std::bucket_count and std::bucket_size in unordered_map in C++

as we know, internally unordered_map is implemented using hash table so, a bucket is a slot in the internal hash Table to which elements are assigned based on the hash value of their key. Buckets are numbered from 0 to (bucket_count-1). Hence this function returns the bucket no. where element with key is located in unordered_map.
Time Complexity: O(1).

Syntax:

unordered_map.bucket(k);
k is the key corresponds to which we want to know bucket number.
Returns: The order number of the bucket corresponding to key k.

There are two more functions regarding bucket:
1. std::bucket_count: This function is used to count the total no. of buckets in the unordered_map. No parameter is required to pass into this function.
Time Complexity: O(1).

Syntax:

unordered_map.bucket_count();
Returns: The number of the bucket present in hash table of unordered_map.

2. std::bucket_size: This function count the number of elements present in each bucket of the unordered_map.
Time Complexity: Linear in the bucket size.



Syntax:

unordered_map.bucket_size(i);
where 'i' is the bucket number in which we want 
to find no. of elements. (i < bucket_count)
Returns: The number of elements present in bucket 'i'.
// C++ program to demonstrate the use of std::bucket
#include <bits/stdc++.h>
using namespace std;

int main()
{
    // Declaring umap to be of <string, double> type
    // key will be of string type and mapped value will
    // be of double type
    unordered_map<string, double> umap;

    // inserting values by using [] operator
    umap["PI"] = 3.14;
    umap["root2"] = 1.414;
    umap["log10"] = 2.302;
    umap["loge"] = 1.0;
    umap["e"] = 2.718;

    // Display bucket no. where key, value pair is located
    // using bucket(key)
    for (auto& x : umap) {
        cout << "(" << x.first << ", " << x.second << ")";
        cout << " is in bucket= " 
            << umap.bucket(x.first) << endl;
    }
    cout << endl;

    // Count the no.of buckets in the unordered_map 
    // using bucket_count()
    int n = umap.bucket_count();
    cout << "umap has " << n << " buckets.\n\n";

    // Count no. of elements in each bucket using 
    // bucket_size(position)
    for (int i = 0; i < n; i++) {
        cout << "Bucket " << i << " has " 
             << umap.bucket_size(i) << " elements.\n";
    }

    return 0;
}

Output:

(PI, 3.14) is in bucket= 5
(e, 2.718) is in bucket= 1
(root2, 1.414) is in bucket= 1
(log10, 2.302) is in bucket= 10
(loge, 1) is in bucket= 7

umap has 11 buckets.

Bucket 0 has 0 elements.
Bucket 1 has 2 elements.
Bucket 2 has 0 elements.
Bucket 3 has 0 elements.
Bucket 4 has 0 elements.
Bucket 5 has 1 elements.
Bucket 6 has 0 elements.
Bucket 7 has 1 elements.
Bucket 8 has 0 elements.
Bucket 9 has 0 elements.
Bucket 10 has 1 elements.

We can also print all the elements present in each bucket of the unordered_map.

// C++ program to print all elements present in each bucket
#include <bits/stdc++.h>
using namespace std;

int main()
{
    // Declaring umap to be of <string, double> type
    // key will be of string type and mapped value 
    // will be of double type
    unordered_map<string, double> umap;

    // inserting values by using [] operator
    umap["PI"] = 3.14;
    umap["root2"] = 1.414;
    umap["log10"] = 2.302;
    umap["loge"] = 1.0;
    umap["e"] = 2.718;

    unsigned n = umap.bucket_count();

    // Prints elements present in each bucket
    for (unsigned i = 0; i < n; i++) {
        cout << "Bucket " << i << " contains: ";
        for (auto it = umap.begin(i); it != umap.end(i); it++)
            cout << "(" << it->first << ", " 
                 << it->second << ") ";
        cout << "\n";
    }
    return 0;
}

Output:

Bucket 0 contains: 
Bucket 1 contains: (e, 2.718) (root2, 1.414) 
Bucket 2 contains: 
Bucket 3 contains: 
Bucket 4 contains: 
Bucket 5 contains: (PI, 3.14) 
Bucket 6 contains: 
Bucket 7 contains: (loge, 1) 
Bucket 8 contains: 
Bucket 9 contains: 
Bucket 10 contains: (log10, 2.302) 

Use of bucket in std::unordered_map: There is a number of algorithms which require the objects to be hashed into some number of buckets, and then each bucket is processed. Let say, you want to find duplicates in a collection. You hash all items in the collection, then in each bucket you compare items pairwise. A bit less trivial example is Apriori algorithm for finding frequent itemsets.

This article is contributed by Akash Gupta. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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