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Insertion Sort in Java: Iterative & Recursive Approach (with code)

  • May 29, 2023
  • 7 Minutes Read
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  • By Harshita Rajput
Insertion Sort in Java: Iterative & Recursive Approach (with code)

Sorting techniques help you arrange the given objects in a specific order. Today we are going to learn one of these sorting techniques, Insertion Sort in Java. While playing a game of cards with your friends, after the cards get distributed, you arrange them in order. How do you do that? Which sorting technique do you use? 

What is Insertion Sort?

Insertion sort is a comparison-based sorting algorithm that inserts each element at its correct position in a single journey or iteration. First, the sorted part contains only the first element, while the rest of them are in the sorted part. We then iterate through the unsorted by taking one element at a time and inserting it into its correct position in the sorted part.

This works in the same way as how we arrange the cards in the game. We can also understand from the example of boxes where you need to arrange boxes having different numbers written on them according to the numbers on them. We will use the insertion sort algorithm here.

We assume that the first box is already at its position (or sorted) and then we start from other boxes which aren't at their position according to the prescribed position. If the number on that box is greater than the no. on the box which is sorted it is placed on the right of that box otherwise left. This way, each unsorted box is compared with sorted boxes and put in its right position.

How does Insertion Sort work?

Given with an array say arr of size 4. We need to sort this array using the insertion sort algorithm. Remember that in each journey we put each element at its correct position. And we already assume that the first element i.e. element at the 0th index is already at its correct position.

insertion java 1

Journey 1: The element next to the last sorted element, i.e. element at index 1 is compared with each element in the sorted part, so as only 1 element exists in the sorted part, the Element at index1 will be compared to the Element at index0.

Pass 1:

insertion java 2

We store the element to be compared as Ele. Here Ele (or 10) < 20, therefore we swap 10 and 20. But as there is no other element except 20 in the sorted part, we stop. 

insertion java 3

And 2 elements have been sorted. Notice for each Ele or element in an unsorted array, at max no. of passes = no. of comparisons with Ele with sorted elements.

Journey 2:

Pass 1:

insertion java 4

As 90 > 20, implies that 90 is at its correct position and no more comparisons are done. Therefore no. of comparisons = 1.

insertion java 5

3 elements have been sorted after journey 3. We notice that in each journey we place an element at its correct position. Therefore journeys needed to put these 4 elements at the correct position is 3, i.e. no. of elements in array -1. We used -1 as we initially assumed 1 element to be sorted.

Journey 2:

Pass 1:

insertion java 6

The next element to be compared is 4. As 4 < 90, swapping happens.

Pass 2:

insertion java 7

 As 4 < 20, swapping happens.

Pass 3:

insertion java 8

As 4 < 10, swapping happens.

insertion java 9

Now all the journeys and no. of passes have been completed and we notice that we are left with the sorted array. Please note these 3 points every time:

  1. We assume 1st element to always be sorted.
  2. No. of Journeys = no. of elements - 1.
  3. No. of passes = Elements in sorted part.

Iterative Insertion Sort Algorithm in Java

Here is the algorithm for insertion sort with the iterative approach in ascending order:

  1. Run an outer for loop, say i, that keeps track of no. of journeys. It basically points to an element that has to be shifted to its correct position.
  2. The inner for loop, say j, starts from the last element of the sorted part i.e. i -1, and goes till the first element.
  3. If the element at i th index < element at j th index, we swap. Else break the loop.

