Solve almostIncreasingSequence (Codefights)

Your algorithm is much too simplistic. You have a right idea, checking consecutive pairs of elements that the earlier element is less than the later element, but more is required.

Make a routine first_bad_pair(sequence) that checks the list that all pairs of elements are in order. If so, return the value -1. Otherwise, return the index of the earlier element: this will be a value from 0 to n-2. Then one algorithm that would work is to check the original list. If it works, fine, but if not try deleting the earlier or later offending elements. If either of those work, fine, otherwise not fine.

I can think of other algorithms but this one seems the most straightforward. If you do not like the up-to-two temporary lists that are made by combining two slices of the original list, the equivalent could be done with comparisons in the original list using more if statements.

Here is Python code that passes all the tests you show.

def first_bad_pair(sequence):
    """Return the first index of a pair of elements where the earlier
    element is not less than the later elements. If no such pair
    exists, return -1."""
    for i in range(len(sequence)-1):
        if sequence[i] >= sequence[i+1]:
            return i
    return -1

def almostIncreasingSequence(sequence):
    """Return whether it is possible to obtain a strictly increasing
    sequence by removing no more than one element from the array."""
    j = first_bad_pair(sequence)
    if j == -1:
        return True  # List is increasing
    if first_bad_pair(sequence[j-1:j] + sequence[j+1:]) == -1:
        return True  # Deleting earlier element makes increasing
    if first_bad_pair(sequence[j:j+1] + sequence[j+2:]) == -1:
        return True  # Deleting later element makes increasing
    return False  # Deleting either does not make increasing

If you do want to avoid those temporary lists, here is other code that has a more complicated pair-checking routine.

def first_bad_pair(sequence, k):
    """Return the first index of a pair of elements in sequence[]
    for indices k-1, k+1, k+2, k+3, ... where the earlier element is
    not less than the later element. If no such pair exists, return -1."""
    if 0 < k < len(sequence) - 1:
        if sequence[k-1] >= sequence[k+1]:
            return k-1
    for i in range(k+1, len(sequence)-1):
        if sequence[i] >= sequence[i+1]:
            return i
    return -1

def almostIncreasingSequence(sequence):
    """Return whether it is possible to obtain a strictly increasing
    sequence by removing no more than one element from the array."""
    j = first_bad_pair(sequence, -1)
    if j == -1:
        return True  # List is increasing
    if first_bad_pair(sequence, j) == -1:
        return True  # Deleting earlier element makes increasing
    if first_bad_pair(sequence, j+1) == -1:
        return True  # Deleting later element makes increasing
    return False  # Deleting either does not make increasing

And here are the tests I used.

print('\nThese should be True.')
print(almostIncreasingSequence([]))
print(almostIncreasingSequence([1]))
print(almostIncreasingSequence([1, 2]))
print(almostIncreasingSequence([1, 2, 3]))
print(almostIncreasingSequence([1, 3, 2]))
print(almostIncreasingSequence([10, 1, 2, 3, 4, 5]))
print(almostIncreasingSequence([0, -2, 5, 6]))
print(almostIncreasingSequence([1, 1]))
print(almostIncreasingSequence([1, 2, 3, 4, 3, 6]))
print(almostIncreasingSequence([1, 2, 3, 4, 99, 5, 6]))
print(almostIncreasingSequence([1, 2, 2, 3]))

print('\nThese should be False.')
print(almostIncreasingSequence([1, 3, 2, 1]))
print(almostIncreasingSequence([3, 2, 1]))
print(almostIncreasingSequence([1, 1, 1]))

The solution is close to the intuitive one, where you check if the current item in the sequence is greater than the current maximum value (which by definition is the previous item in a strictly increasing sequence).

The wrinkle is that in some scenarios you should remove the current item that violates the above, whilst in other scenarios you should remove previous larger item.

For example consider the following:

[1, 2, 5, 4, 6]

You check the sequence at item with value 4 and find it breaks the increasing sequence rule. In this example, it is obvious you should remove the previous item 5, and it is important to consider why. The reason why is that the value 4 is greater than the "previous" maximum (the maximum value before 5, which in this example is 2), hence the 5 is the outlier and should be removed.

Next consider the following:

[1, 4, 5, 2, 6]

You check the sequence at item with value 2 and find it breaks the increasing sequence rule. In this example, 2 is not greater than the "previous" maximum of 4 hence 2 is the outlier and should be removed.

Now you might argue that the net effect of each scenario described above is the same - one item is removed from the sequence, which we can track with a counter.
The important distinction however is how you update the maximum and previous_maximum values:

• For [1, 2, 5, 4, 6], because 5 is the outlier, 4 should become the new maximum.

• For [1, 4, 5, 2, 6], because 2 is the outlier, 5 should remain as the maximum.

This distinction is critical in evaluating further items in the sequence, ensuring we correctly ignore the previous outlier.

Here is the solution based upon the above description (O(n) complexity and O(1) space):

def almostIncreasingSequence(sequence):
    removed = 0
    previous_maximum = maximum = float('-infinity')
    for s in sequence:
        if s > maximum:
            # All good
            previous_maximum = maximum
            maximum = s
        elif s > previous_maximum:
            # Violation - remove current maximum outlier
            removed += 1
            maximum = s
        else:
            # Violation - remove current item outlier
            removed += 1
        if removed > 1:
            return False
    return True

We initially set the maximum and previous_maximum to -infinity and define a counter removed with value 0.

The first test case is the "passing" case and simply updates the maximum and previous_maximum values.

The second test case is triggered when s <= maximum and checks if s > previous_maximum - if this is true, then the previous maximum value is the outlier and is removed, with s being updated to the new maximum and the removed counter incremented.

The third test case is triggered when s <= maximum and s <= previous_maximum - in this case, s is the outlier, so s is removed (no changes to maximum and previous_maximum) and the removed counter incremented.

One edge case to consider is the following:

[10, 1, 2, 3, 4]

For this case, the first item is the outlier, but we only know this once we examine the second item (1). At this point, maximum is 10 whilst previous_maximum is -infinity, so 10 (or any sequence where the first item is larger than the second item) will be correctly identified as the outlier.