdown(u)
  mid := (t.tr[u].l + t.tr[u].r) >> 1
  if l <= mid {
    t.modify(u<<1, l, r)
  }
  if r > mid {
    t.modify(u<<1|1, l, r)
  }
  t.pushup(u)
}

func (t *segmentTree) query(u, l, r int) int {
  if t.tr[u].l >= l && t.tr[u].r <= r {
    return t.tr[u].s
  }
  t.pushdown(u)
  mid := (t.tr[u].l + t.tr[u].r) >> 1
  res := 0
  if l <= mid {
    res += t.query(u<<1, l, r)
  }
  if r > mid {
    res += t.query(u<<1|1, l, r)
  }
  return res
}

func (t *segmentTree) pushup(u int) {
  t.tr[u].s = t.tr[u<<1].s + t.tr[u<<1|1].s
}

func (t *segmentTree) pushdown(u int) {
  if t.tr[u].lazy == 1 {
    mid := (t.tr[u].l + t.tr[u].r) >> 1
    t.tr[u<<1].s = mid - t.tr[u].l + 1 - t.tr[u<<1].s
    t.tr[u<<1].lazy ^= 1
    t.tr[u<<1|1].s = t.tr[u].r - mid - t.tr[u<<1|1].s
    t.tr[u<<1|1].lazy ^= 1
    t.tr[u].lazy ^= 1
  }
}

func handleQuery(nums1 []int, nums2 []int, queries [][]int) (ans []int64) {
  tree := newSegmentTree(nums1)
  var s int64
  for _, x := range nums2 {
    s += int64(x)
  }
  for _, q := range queries {
    if q[0] == 1 {
      tree.modify(1, q[1]+1, q[2]+1)
    } else if q[0] == 2 {
      s += int64(q[1] * tree.query(1, 1, len(nums1)))
    } else {
      ans = append(ans, s)
    }
  }
  return
}