How To Find Insertion Point Using Binary Search

Software engineering interviews revolve around tweaking well-known algorithms and then that they can be used to solve complex coding problems. Binary insertion sort is one such topic that involves tweaking the well-known insertion sort algorithm. This commodity is dedicated to analyzing this. We'll encompass:

• What Is Insertion Sort?
• What is Binary Insertion Sort?
• Applications of Binary Insertion Sort
• How Does Binary Insertion Sort Piece of work?
• Binary Insertion Sort Algorithm
• Binary Insertion Sort Pseudocode
• Binary Insertion Sort Code
• Binary Insertion Sort Complexities
• Strengths and Weaknesses of Binary Insertion Sort
• Insertion Sort vs. Binary Insertion Sort
• FAQs on Binary Insertion Sort

What Is Insertion Sort?

In a simple insertion sort algorithm, we maintain a sorted and an unsorted subarray of the given array. In each iteration, one value from the unsorted part is picked and is inserted in its correct position in the sorted function. To achieve this, for every element, nosotros iterate over the sorted part of the array to find the position to insert the element.

And so, information technology takes O(N) comparisons and O(N) swaps for inserting one chemical element in the worst case (last element). How tin we optimize it?

For a detailed explanation of the insertion sort algorithm, check out "Learn the Insertion Sort Algorithm."

What Is Binary Insertion Sort?

Binary insertion sort is a sorting algorithm similar to insertion sort, but instead of using linear search to find the position where the element should be inserted, nosotros use binary search. Thus, we reduce the number of comparisons for inserting one element from O(N) to O(log Northward).

It is an adaptive algorithm, which means that it works faster when the given array is already substantially sorted, i.east., the electric current position of the element is near its actual position in the sorted assortment.

It is a stable sorting algorithm — the elements with the same values appear in the same order in the final array every bit they were in the initial array.

Applications of Binary Insertion Sort

Binary insertion sort works efficiently when the array has a depression number of elements.

While performing quicksort or merge sort, when the subarray'south size becomes small (say <=25 elements), it's more efficient to use binary insertion sort.

This algorithm is also efficient when the cost of comparison between keys is sufficiently high. For case, if nosotros want to sort an assortment of strings, the comparison operation of ii strings will be high.

How Does Binary Insertion Sort Work?

In binary insertion sort, we dissever the assortment into two subarrays — sorted and unsorted. The starting time element of the array is in the sorted subarray, and the rest of the elements are in the unsorted one.

We and so iterate from the 2d chemical element to the last element. For the i-th iteration, we make the current element our "key." This fundamental is the chemical element that we take to add to our existing sorted subarray.

To do this, nosotros first utilise binary search on the sorted subarray to find the position of the element that is just greater than our cardinal. Let's call this position "pos." We and then right shift all elements from position pos to i-1 and then make Array[pos] = key.

We can notation that for every i-thursday iteration, the left office of the array till (i-1) is ever sorted.

Binary Insertion Sort Example

Suppose we need to sort the post-obit array:

1. We assume that the first chemical element is already sorted.
2. We take the 2nd element and store it in a variable (primal).
3. Now, nosotros use binary search to find the chemical element on the left of the current element, which is merely greater than it.
4. In this case, nosotros have only ane element, viii, and it is greater than vi. So, we shift viii 1 alphabetize towards the correct and place 6 at its position.

The array now looks like this:

5. Now, nosotros accept the 3rd element, 1. Note that all the elements earlier the current element are sorted.
6. We store 1 in key and find the element only greater than 1 in the sorted office using binary search.
7. Hither, the required element is 6. So, we shift half dozen and 8 one index towards the right and identify 1 at the position of 6 before shifting.

The array now looks like this:

8. We at present take the 4th element, 5, and shop information technology in central.
9. Using binary search, we find the element only greater than 5 in the sorted office. In this case, the required chemical element is half-dozen.
10. Over again, we shift 6 and 8 1 alphabetize towards the correct and identify 5 at the position of 6 before shifting.

The assortment now looks like this:

11. Nosotros now take the last (5th) element, which is 3, and find the element just greater than it in the sorted part.
12. The required element is five. Nosotros shift 5, 6, and 8 ane index towards the right and place 3 at the position of 5 before shifting.

The resulting array is:

Nosotros have sorted the given array using binary insertion sort.

Binary Insertion Sort Algorithm

Binary insertion sort for array A:

• Step ane: Iterate the array from the 2nd element to the concluding element.
• Step 2: Store the electric current element A[i] in a variable key.
• Step 3: Find the position of the element but greater than A[i] in the subarray from A[0] to A[i-1] using binary search. Say this element is at index pos.
• Step 4: Shift all the elements from index pos to i-1 towards the right.
• Step 5: A[pos] = key.

