Maximum Number of Robots Within Budget
<p>You have <code>n</code> robots. You are given two <strong>0-indexed</strong> integer arrays, <code>chargeTimes</code> and <code>runningCosts</code>, both of length <code>n</code>. The <code>i<sup>th</sup></code> robot costs <code>chargeTimes[i]</code> units to charge and costs <code>runningCosts[i]</code> units to run. You are also given an integer <code>budget</code>.</p>

<p>The <strong>total cost</strong> of running <code>k</code> chosen robots is equal to <code>max(chargeTimes) + k * sum(runningCosts)</code>, where <code>max(chargeTimes)</code> is the largest charge cost among the <code>k</code> robots and <code>sum(runningCosts)</code> is the sum of running costs among the <code>k</code> robots.</p>

<p>Return<em> the <strong>maximum</strong> number of <strong>consecutive</strong> robots you can run such that the total cost <strong>does not</strong> exceed </em><code>budget</code>.</p>

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

<pre>
<strong>Input:</strong> chargeTimes = [3,6,1,3,4], runningCosts = [2,1,3,4,5], budget = 25
<strong>Output:</strong> 3
<strong>Explanation:</strong> 
It is possible to run all individual and consecutive pairs of robots within budget.
To obtain answer 3, consider the first 3 robots. The total cost will be max(3,6,1) + 3 * sum(2,1,3) = 6 + 3 * 6 = 24 which is less than 25.
It can be shown that it is not possible to run more than 3 consecutive robots within budget, so we return 3.
</pre>

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

<pre>
<strong>Input:</strong> chargeTimes = [11,12,19], runningCosts = [10,8,7], budget = 19
<strong>Output:</strong> 0
<strong>Explanation:</strong> No robot can be run that does not exceed the budget, so we return 0.
</pre>

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

<ul>
	<li><code>chargeTimes.length == runningCosts.length == n</code></li>
	<li><code>1 &lt;= n &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= chargeTimes[i], runningCosts[i] &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= budget &lt;= 10<sup>15</sup></code></li>
</ul>
Queue; Array; Binary Search; Prefix Sum; Sliding Window; Monotonic Queue; Heap (Priority Queue)
class Solution {
public:
    int maximumRobots(vector<int>& chargeTimes, vector<int>& runningCosts, long long budget) {
        deque<int> q;
        long long s = 0;
        int ans = 0;
        int n = chargeTimes.size();
        for (int l = 0, r = 0; r < n; ++r) {
            s += runningCosts[r];
            while (q.size() && chargeTimes[q.back()] <= chargeTimes[r]) {
                q.pop_back();
            }
            q.push_back(r);
            while (q.size() && (r - l + 1) * s + chargeTimes[q.front()] > budget) {
                if (q.front() == l) {
                    q.pop_front();
                }
                s -= runningCosts[l++];
            }
            ans = max(ans, r - l + 1);
        }
        return ans;
    }
};
