Create furthest_building_reach_possible.cpp

Author: avidLearnerInProgress

furthest_building_reach_possible.cpp

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+class Solution {
+public:
+    int furthestBuilding(vector<int>& heights, int bricks, int ladders) {
+        
+        /*
+            The idea is :-
+            
+            - Ladder allows infinite jumps and bricks have h[i+1] - h[i] possible jumps at given i.
+            We have to use ladder juidiciously.
+            
+            - This means, greddily picking up bricks and ensuring that we run out of bricks first. then if ladders are remaining, use them.
+            
+            - Also, we have to find the place where we used maximum number of bricks and change it to ladder so that we can regain those many bricks that can be used ahead. => For this purpose we use a maxHeap to maintain the count of bricks used.
+
+        */
+        
+        if(heights.size() == 0) return 0;
+        
+        priority_queue<int> bricksUsed; //maxHeap;
+
+        for(int i = 0; i < heights.size() - 1; ++i) {
+            
+            if(heights[i] >= heights[i+1]) continue; //prev >= next
+            
+            int difference = heights[i+1] - heights[i];
+            bricksUsed.push(difference);
+            bricks -= difference;
+            
+            if(bricks >= 0) continue; //more usage of bricks possible
+            
+            if(ladders == 0) return i; //if we are out of ladders, that's the furthest we can reach
+            
+            bricks += bricksUsed.top(); //pick the most used bricks(most difference) and then replace it with ladder; thereby regaining the bricks
+            bricksUsed.pop(); 
+            ladders--;
+        }
+        
+        return heights.size() - 1;
+        
+    }
+};