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Do equal elements preserve their order in Insertion Sort algorithm?


Do equal elements preserve their order in Insertion Sort algorithm?

By : Samuel.T
Date : November 21 2020, 03:00 PM
Hope that helps You are absolutely right. This here is a snippet of the Insertion Sort Algorithm from the popular book "Introduction to Algorithms" by Thomas H. Cormen.
code :
INSERTION-SORT(A)
1. for j=2 to A.length
2. key = A[j]
3. // Insert A[j] into the sorted sequence A[1..j-1].
4. i = j-1
5. while i > 0 and A[i] > key
6. A[i+1] = A[i]
7. i = i-1
8. A[i+1] = key


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Will std::multimap preserve the insert order if the key of 2 elements equal to each other?

Will std::multimap preserve the insert order if the key of 2 elements equal to each other?


By : Ben Swanson
Date : March 29 2020, 07:55 AM
wish of those help The elements in a std::map must have unique keys, so... no.
The std::multimap container allows multiple values mapped to one key. When iterating over a std::multimap the elements are ordered by key, but the order of elements having the same key is not specified.
Maintain insertion order for equal elements in std::multiset

Maintain insertion order for equal elements in std::multiset


By : ghoswheel
Date : March 29 2020, 07:55 AM
seems to work fine Here's a wild idea: As @Jerry suggests, maintain a counter. Since the pair of object and counter is unique, we can now use a set (ordered lexicographically). Then we use lower_bound to find the insertion point and compute the next counter value:
code :
unsigned int const uimax = std::numeric_limits<unsigned int>::max();

typedef std::set<std::pair<T, unsigned int>> pair_set;

void counted_insert(pair_set & s, T const & t)
{
    pair_set::iterator it = s.lower_bound(std::make_pair(t, uimax));

    if (it == s.begin() || !(*--it == t))
    {
        s.insert(it, std::make_pair(t, 0));
    }
    else
    {
        s.insert(it, std::make_pair(t, it->first + 1));
    }
}
Preserve insertion order of elements in a data member for Neo4j

Preserve insertion order of elements in a data member for Neo4j


By : Vaishali Aghera
Date : March 29 2020, 07:55 AM
I hope this helps . There is a special implementation of Set, the class TreeSet keeps the elements in the set sorted, either by their natural ordering or by asking a Comparator how they should be ordered. TreeSets reorders the set whenever you add/remove elements.
There is also the LinkedHashSet implementation which keeps the items according to the insertion order.
Preserve original order if numeric value is equal in coreutils sort?

Preserve original order if numeric value is equal in coreutils sort?


By : SADIK KUZU
Date : March 29 2020, 07:55 AM
seems to work fine Just add the -s (stable sort) flag, this disables last-resort comparison
code :
echo '7 a
3 c
3 b
2 first
2 second
2 third
2 fourth
2 fifth
9 d
2 sixth
' | sort  -k 1,1n -s

2 first
2 second
2 third
2 fourth
2 fifth
2 sixth
3 c
3 b
7 a
9 d
Stable sort a C++ hash map - preserve the insertion order for equal elements

Stable sort a C++ hash map - preserve the insertion order for equal elements


By : MariusVT
Date : March 29 2020, 07:55 AM
may help you . Here is a possible solution I came up with using an std::unordered_map and tracking the order of inserting using a std::vector.
Create a hash map with the string as key and count as value.
code :
    unordered_map<std::string, int> map;
    std::vector<std::unordered_map<std::string,int>::iterator> order;

    // Lets assume this is my string stream
    std::vector<std::string> words = {"a","b","a" ... };

    // Insert elements to map and the corresponding iterator to order
    for (auto& word : words){
        auto it = map.emplace(word,1);
        if (!it.second){
            it.first->second++;
        }
        else {
            order.push_back(it.first);
        }
        max_count = std::max(max_count,it.first->second);
    }

    //  Bucket Sorting
    /*  We are iterating over the vector and not the map
        this ensures we are iterating by the order they got inserted */
    std::vector<std::vector<std::string>> buckets(max_count);
    for (auto o : order){
        int count = o->second;
        buckets[count-1].push_back(o->first);
    }

    std::vector<std::string> res;
    for (auto it = buckets.rbegin(); it != buckets.rend(); ++it)
        for (auto& str : *it)
            res.push_back(str);
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