Binary Insertion Sort Pseudocode

                          procedure binarySearch(Assortment, N, key)     L = 0     R = N     while L < R:         mid = (L + R)/2         if Assortment[mid] <= central:             Fifty = mid + i         else:             R = mid     return L end procedure   procedure binaryInsertionSort(Array)     for i = 1 to length(Array) do:         key = Array[i]         pos = binarySearch(Array, key, 0, i-1)         j = i         while j > pos             Assortment[j] = Array[j-1]             j = j-1         Assortment[pos] = key     end for finish procedure                      

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Binary Insertion Sort Lawmaking

                          # This role will return the index of element  # just greater than 'key' in Array from 0-N def binarySearch(Array, N, key):     L = 0     R = Due north     while(Fifty < R):         mid = (L + R)//two         if (Array[mid] <= primal):             50 = mid + one         else:             R = mid     render L      def binaryInsertionSort(Array):     # We assume the 1st element of Array to be already sorted.     # Now we offset iterating from the 2nd element to the last element.     for i in range (one,len(Array)):         central = Array[i]         pos = binarySearch(Array, i, key)         # 'pos' will at present incorporate the alphabetize where 'fundamental' should be inserted.         j = i         # Shifting every chemical element from 'pos' to 'i' towards right.         while(j > pos):             Assortment[j] = Assortment[j-one]             j = j-1         # Inserting 'primal' in its correct position.         Array[pos] = key         print("The assortment later on",i,"iterations =", *Array)   Assortment = [8, half dozen, one, 5, 3] binaryInsertionSort(Array)                      

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Output

The assortment after ane iterations = six eight 1 5 3

The assortment after 2 iterations = 1 half dozen viii 5 iii

The assortment after iii iterations = i 5 6 8 3

The assortment after 4 iterations = ane 3 five 6 8

Binary Insertion Sort Complexities

Fourth dimension Complexity Analysis

Worst Example

For inserting the i-th element in its correct position in the sorted, finding the position (pos) will take O(log i) steps. However, to insert the element, we need to shift all the elements from pos to i-1. This will take i steps in the worst case (when we have to insert at the starting position).

Nosotros brand a total of N insertions —  so, the worst-case time complexity of binary insertion sort is O(N^2).

This occurs when the assortment is initially sorted in descending order.

All-time Instance

The best case volition exist when the element is already in its sorted position. In this example, we don't have to shift any of the elements; nosotros can insert the element in O(1).

But we are using binary search to notice the position where we demand to insert. If the element is already in its sorted position, binary search will take (log i) steps. Thus, for the i-th element, nosotros make (log i) operations, and so its all-time-case time complexity is Ω(Northward log Due north).

This occurs when the assortment is initially sorted in ascending order.

Average Case

For average-case time complication, we assume that the elements of the array are jumbled. Thus, on average, we will need O(i /ii) steps for inserting the i-th element, and then the average fourth dimension complexity of binary insertion sort is θ(N^two).

Infinite Complication Assay

Binary insertion sort is an in-identify sorting algorithm. This ways that it only requires a constant corporeality of boosted space. Nosotros sort the given array by shifting and inserting the elements.

Therefore, the space complexity of this algorithm is O(1) if we utilise iterative binary search. Information technology will be O(logN) if we utilise recursive binary search because of O(log N) recursive calls.

Strengths and Weaknesses of Binary Insertion Sort

Binary insertion sort works efficiently for smaller arrays (<= 25 elements). This algorithm also works well for most-sorted arrays, where the elements are nigh their position in the sorted array.

However, when the size of the array is large, the binary insertion sort doesn't perform well. We can use other sorting algorithms like merge sort or quicksort in such cases.

Making fewer comparisons is also ane of the strengths of this sorting algorithm; therefore, information technology is efficient to use it when the cost of comparing is high.

Insertion Sort vs. Binary Insertion Sort

As stated earlier, binary insertion sort is an improvement of insertion sort. We reduce the number of comparisons in insertion sort by using binary search instead of linear search.

Note: Both the algorithms are in place and use O(1) space complexity. All the same, if we use recursive binary search in binary insertion sort, its infinite complexity volition become O(log Northward) due to O(log Due north) recursive calls.

FAQs on Binary Insertion Sort

Question 1. Is Binary insertion sort a stable sorting algorithm?

Answer: Yeah, binary insertion sort is a stable sorting algorithm. This means that two different elements with the same value will appear in the same society in the terminal sorted array equally they appeared in the initial assortment.

In each iteration, we find the element'southward position just greater than our current element in the sorted subarray and insert information technology in that location. Thus, if there were any other elements with the same value before our current chemical element in the initial array, they will be present before information technology in the last sorted array.

Question ii. Which sorting algorithm betwixt binary insertion sort and bubble sort uses fewer swaps?

Respond: Both binary insertion sort and bubble sort apply the same number of swaps. For an element at index "i" in the initial array, if its position in the sorted array is "j," both the algorithms will have abs(i-j) swaps to place it in its sorted position. The total number of swaps used in both the algorithms is equal to the inversion count of the array.

Question iii. How many maximum comparisons volition be made in binary insertion sort in ane iteration?

Respond: O(Log Due north). This will happen when we are in the n-th iteration, and the position where the current element should be inserted is such that the binary search takes O(log N) steps.

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