Here is the code to implement iterative insertion sort in Java:

import java.util.*;
public class InsertionSort {
	public static void main(String[] args) {
		
		int [] arr = {20, 10, 90, 4};
		
		insertionSort(arr);
		System.out.println("Sorted Array is ->");
		for(int var:arr) {
			System.out.print(var + " ");
		}
	}
	
	public static int [] insertionSort(int [] arr) {
		
		// outer loop keeping a track of no. of journeys
		for (int i = 1; i < arr.length; i++) {
			System.out.println("Journey "+ i);
			
			// maximum no. of passes for each element
			for(int j = i-1;j >= 0; j--) {
				
				// to define the pass number
				int pass = i-j;
				System.out.println("Pass "+ pass);
				System.out.println("Comparing " +  arr[i]+ " and "+arr[j]);
				
				// if element is not at its correct position
				if(arr[j] > arr[j+1]) {
					swap(arr, j, j+1);
					
				}else {
					// the element is already at its correct position
					break;
				}
			}
			System.out.println();
		}
		return arr;
	}
	
	public static void swap(int [] arr, int i ,int j) {
		
		System.out.println("Swapping " +  arr[i]+ " and "+arr[j]);
		int temp = arr[i];
		arr[i] = arr[j];
		arr[j] = temp;
	}
}

 

Output:

Journey 1
Pass 1
Comparing 10 and 20
Swapping 20 and 10

Journey 2
Pass 1
Comparing 90 and 20

Journey 3
Pass 1
Comparing 4 and 90
Swapping 90 and 4
Pass 2
Comparing 90 and 20
Swapping 20 and 4
Pass 3
Comparing 90 and 10
Swapping 10 and 4

Sorted Array is ->
4 10 20 90 

 

Now, to arrange the array in descending order we only change the condition of comparing the elements for swapping. If the element at j is less than the element at j+1 swapping occurs. Here is the Java code to implement iterative insertion sort in descending order:

import java.util.*;
public class InsertionSort {
	public static void main(String[] args) {
		
		int [] arr = {20, 10, 90, 40, 30};
		
		insertionSort(arr);
		System.out.println("Sorted Array is ->");
		for(int var:arr) {
			System.out.print(var + " ");
		}
	}
	
	public static int [] insertionSort(int [] arr) {
		
		// outer loop keeping a track of no. of journeys
		for (int i = 1; i < arr.length; i++) {
			System.out.println("Journey "+ i);
			
			// maximum no. of passes for each element
			for(int j = i-1;j >= 0; j--) {
				
				// to define the pass number
				int pass = i-j;
				System.out.println("Pass "+ pass);
				System.out.println("Comparing " +  arr[i]+ " and "+arr[j]);
				
				// if element is not at its correct position
				if(arr[j] < arr[j+1]) {
					swap(arr, j, j+1);
					
				}else {
					// the element is already at its correct position
					break;
				}
			}
			System.out.println();
		}
		return arr;
	}
	
	public static void swap(int [] arr, int i ,int j) {
		
		System.out.println("Swapping " +  arr[i]+ " and "+arr[j]);
		int temp = arr[i];
		arr[i] = arr[j];
		arr[j] = temp;
	}
}

 

Output:

Journey 1
Pass 1
Comparing 10 and 20

Journey 2
Pass 1
Comparing 90 and 10
Swapping 10 and 90
Pass 2
Comparing 10 and 20
Swapping 20 and 90

Journey 3
Pass 1
Comparing 40 and 10
Swapping 10 and 40
Pass 2
Comparing 10 and 20
Swapping 20 and 40
Pass 3
Comparing 10 and 90

Journey 4
Pass 1
Comparing 30 and 10
Swapping 10 and 30
Pass 2
Comparing 10 and 20
Swapping 20 and 30
Pass 3
Comparing 10 and 40

Sorted Array is ->
90 40 30 20 10 

 

Time and Space Complexity

  • Worst Case Scenario: When the whole array is in reverse order, to sort the array or for iteration of N-1, the total no. of passes goes like 1 + 2+ 3, and so on. This forms an arithmetic progression that sums up to n(n+1)/2. Therefore the time complexity is O( N^2).
  • Best Case Scenario: When the complete array is sorted already, for the iteration of N-1, the no. of passes comes out to be 1 + 1+ 1 ---- N times. This Sums to N. Therefore time complexity is O(N).
  • Average Case scenario: When the array elements are in random order, it still takes O(N^2) time complexity.

