lower-bound

I'm converting some C++ code to C# and it calls std::map::lower_bound(k) to find an entry in the map whose key is equal to or greater than k. However, I don't see any way to do the same thing with .NET's SortedDictionary. I suspect I could implement a workaround using SortedList, but unfortunately SortedList is too slow (O(n) for inserting and deleting keys). What can I do? Note: I found a workaround using that takes advantage of my particular scenario... Specifically, my keys are a dense population of integers starting at just over 0, so I used a List<TValue> as my dictionary with the list

Based on the following definition found here Returns an iterator pointing to the first element in the sorted range [first,last) which does not compare less than value. The comparison is done using either operator< for the first version, or comp for the second. What would be the C equivalent implementation of lower_bound(). I understand that it would be a modification of binary search, but can't seem to quite pinpoint to exact implementation. int lower_bound(int a[], int lowIndex, int upperIndex, int e); Sample Case: int a[]= {2,2, 2, 7 }; lower_bound(a, 0, 1,2) would return 0 --> upperIndex is

I have a List of object sorted and I want to find the first occurrence and the last occurrence of an object. In C++, I can easily use std::equal_range (or just one lower_bound and one upper_bound). For example: bool mygreater (int i,int j) { return (i>j); } int main () { int myints[] = {10,20,30,30,20,10,10,20}; std::vector<int> v(myints,myints+8); // 10 20 30 30 20 10 10 20 std::pair<std::vector<int>::iterator,std::vector<int>::iterator> bounds; // using default comparison: std::sort (v.begin(), v.end()); // 10 10 10 20 20 20 30 30 bounds=std::equal_range (v.begin(), v.end(), 20); // ^ ^ //

It's well-known that the worst-case runtime for heapsort is Ω(n lg n), but I'm having trouble seeing why this is. In particular, the first step of heapsort (making a max-heap) takes time Θ(n). This is then followed by n heap deletions. I understand why each heap deletion takes time O(lg n); rebalancing the heap involves a bubble-down operation that takes time O(h) in the height of the heap, and h = O(lg n). However, what I don't see is why this second step should take Ω(n lg n). It seems like any individual heap dequeue wouldn't necessarily cause the node moved to the top to bubble all the way

I wrote a simple program that sorts in O(n). It is highly memory inefficient, but that's not the point. It uses the principle behind a HashMap for sorting: public class NLogNBreak { public static class LinkedListBack { public LinkedListBack(int val){ first = new Node(); first.val = val; } public Node first = null; public void insert(int i){ Node n = new Node(); n.val = i; n.next = first; first = n; } } private static class Node { public Node next = null; public int val; } //max > in[i] > 0 public static LinkedListBack[] sorted(int[] in, int max){ LinkedListBack[] ar = new LinkedListBack[max +