# 2569. Handling Sum Queries After Update

## 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 {
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;
}
};

• 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


• 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
}