@mvouzikis thanks for your contribution, the comments look good. However, there is a bug in the existing/original code. Since you have it fresh - would you like to fix it?

As far as I see, the current code simply returns the maximum of the input array (the test cases are wrong!). Consider the input {1, 1, 5, 1}. The sub arrays are:

{1} -> min: 1,
{1, 5} -> min: 1,
{1, 5, 1} -> min: 1,

So the result should be max{1, 1, 1} = 1.

@siriak, @BamaCharanChhandogi: please have a look

Isn't it always equal to the first element? If not, please provide a counterexample.

I probably don't understand the problem statement.

If no progress is made, I can check it tomorrow run some examples and see where the bug lies

@mvouzikis thanks for your contribution, the comments look good. However, there is a bug in the existing/original code. Since you have it fresh - would you like to fix it?

As far as I see, the current code simply returns the maximum of the input array (the test cases are wrong!). Consider the input {1, 1, 5, 1}. The sub arrays are:

{1} -> min: 1,
{1, 5} -> min: 1,
{1, 5, 1} -> min: 1,

So the result should be max{1, 1, 1} = 1.

@siriak, @BamaCharanChhandogi: please have a look

The code is correct.
{1,1,5,1}

[1] -> 1
[1, 1] -> 1
[1, 1, 5]  -> 1
[1, 1, 5, 1]  -> 1
[1, 1, 5, 1]  -> 1
[1, 5]  -> 1
[1, 5, 1] -> 1
[5] -> 5
[5, 1] -> 1
[1] -> 1

Output will be: 5(max)

I think this method is finding the maximum of minima of all subarrays (of any length). In this case it is simply the max (because it is enough to consider arrays of length 1), so @BamaCharanChhandogi is correct.

@siriak: if this problem would only consider arrays of the type (python notation) arr[0:1], arr[0:2], arr[0:3], ..., then the result will be always arr[0] (because min(arr[0:n]) >= min(arr[0:n], arr[n]) = min(arr[0:n+1])).

I think that the real problem should be finding maximum of minima of sub arrays of given length.

I suggest to remove current implementation and create a new PR with a correct one. @mvouzikis would you like to do that?

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