Selection Sort in Java - Stack Abuse

# Selection Sort in Java

### Introduction

Sorting data is a frequent problem in computer science. Given a collection of elements, the goal is to rearrange them in some order. Common examples are sorting an array alphabetically or from smallest to largest.

Sorted data is a lot easier to manipulate. Finding the largest or smallest element of an array can be done in constant time if the array is sorted. Searching for an element is a lot faster using algorithms such as Binary Search which rely on the assumption that the array is already sorted.

One of the simplest algorithms for sorting data is Selection Sort. It's usually taught in beginner programming classes and tutorials to explain the concept of sorting, so we'll keep this article very beginner-friendly.

### Selection Sort

Selection sort is an in-place comparison sorting algorithm that uses brute force to sort an array.

In-place means that the algorithm uses a small constant amount of space for extra storage.

It's called a "brute force" algorithm because it uses the simplest and most ineffective way of calculating the solution. However, it does makes up for it with its straightforward implementation.

The algorithm divides the array into two subarrays:

• A sorted subarray
• An unsorted subarray

The sorted subarray is empty in the beginning. In every iteration, the smallest element of the unsorted array will be appended to the end of the sorted array by swapping. This way, the sorted array will eventually contain all the elements of the original array.

An example array we want to sort in ascending order:

 Sorted array Unsorted array Minimal element of the unsorted array [] [16, 5, 30, 6, 2, 7] 2 [2] [16, 5, 20, 6, 7] 5 [2, 5] [16, 20, 6, 7] 6 [2, 5, 6] [16, 7, 20] 7 [2, 5, 6, 7] [16, 20] 16 [2, 5, 6, 7, 16] [20] 20 [2, 5, 6, 7, 16, 20] []

### Implementation

The selectionSort() method takes just one argument, the array that needs to be sorted. We'll iterate trough the unsorted array, which will be between indexes i and j, find it's minimum and place it into the sorted array by swapping:

public static void selectionSort(int[] nums) {
for (int i = 0; i < nums.length; i++) {
// min is the index of the smallest element with an index greater or equal to i
int min = i;
for (int j = i + 1; j < nums.length; j++) {
if (nums[j] < nums[min]) {
min = j;
}
}
// Swapping i-th and min-th elements
int swap = nums[i];
nums[i] = nums[min];
nums[min] = swap;
}
}


Lets test out the code:

int[] array = new int[]{16, 5, 30, 6, 7, 2};
selectionSort(array);
System.out.println(Arrays.toString(array));


This will print out:

[2, 5, 6, 7, 16, 30]


### Selection Sort Time Complexity

Time complexity is a way to describe how much time an algorithm needs to finish executing relative to the size of the input. Analyzing the time it takes for an algorithm to give output is of crucial importance. Imagine a telephone book application that would take a day to sort all the numbers after a new number was added. That would be way less useful than the same app that would do it almost instantly.

Performance depends on the hardware as well as software, but the same program can be run on many different types of hardware. The Big-O notation makes it easier to approximate the time needed for a program to execute, regardless of software.

The average and worst-case time complexity of Selection Sort is O(n2). This makes Selection Sort a lot slower than many other comparison sorting algorithms like Merge Sort or Insertion Sort which have the worst-case time complexity (O(nlogn)). Interestingly, O(nlogn) is the best that can be achieved by any comparison sorting algorithm.

### Time Complexity Analysis

Showing that Selection Sort has quadratic time complexity comes down to calculating the number of times the inner loop will be iterated. We can see this if we go through the code line by line and try to approximate the time it takes to execute each line of code:

for (int i = 0; i < nums.length; i++) {


Everything in the inner block of the loop will be executed n times, where n is the length of a given array:

int min = i;


min will be initialized to i exactly n times. Now comes the tricky part:

for (int j = i + 1; j < nums.length; j++)


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Since this loop is nested, it takes a bit of math to calculate the number of times the block of code inside it will execute. Let's work it out.

When i is equal to 0, j will go from 1 to n, meaning every instruction in the inner block will execute n times. When i increases to 1, j will stay between 2 and n, which implies the inner block will execute n-2 times. Summing this up:

(n - 1) + (n - 2) + ... + 1


The sum of a sequence of natural numbers is calculated using something called Gauss's trick, and it results in (n2 - n)/2. Simplifying this, results in O(n2) time complexity.

Put simply, when calculating the complexity of an algorithm O(f(n)), we need to look for the highest power of n in the function f(n) and isolate it. This is because any part of the equation that has a lower power will not affect the result in any significant way.

For example, we have the function f(x) = x2+13x+23

O(f(x)) would be the highest power of x in the equation, which in this case is x2.

Here's how it performed after sorting an array containing 10,000 integers in random order:

public static void main(String[] args) {
int[] array = new int[10000];
for (int i = 0; i < array.length; i++) {
array[i] = i;
}

// Shuffle array
Collections.shuffle(Arrays.asList(array));

// Print shuffled collection
System.out.println(Arrays.toString(array));

long startTime = System.nanoTime();
selectionSort(array);
long endTime = System.nanoTime();

// Print sorted collection
System.out.println(Arrays.toString(array));

// Print runtime in seconds
System.out.println("Selection Sort runtime: " + (endTime - startTime)/1000000000);
}


Running it 10 times, this code produced the following results:

 Time(s) Selection Sort First Run 0.024 Second Run 0.020 Third Run 0.022 Fourth Run 0.020 Fifth Run 0.025 Sixth Run 0.022 Seventh Run 0.021 Eight Run 0.031 Ninth Run 0.022 Tenth Run 0.029

The average run time was 0.0236 seconds, though, this will majorly depend on your machine as well.

### Selection Sort Space Complexity

Space Complexity is also a big factor in algorithm design. Our programs are bound, not only by the time that they need to execute but also by memory usage. There is a limited amount of memory on any computer, so a programmer should keep an eye on that too.

The space complexity of Selection Sort is constant(O(1)) because it is in-place, which is great. Worst case complexity of Selection Sort is, unfortunately, O(n2) as well, which means that even if the algorithm gets an already sorted array as input, it will still take a lot of time to return the unchanged array.

This algorithm has decent performance if the collection doesn't have a lot of elements. If the array has ~10 elements, the difference in performance between different sorting algorithms shouldn't be that noticeable, and Selection Sort might even outperform other divide-and-conquer algorithms.

Where Selection Sort shines, is when the number of swaps needs to be minimal. In the worst case, there will only be n-1 swaps, which is the minimal possible number of swaps that need to be performed. This is quite intuitive if consider that every element will be placed in its right spot in the sorted array right away.

### Conclusion

Selection Sort is a brute force in-place comparison sort which continuously finds the minimum of an unsorted subarray and places it in the correct position in the sorted subarray. Due to its simplicity, it's often one of the first algorithms that are taught in computer science courses all around the world.

Even if more efficient algorithms come built-it, it's still important to understand the underlying logic and complexity analysis to avoid common issues and to make sure that the tool being used is the one that's best suited for the job at hand.

Last Updated: November 23rd, 2020

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Kristina PopovicAuthor

CS student with a passion for juggling and math.

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