Bucket sort is mainly useful when input is uniformly distributed over a range. For example, consider the following problem.
Sort a large set of floating point numbers which are in range from 0.0 to 1.0 and are uniformly distributed across the range. How do we sort the numbers efficiently?
A simple way is to apply a comparison based sorting algorithm. The lower bound for Comparison based sorting algorithm (Merge Sort, Heap Sort, Quick-Sort .. etc) is Ω(n Log n), i.e., they cannot do better than nLogn.
Can we sort the array in linear time? Counting sort can not be applied here as we use keys as index in counting sort. Here keys are floating point numbers.
The idea is to use bucket sort. Following is bucket algorithm.
bucketSort(arr[], n) 1) Create n empty buckets (Or lists). 2) Do following for every array element arr[i]. .......a) Insert arr[i] into bucket[n*array[i]] 3) Sort individual buckets using insertion sort. 4) Concatenate all sorted buckets.
Time Complexity: If we assume that insertion in a bucket takes O(1) time then steps 1 and 2 of the above algorithm clearly take O(n) time. The O(1) is easily possible if we use a linked list to represent a bucket (In the following code, C++ vector is used for simplicity). Step 4 also takes O(n) time as there will be n items in all buckets.
The main step to analyze is step 3. This step also takes O(n) time on average if all numbers are uniformly distributed (please refer CLRS book for more details)
Following is the implementation of the above algorithm.
C++
// C++ program to sort an// array using bucket sort#include <algorithm>#include <iostream>#include <vector>using namespace std;// Function to sort arr[] of// size n using bucket sortvoid bucketSort(float arr[], int n){ // 1) Create n empty buckets vector<float> b[n]; // 2) Put array elements // in different buckets for (int i = 0; i < n; i++) { int bi = n * arr[i]; // Index in bucket b[bi].push_back(arr[i]); } // 3) Sort individual buckets for (int i = 0; i < n; i++) sort(b[i].begin(), b[i].end()); // 4) Concatenate all buckets into arr[] int index = 0; for (int i = 0; i < n; i++) for (int j = 0; j < b[i].size(); j++) arr[index++] = b[i][j];}/* Driver program to test above function */int main(){ float arr[] = { 0.897, 0.565, 0.656, 0.1234, 0.665, 0.3434 }; int n = sizeof(arr) / sizeof(arr[0]); bucketSort(arr, n); cout << "Sorted array is \n"; for (int i = 0; i < n; i++) cout << arr[i] << " "; return 0;} |
Java
// Java program to sort an array// using bucket sortimport java.util.*;import java.util.Collections;class GFG { // Function to sort arr[] of size n // using bucket sort static void bucketSort(float arr[], int n) { if (n <= 0) return; // 1) Create n empty buckets @SuppressWarnings("unchecked") Vector<Float>[] buckets = new Vector[n]; for (int i = 0; i < n; i++) { buckets[i] = new Vector<Float>(); } // 2) Put array elements in different buckets for (int i = 0; i < n; i++) { float idx = arr[i] * n; buckets[(int)idx].add(arr[i]); } // 3) Sort individual buckets for (int i = 0; i < n; i++) { Collections.sort(buckets[i]); } // 4) Concatenate all buckets into arr[] int index = 0; for (int i = 0; i < n; i++) { for (int j = 0; j < buckets[i].size(); j++) { arr[index++] = buckets[i].get(j); } } } // Driver code public static void main(String args[]) { float arr[] = { (float)0.897, (float)0.565, (float)0.656, (float)0.1234, (float)0.665, (float)0.3434 }; int n = arr.length; bucketSort(arr, n); System.out.println("Sorted array is "); for (float el : arr) { System.out.print(el + " "); } }}// This code is contributed by Himangshu Shekhar Jha |
Python3
