Handling Sum Queries After Update
<p>You are given two <strong>0-indexed</strong> arrays <code>nums1</code> and <code>nums2</code> and a 2D array <code>queries</code> of queries. There are three types of queries:</p>

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

<p>Return <em>an array containing all the answers to the third type&nbsp;queries.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [1,0,1], nums2 = [0,0,0], queries = [[1,1,1],[2,1,0],[3,0,0]]
<strong>Output:</strong> [3]
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>

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

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums1.length,nums2.length &lt;= 10<sup>5</sup></code></li>
	<li><code>nums1.length = nums2.length</code></li>
	<li><code>1 &lt;= queries.length &lt;= 10<sup>5</sup></code></li>
	<li><code><font face="monospace">queries[i].length = 3</font></code></li>
	<li><code><font face="monospace">0 &lt;= l &lt;= r &lt;= nums1.length - 1</font></code></li>
	<li><code><font face="monospace">0 &lt;= p &lt;= 10<sup>6</sup></font></code></li>
	<li><code>0 &lt;= nums1[i] &lt;= 1</code></li>
	<li><code>0 &lt;= nums2[i] &lt;= 10<sup>9</sup></code></li>
</ul>
Segment Tree; Array
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
