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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