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Bucket Sort
  • Difficulty Level : Easy
  • Last Updated : 15 Feb, 2021

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.

BucketSort

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 sort
void 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 sort
import 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 sort
def 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 Code
x = [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


Output

Sorted array is 
0.1234 0.3434 0.565 0.656 0.665 0.897 

Output:  



Sorted array is
0.1234 0.3434 0.565 0.656 0.665 0.897 

Bucket Sort for numbers having integer part:

Algorithm

  1. Find maximum element and minimum of the array
  2. Calculate the range of each bucket
          range = (max - min) / n
          n is the number of buckets

        3. 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 part
def 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 Code
arr = [9.8, 0.6, 10.1, 1.9, 3.07, 3.04, 5.0, 8.0, 4.8, 7.68]
noOfBuckets = 5
bucketSort(arr, noOfBuckets)
print("Sorted array: ", arr)
 
# This code is contributed by
# Vinita Yadav


Output

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: 

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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

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