# Python3 program to sort an array# using bucket sortdef insertionSort(b): for i in range(1, len(b)): up = b[i] j = i - 1 while j >= 0 and b[j] > up: b[j + 1] = b[j] j -= 1 b[j + 1] = up return b def bucketSort(x): arr = [] slot_num = 10 # 10 means 10 slots, each # slot's size is 0.1 for i in range(slot_num): arr.append([]) # Put array elements in different buckets for j in x: index_b = int(slot_num * j) arr[index_b].append(j) # Sort individual buckets for i in range(slot_num): arr[i] = insertionSort(arr[i]) # concatenate the result k = 0 for i in range(slot_num): for j in range(len(arr[i])): x[k] = arr[i][j] k += 1 return x# Driver Codex = [0.897, 0.565, 0.656, 0.1234, 0.665, 0.3434]print("Sorted Array is")print(bucketSort(x))# This code is contributed by# Oneil Hsiao |
C#
// C# program to sort an array// using bucket sortusing System;using System.Collections;using System.Collections.Generic;class GFG { // Function to sort arr[] of size n // using bucket sort static void bucketSort(float []arr, int n) { if (n <= 0) return; // 1) Create n empty buckets List<float>[] buckets = new List<float>[n]; for (int i = 0; i < n; i++) { buckets[i] = new List<float>(); } // 2) Put array elements in different buckets for (int i = 0; i < n; i++) { float idx = arr[i] * n; buckets[(int)idx].Add(arr[i]); } // 3) Sort individual buckets for (int i = 0; i < n; i++) { buckets[i].Sort(); } // 4) Concatenate all buckets into arr[] int index = 0; for (int i = 0; i < n; i++) { for (int j = 0; j < buckets[i].Count; j++) { arr[index++] = buckets[i][j]; } } } // Driver code public static void Main() { float []arr = { (float)0.897, (float)0.565, (float)0.656, (float)0.1234, (float)0.665, (float)0.3434 }; int n = arr.Length; bucketSort(arr, n); Console.WriteLine("Sorted array is "); foreach(float el in arr) { Console.Write(el + " "); } }}// This code is contributed by rutvik_56 |
Sorted array is 0.1234 0.3434 0.565 0.656 0.665 0.897
Bucket Sort for numbers having integer part:
Algorithm :
- Find maximum element and minimum of the array
- Calculate the range of each bucket
range = (max - min) / n
n is the number of buckets3. Create n buckets of calculated range
4. Scatter the array elements to these buckets
BucketIndex = ( arr[i] - min ) / range
6. Now sort each bucket individually
7. Gather the sorted elements from buckets to original array
Input : Unsorted array: [ 9.8 , 0.6 , 10.1 , 1.9 , 3.07 , 3.04 , 5.0 , 8.0 , 4.8 , 7.68 ] No of buckets : 5 Output : Sorted array: [ 0.6 , 1.9 , 3.04 , 3.07 , 4.8 , 5.0 , 7.68 , 8.0 , 9.8 , 10.1 ]
Input : Unsorted array: [0.49 , 5.9 , 3.4 , 1.11 , 4.5 , 6.6 , 2.0] No of buckets: 3 Output : Sorted array: [0.49 , 1.11 , 2.0 , 3.4 , 4.5 , 5.9 , 6.6]
Code :
Python3
# Python program for the above approach# Bucket sort for numbers# having interger partdef bucketSort(arr, noOfBuckets): max_ele = max(arr) min_ele = min(arr) # range(for buckets) rnge = (max_ele - min_ele) / noOfBuckets temp = [] # create empty buckets for i in range(noOfBuckets): temp.append([]) # scatter the array elements # into the correct bucket for i in range(len(arr)): diff = (arr[i] - min_ele) / rnge - int((arr[i] - min_ele) / rnge) # append the boundary elements to the lower array if(diff == 0 and arr[i] != min_ele): temp[int((arr[i] - min_ele) / rnge) - 1].append(arr[i]) else: temp[int((arr[i] - min_ele) / rnge)].append(arr[i]) # Sort each bucket individually for i in range(len(temp)): if len(temp[i]) != 0: temp[i].sort() # Gather sorted elements # to the original array k = 0 for lst in temp: if lst: for i in lst: arr[k] = i k = k+1# Driver Codearr = [9.8, 0.6, 10.1, 1.9, 3.07, 3.04, 5.0, 8.0, 4.8, 7.68]noOfBuckets = 5bucketSort(arr, noOfBuckets)print("Sorted array: ", arr)# This code is contributed by# Vinita Yadav |
Sorted array: [0.6, 1.9, 3.04, 3.07, 4.8, 5.0, 7.68, 8.0, 9.8, 10.1]
Bucket Sort To Sort an Array with Negative Numbers
References:
Introduction to Algorithms 3rd Edition by Clifford Stein, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest
http://en.wikipedia.org/wiki/Bucket_sort
https://youtu.be/VuXbEb5ywrU
Snapshots:
Quiz on Bucket Sort
Other Sorting Algorithms on GeeksforGeeks/GeeksQuiz:
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
