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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