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).
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).
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.
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'.
(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.
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.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.