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Insertion Sort

Estimated reading time: 3 mins

Question: Suppose you have an algorithm that you are running that has multiple steps. When you are approximating, when do you add the runtimes and when do you multiply them?

  • if do this, then when you are done with this, do that, then you add runtimes
  • if do this for each time you do that, then you multiply runtimes

The Insertion-Sort Algorithm

The goal of this algorithm is simple: take an array of elements and sort it in an increasing order. It does so by inserting one element at a time and by comparing elements during insertion.

Algorithm InsertionSort(A):

Input: An array A of n comparable elements

Output: The array A with elements rearranged in nondecreasing order

We basically walk one element at a time, from 0 to n:

for k from 1 to n do

Insert element A[k] at its proper location in the array at position A[0], A[1], A[2], …, A[k]

or as you would like to call it — increasing order.

Question: When will this algorithm behave the best?

Answer: When the elements in the array are sorted in order.

Question: When will this algorithm have the worst of times?

Answer: When the elements in the array are placed in the reversed order (from high to low - a decreasing order). Our little guy will have to swap every element and move it to the other side.

Guided with the following from textbook

using an outer loop to consider each element in turn, and an inner loop that moves a newly considered element to its proper location relative to the (sorted) subarray of elements that are to its left

I went on to do my own implementation — because why not and I could use some review.

public static char[] insertionSort(char[] input) {
        if (input.length == 0) {
            return input;

        Assumption: position 0 is sorted by definition

        Start with position 1
            Grab the value at the position
                Question: is the value at previous position 0 (n-1) bigger than current value
                    if yes
                        do a loop
                            place previous value that is bigger in the current place
                    if no
                        do no loop

               Place the current value in the position based on the answer to the question

        for (int i = 1; i < input.length; i++) {
            char current = input[i];
            int insertPosition = i;

                We are walking from the current position to the front of the array at position 0

            for (; insertPosition > 0; ) {
                if (input[insertPosition - 1] > current) {
                    input[insertPosition] = input[insertPosition - 1];
                } else {

            input[insertPosition] = current;

        return input;

See you tomorrow.

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