Maximum Sum Queries
<p>You are given two <strong>0-indexed</strong> integer arrays <code>nums1</code> and <code>nums2</code>, each of length <code>n</code>, and a <strong>1-indexed 2D array</strong> <code>queries</code> where <code>queries[i] = [x<sub>i</sub>, y<sub>i</sub>]</code>.</p>

<p>For the <code>i<sup>th</sup></code> query, find the <strong>maximum value</strong> of <code>nums1[j] + nums2[j]</code> among all indices <code>j</code> <code>(0 &lt;= j &lt; n)</code>, where <code>nums1[j] &gt;= x<sub>i</sub></code> and <code>nums2[j] &gt;= y<sub>i</sub></code>, or <strong>-1</strong> if there is no <code>j</code> satisfying the constraints.</p>

<p>Return <em>an array </em><code>answer</code><em> where </em><code>answer[i]</code><em> is the answer to the </em><code>i<sup>th</sup></code><em> query.</em></p>

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

<pre>
<strong>Input:</strong> nums1 = [4,3,1,2], nums2 = [2,4,9,5], queries = [[4,1],[1,3],[2,5]]
<strong>Output:</strong> [6,10,7]
<strong>Explanation:</strong> 
For the 1st query <code node="[object Object]">x<sub>i</sub> = 4</code>&nbsp;and&nbsp;<code node="[object Object]">y<sub>i</sub> = 1</code>, we can select index&nbsp;<code node="[object Object]">j = 0</code>&nbsp;since&nbsp;<code node="[object Object]">nums1[j] &gt;= 4</code>&nbsp;and&nbsp;<code node="[object Object]">nums2[j] &gt;= 1</code>. The sum&nbsp;<code node="[object Object]">nums1[j] + nums2[j]</code>&nbsp;is 6, and we can show that 6 is the maximum we can obtain.

For the 2nd query <code node="[object Object]">x<sub>i</sub> = 1</code>&nbsp;and&nbsp;<code node="[object Object]">y<sub>i</sub> = 3</code>, we can select index&nbsp;<code node="[object Object]">j = 2</code>&nbsp;since&nbsp;<code node="[object Object]">nums1[j] &gt;= 1</code>&nbsp;and&nbsp;<code node="[object Object]">nums2[j] &gt;= 3</code>. The sum&nbsp;<code node="[object Object]">nums1[j] + nums2[j]</code>&nbsp;is 10, and we can show that 10 is the maximum we can obtain. 

For the 3rd query <code node="[object Object]">x<sub>i</sub> = 2</code>&nbsp;and&nbsp;<code node="[object Object]">y<sub>i</sub> = 5</code>, we can select index&nbsp;<code node="[object Object]">j = 3</code>&nbsp;since&nbsp;<code node="[object Object]">nums1[j] &gt;= 2</code>&nbsp;and&nbsp;<code node="[object Object]">nums2[j] &gt;= 5</code>. The sum&nbsp;<code node="[object Object]">nums1[j] + nums2[j]</code>&nbsp;is 7, and we can show that 7 is the maximum we can obtain.

Therefore, we return&nbsp;<code node="[object Object]">[6,10,7]</code>.
</pre>

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

<pre>
<strong>Input:</strong> nums1 = [3,2,5], nums2 = [2,3,4], queries = [[4,4],[3,2],[1,1]]
<strong>Output:</strong> [9,9,9]
<strong>Explanation:</strong> For this example, we can use index&nbsp;<code node="[object Object]">j = 2</code>&nbsp;for all the queries since it satisfies the constraints for each query.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [2,1], nums2 = [2,3], queries = [[3,3]]
<strong>Output:</strong> [-1]
<strong>Explanation:</strong> There is one query in this example with <code node="[object Object]">x<sub>i</sub></code> = 3 and <code node="[object Object]">y<sub>i</sub></code> = 3. For every index, j, either nums1[j] &lt; <code node="[object Object]">x<sub>i</sub></code> or nums2[j] &lt; <code node="[object Object]">y<sub>i</sub></code>. Hence, there is no solution. 
</pre>

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

<ul>
	<li><code>nums1.length == nums2.length</code>&nbsp;</li>
	<li><code>n ==&nbsp;nums1.length&nbsp;</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums1[i], nums2[i] &lt;= 10<sup>9</sup>&nbsp;</code></li>
	<li><code>1 &lt;= queries.length &lt;= 10<sup>5</sup></code></li>
	<li><code>queries[i].length ==&nbsp;2</code></li>
	<li><code>x<sub>i</sub>&nbsp;== queries[i][1]</code></li>
	<li><code>y<sub>i</sub> == queries[i][2]</code></li>
	<li><code>1 &lt;= x<sub>i</sub>, y<sub>i</sub> &lt;= 10<sup>9</sup></code></li>
</ul>
Stack; Binary Indexed Tree; Segment Tree; Array; Binary Search; Sorting; Monotonic Stack
class BinaryIndexedTree {
private:
    int n;
    vector<int> c;

public:
    BinaryIndexedTree(int n) {
        this->n = n;
        c.resize(n + 1, -1);
    }

    void update(int x, int v) {
        while (x <= n) {
            c[x] = max(c[x], v);
            x += x & -x;
        }
    }

    int query(int x) {
        int mx = -1;
        while (x > 0) {
            mx = max(mx, c[x]);
            x -= x & -x;
        }
        return mx;
    }
};

class Solution {
public:
    vector<int> maximumSumQueries(vector<int>& nums1, vector<int>& nums2, vector<vector<int>>& queries) {
        vector<pair<int, int>> nums;
        int n = nums1.size(), m = queries.size();
        for (int i = 0; i < n; ++i) {
            nums.emplace_back(-nums1[i], nums2[i]);
        }
        sort(nums.begin(), nums.end());
        sort(nums2.begin(), nums2.end());
        vector<int> idx(m);
        iota(idx.begin(), idx.end(), 0);
        sort(idx.begin(), idx.end(), [&](int i, int j) { return queries[j][0] < queries[i][0]; });
        vector<int> ans(m);
        int j = 0;
        BinaryIndexedTree tree(n);
        for (int i : idx) {
            int x = queries[i][0], y = queries[i][1];
            for (; j < n && -nums[j].first >= x; ++j) {
                int k = nums2.end() - lower_bound(nums2.begin(), nums2.end(), nums[j].second);
                tree.update(k, -nums[j].first + nums[j].second);
            }
            int k = nums2.end() - lower_bound(nums2.begin(), nums2.end(), y);
            ans[i] = tree.query(k);
        }
        return ans;
    }
};