You can sort of follow this GIFFY

https://upload.wikimedia.org/wikipedia/commons/0/0f/Insertion-sort-example-300px.gif

https://github.com/shivasurya/cc150/blob/master/sorting/insertions%20sort/C%20lang/insertion.c

check this with algorithm description

agreed, it's a shame that it assumes some iterations that print the array even if nothing was moved, i.e.:

Input
6
1 4 3 5 6 2

Expected Output

1 4 3 5 6 2
1 3 4 5 6 2
1 3 4 5 6 2
1 3 4 5 6 2
1 2 3 4 5 6

And solutions like:

void insertion_sort(vector<int> values)
{
    print_values(values);
    
    while(true)
    {
        auto what = std::adjacent_find(values.begin(), values.end(), std::greater<int>());
        if(what == values.end())
        {
            return;
        }

        auto last = what+1;
        auto v = *last;
        auto tmp = *last;
        
        auto where = std::lower_bound(values.begin(), what, v);
        for(auto x = last; x != where; --x)
        {
            auto prev = x-1;
            *x = *prev;
        }
        *where = tmp;
        print_values(values);
    }
}

that are insertion sort, would print it only when it actually changed, i.e.:

Your Output (stdout)

1 4 3 5 6 2 
1 3 4 5 6 2 
1 2 3 4 5 6 

I agreed, why do we have to print out in that weird way???

The problem seems to be designed in a very bad way.But after finally solving it you kindof get the idea what insertion sort does at every step.

loved the usage of *a. Learnt something new instead of using join()

I am glad you learned something new..Do let me know if i can help you with something.

Insertion swapping

I did about the same, but using a sentinel at the front of the array:

n, amin = int(input()), -10**4
A = [2*amin] + list(map(int, input().split()))
for j in range(2, n+1):
    i, a = j, A[j]
    while a < A[i-1]:
        A[i] = A[i-1]
        i -= 1
    A[i] = a
    print(*A[1:])

def insertion_sort(ar):

for i in range(1, len(ar)):

    for j in range(i):
        if (ar[i]<ar[j]):
            k = ar[i]
            ar[i] = ar[j]
            ar[j] = k
    print(*ar)

Thank you. that's exactly what i need to fix my submission. my solution was string based sorting.

arr.map(Number) works just as well. Still, this was my bug so good spot.

exactly this, saved me some frustration, thank you!

arr.map(Number) or arr.map(x => +x) are cuter though :)

Life saver.

Hey, converted your answer to JS, but it will only pass first 2 test, thought?

for (var i = 1; i < array.length; i++) {
  var j = i;
  var value = array[i];
  while (j >= 1 && (array[j-1] > value)) {
    array[j] = array[j-1];
    j--;
  }
  array[j] = value;
  console.log(array.join(' '));
}

Hi, I've written a similar code but it failed 3/5 test cases reason being error with ArrayOutOfBound exception. Can you please tell me what wrong did I do? Thanks.

for(int j=1;j<ar.length;j++){
                int temp=ar[j];
                int k = j-1;
                while(ar[k]>temp&&k>=0)
                {
                    ar[k+1]=ar[k];
                    k--;
                }
                ar[k+1]=temp;
                printArray(ar);
            }

is there any need to use "value"..? Can't we simply write array[j-1]>array[i] ..?

@briancmpbll did you get the answer for this?