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Recursive solution is not preferred, they are rarely used in production code.- Poison September 04, 2014
1. Use two max heaps, one for storing all remaining capacities of servers and one for storing tasks.
2. Pop max. task from tasks' heap and subtract max. task length from the max. server. If remaining capacity of server becomes zero, delete the element.
Max heapify both heaps.
3. Do this untill the tasks heap is finished or max. task length is greater than max. server capacity.
Ofcourse, you have to prove this greedy algo. is optimal.
Complexity: O(nlogn + mlogm)