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 {
    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;
    }
}