Hello, in this article I would be telling an alternative solution to the problem [problem:580B]↵
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Most of the solutions , which I have seen use Binary search to solve the problem after sorting of course, so the overall complexity of the solution is O(nlogn) + O(nlogn). I am discussing approach with the complexity of O(nlogn) + O(n). Here , the first O(nlogn) is due to sorting. So, after sorting there is a great improvement in the algorithm complexity. My approach is similar to Z-algorithm for string matching.↵
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Just like Z-algorithm, we maintain a Z[i] array which will contain two things (ie a pair<int,int>), first the maximum index till which we can invite friends starting at ith position and second the total cost of doing so. ↵
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Now, while iterating over the entire array, we check whether the current index is less than the Z[i-1].first, if it is so then we immediately know that Z[i].first is at least Z[i-1].first and we iterate from there and update the Z array. ↵
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For any clarification, please see my following submission [submission:13210461753]
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Most of the solutions , which I have seen use Binary search to solve the problem after sorting of course, so the overall complexity of the solution is O(nlogn) + O(nlogn). I am discussing approach with the complexity of O(nlogn) + O(n). Here , the first O(nlogn) is due to sorting. So, after sorting there is a great improvement in the algorithm complexity. My approach is similar to Z-algorithm for string matching.↵
↵
Just like Z-algorithm, we maintain a Z[i] array which will contain two things (ie a pair<int,int>), first the maximum index till which we can invite friends starting at ith position and second the total cost of doing so. ↵
↵
Now, while iterating over the entire array, we check whether the current index is less than the Z[i-1].first, if it is so then we immediately know that Z[i].first is at least Z[i-1].first and we iterate from there and update the Z array. ↵
↵
For any clarification, please see my following submission [submission:13210
