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Sorting is ordering a list of objects. We can distinguish two types of sorting. If the number of objects is small enough to fits into the main memory, sorting is called internal sorting. If the number of objects is so large that some of them reside on external storage during the sort, it is called external sorting. In this chapter we consider the following internal sorting algorithms

  • Bucket sort
  • Bubble sort
  • Insertion sort
  • Selection sort
  • Heapsort
  • Mergesort

O(n) algorithms

Bucket Sort

Suppose we need to sort an array of positive integers {3,11,2,9,1,5}. A bucket sort works as follows: create an array of size 11. Then, go through the input array and place integer 3 into a second array at index 3, integer 11 at index 11 and so on. We will end up with a sorted list in the second array.

Suppose we are sorting a large number of local phone numbers, for example, all residential phone numbers in the 412 area code region (about 1 million) We sort the numbers without use of comparisons in the following way. Create an a bit array of size 107. It takes about 1Mb. Set all bits to 0. For each phone number turn-on the bit indexed by that phone number. Finally, walk through the array and for each bit 1 record its index, which is a phone number.

We immediately see two drawbacks to this sorting algorithm. Firstly, we must know how to handle duplicates. Secondly, we must know the maximum value in the unsorted array.. Thirdly, we must have enough memory – it may be impossible to declare an array large enough on some systems.

The first problem is solved by using linked lists, attached to each array index. All duplicates for that bucket will be stored in the list. Another possible solution is to have a counter. As an example let us sort 3, 2, 4, 2, 3, 5. We start with an array of 5 counters set to zero.


  0   1   2   3   4   5
 0  0  0  0  0  0

Moving through the array we increment counters:


  0   1   2   3   4   5
 0  0  2  2  1  1

Next,we simply read off the number of each occurrence: 2 2 3 3 4 5.

O(n2) algorithms

Bubble Sort

The algorithm works by comparing each item in the list with the item next to it, and swapping them if required. In other words, the largest element has bubbled to the top of the array. The algorithm repeats this process until it makes a pass all the way through the list without swapping any items.

void bubbleSort(int ar[])
   for (int i = (ar.length - 1); i >= 0; i--)
      for (int j = 1; j ≤ i; j++)
         if (ar[j-1] > ar[j])
              int temp = ar[j-1];
              ar[j-1] = ar[j];
              ar[j] = temp;
   } } } }

Example. Here is one step of the algorithm. The largest element – 7 – is bubbled to the top:

7, 5, 2, 4, 3, 9
5, 7, 2, 4, 3, 9
5, 2, 7, 4, 3, 9
5, 2, 4, 7, 3, 9
5, 2, 4, 3, 7, 9
5, 2, 4, 3, 7, 9

The worst-case runtime complexity is O(n2). See explanation below

Selection Sort

The algorithm works by selecting the smallest unsorted item and then swapping it with the item in the next position to be filled.The selection sort works as follows: you look through the entire array for the smallest element, once you find it you swap it (the smallest element) with the first element of the array. Then you look for the smallest element in the remaining array (an array without the first element) and swap it with the second element. Then you look for the smallest element in the remaining array (an array without first and second elements) and swap it with the third element, and so on. Here is an example,

void selectionSort(int[] ar){
   for (int i = 0; i ‹ ar.length-1; i++)
      int min = i;
      for (int j = i+1; j ‹ ar.length; j++)
            if (ar[j] ‹ ar[min]) min = j;
      int temp = ar[i];
      ar[i] = ar[min];
      ar[min] = temp;
} }


29, 64, 73, 34, 20,
20, 64, 73, 34, 29,
20, 29, 7334, 64
20, 29, 34, 7364
20, 29, 34, 64, 73

The worst-case runtime complexity is O(n2).

Insertion Sort

To sort unordered list of elements, we remove its entries one at a time and then insert each of them into a sorted part (initially empty):

void insertionSort(int[] ar)
   for (int i=1; i ‹ ar.length; i++)
      int index = ar[i]; int j = i;
      while (j > 0 && ar[j-1] > index)
           ar[j] = ar[j-1];
      ar[j] = index;
} }

Example. We color a sorted part in green, and an unsorted part in black. Here is an insertion sort step by step. We take an element from unsorted part and compare it with elements in sorted part, moving form right to left.

29, 20, 73, 34, 64
29, 20, 73, 34, 64
20, 29, 73, 34, 64
20, 29, 73, 34, 64
20, 29, 34, 73, 64
20, 29, 34, 64, 73

Let us compute the worst-time complexity of the insertion sort. In sorting the most expensive part is a comparison of two elements. Surely that is a dominant factor in the running time. We will calculate the number of comparisons of an array of N elements:

we need 0 comparisons to insert the first element
we need 1 comparison to insert the second element
we need 2 comparisons to insert the third element

we need (N-1) comparisons (at most) to insert the last element


1 + 2 + 3 + … + (N-1) = O(n2)

The worst-case runtimecomplexity is O(n2).What is the best-case runtime complexity? O(n). The advantage of insertion sort comparing it to the previous two sorting algorithm is that insertion sort runs in linear time on nearly sorted data.

O(n log n) algorithms


Merge-sort is based on the divide-and-conquer paradigm. It involves the following three steps:

  • Divide the array into two (or more) subarrays
  • Sort each subarray (Conquer)
  • Merge them into one (in a smart way!)

Example. Consider the following array of numbers

27  10  12  25  34  16  15  31
      divide it into two parts
27  10  12  25            34  16  15  31
      divide each part into two parts
27  10            12  25            34  16            15  31
      divide each part into two parts
27       10       12       25       34       16       15       31


merge (cleverly-!) parts

10 27 12 25 16 34 15 31
      merge parts
10  12  25  27                 15  16  31  34
      merge parts into one
10  12  15  16  25  27  31  34

How do we merge two sorted subarrays? We define three references at the front of each array.

We keep picking the smallest element and move it to a temporary array, incrementing the corresponding indices.

Complexity of Mergesort

Suppose T(n) is the number of comparisons needed to sort an array of n elements by the MergeSort algorithm. By splitting an array in two parts we reduced a problem to sorting two parts but smaller sizes, namely n/2. Each part can be sort in T(n/2). Finally, on the last step we perform n-1 comparisons to merge these two parts in one. All together, we have the following equation

T(n) = 2*T(n/2) + n - 1

The solution to this equation is beyond the scope of this course. However I will give you a resoning using a binary tree. We visualize the mergesort dividing process as a tree



1 Comment

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