Create 2398.Maximum-Number-of-Robots-Within-Budget_v3.cpp

Deque/2398.Maximum-Number-of-Robots-Within-Budget/2398.Maximum-Number-of-Robots-Within-Budget_v3.cpp

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+using LL = long long;
+class Solution {
+public:
+    int maximumRobots(vector<int>& chargeTimes, vector<int>& runningCosts, long long budget) 
+    {
+        vector<pair<LL,LL>>robots;
+        int n = chargeTimes.size();
+
+        LL left = 0, right = n;
+        while (left < right)
+        {
+            LL mid = right-(right-left)/2;            
+            if (isOK(mid, chargeTimes, runningCosts, budget))
+                left = mid;
+            else
+                right = mid-1;
+        }
+        return left;
+    }
+
+    bool isOK(LL k, vector<int>& chargeTimes, vector<int>& runningCosts, long long budget)
+    {
+        LL n = chargeTimes.size();
+        LL sum = 0;              
+        deque<int>dq;
+
+        for (int i=0; i<n; i++)
+        {            
+            sum += runningCosts[i];
+            while (!dq.empty() && chargeTimes[dq.back()] <= chargeTimes[i])
+                dq.pop_back();
+            dq.push_back(i);
+            while (!dq.empty() && dq.front() <= i-k)
+                dq.pop_front();
+
+            if (i>=k-1)
+            {   
+                LL ret = chargeTimes[dq.front()] + (LL)k * sum;
+                if (ret <= budget) return true;
+                sum -= runningCosts[i-k+1];
+
+            }            
+        }
+
+        return false;
+    }
+};