As no extra space is being used, the space complexity is O(1).

Recursive Insertion Sort Algorithm in Java

Now, we will discuss the algorithm for insertion sort with the recursive approach in ascending order:

  1. We create a function called recursiveInsertionSort() that takes up the array and the index value of the element that is to be placed at its correct position or the value of i.
  2. We implement the for loop of journeys recursively. The inner for loop remains as it is.
  3. As we are assuming the element at the 0th index to be sorted, we pass the index value or i as 1.
  4. We call the function recursiveInsertionSort() by incrementing the value of i by 1 till i = arr.length.

Here is the code to implement recursive insertion sort in Java:

import java.util.*;
public class InsertionSort {
	public static void main(String[] args) {
		
		int [] arr = {20, 10, 90, 40,30};
		
		recursiveinsertionSort(arr, 1);
		System.out.println("Sorted Array is ->");
		for(int var:arr) {
			System.out.print(var + " ");
		}
	}
	
	
	public static void recursiveinsertionSort(int[] arr, int i) {
		
		// if i becomes equal to array's length, this implies all elements have been placed at their correct position
		if(i == arr.length) {
			return;
		}
		
		System.out.println("Journey "+ i);
		
		for(int j = i - 1; j >= 0; j--) {
			
			int pass = i-j;
			System.out.println("Pass "+ pass);
			System.out.println("Comparing " +  arr[i]+ " and "+arr[j]);
			
			// if element is not at its correct position
			if(arr[j] > arr[j+1]) {
				swap(arr, j, j+1);
				
			}else {
				
				// the element is already at its correct position
				break;
			}
		}
		System.out.println();
		recursiveinsertionSort(arr, i + 1);
		
	}
	public static void swap(int [] arr, int i ,int j) {
		
		System.out.println("Swapping " +  arr[i]+ " and "+arr[j]);
		int temp = arr[i];
		arr[i] = arr[j];
		arr[j] = temp;
	}
}

 

Output:

Journey 1
Pass 1
Comparing 10 and 20
Swapping 20 and 10

Journey 2
Pass 1
Comparing 90 and 20

Journey 3
Pass 1
Comparing 40 and 90
Swapping 90 and 40
Pass 2
Comparing 90 and 20

Journey 4
Pass 1
Comparing 30 and 90
Swapping 90 and 30
Pass 2
Comparing 90 and 40
Swapping 40 and 30
Pass 3
Comparing 90 and 20

Sorted Array is 
10 20 30 40 90 

 

Is the Insertion Sort in place and stable?

A stable sorting technique is one in which the relative order of elements is maintained. As swapping takes place in insertion sort, no relative order of elements is changed, hence it is a stable sorting algorithm.

An in-place algorithm is one in which the changes happen in the input array only and no additional memory is used except for the input array. As insertion sort uses no extra space, it is an in-place algorithm.

Advantages & Applications

Insertion sort is best used when a small no. of elements are there in an array or only a few elements are left to be sorted in the array. It is mostly used in computer applications that construct complex programs like data compression or path-finding.

It is an online sorting algorithm, which means it can sort a list as it receives elements one by one. This can be useful in scenarios where new data continuously arrives and needs to be sorted incrementally. Also in situations like when given an array that is nearly sorted, insertion sort performs exceptionally well in such cases.

However, it becomes inefficient when the items in the input sequence are greater than 100 as it takes too much of time.

Conclusion

Despite having a time complexity of O( N^2), the algorithm sorts in place and takes no extra space as seen in the case of merge sort. Both bubble sort and insertion sort have the same time complexity but still insertion is better as it doesn't execute too many swaps as in bubble sort. Insertion sort in Java follows an incremental approach and sorts the array in an efficient way.

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About The Author
Harshita Rajput
I am a highly motivated individual with a deep passion for technology and programming. My exceptional performance in programming contests, and my dedication to solving data structures and algorithmic problems on global platforms, demonstrate my commitment to the field of technology.