创建一个严格递增的序列减去 1 个元素

Question

I am working through a code challenge where I am to check if a list is strictly increasing if I remove no more than 1 element from the list. Here are some input/output examples:

For sequence = [1, 3, 2, 1], the output should be
almostIncreasingSequence(sequence) = false.

There is no one element in this array that can be removed in order to get a strictly increasing sequence.

For sequence = [1, 3, 2], the output should be
almostIncreasingSequence(sequence) = true.

This is the logic I've built. Basically, I've created a copy of list and I'm doing the testing on the copy. I remove one element in each iteration and if it is strictly increasing, I increment the counter by a value of 1 but if it isn't, I decrement it by one. At the end, I check if the counter is greater than or equal to 0 and return a boolean value based on that:

def almostIncreasingSequence(sequence):
    
    check = 0
    sequence_copy = sequence
    
    for element in sequence:
        sequence_copy.remove(element)
        
        if all(i < j for i, j in zip(sequence_copy, sequence_copy[1:])):
            check += 1
        else:
            check -= 1
        
        sequence_copy = sequence  
    
    if check > 0:
        return True
    else:
        return False

However, there are 2 issues with this code. First of all, the logic does not work on some test cases like this one:

Input: sequence: [1, 3, 2]
Output: false
Expected Output: true

And the other issue is that the code takes too long for larger sequences. How can I fix my existing code to make it faster and generic for every kind of input? While I appreciate all help, just posting an answer does not help me at all since this is a code challenge meant for learning. Telling me how to optimize my code helps far more than just an answer.

Answer 1

From the start, find the strictly increasing prefix. From the end find the strictly increasing suffix (strictly decreasing from the end).

If they overlap, return true because the whole sequence is strictly increasing.

If there are any elements between them, then return false.

Otherwise, the prefix and suffix are adjacent. Return true if you can connect them into an increasing sequence by deleting the last element in the prefix, or the first element in the suffix.

Answer 2

Try to reduce the number of times you perform an operation that requires you to iterate over the array. Here's one way where you only need one loop through the array.

def isStrictlyIncreasing(sequence):
    for elem1, elem2 in zip(sequence[:-1], sequence[1:]):
        if elem1 >= elem2:
            return False
    return True

def almostIncreasingSequence(sequence):
    seq_copy1 = sequence[:]
    seq_copy2 = sequence[:]
    for i, (elem1, elem2) in enumerate(zip(sequence[:-1], sequence[1:])):
        if elem1 >= elem2:
            seq_copy1.pop(i)
            seq_copy2.pop(i+1)
            return isStrictlyIncreasing(seq_copy1) or isStrictlyIncreasing(seq_copy2)
    return True

Explanation:

1. isStrictlyIncreasing() checks that a sequence is strictly increasing. Pretty self explanatory

2. almostIncreasingSequence() makes two copies of the sequence.

   1. When it finds that the sequence is not increasing, it removes both the offending elements, one from each copy.
   2. If either of these modified lists is strictly increasing, the original is almost strictly increasing. Otherwise it isn't.
   3. If we reach the end of the function, we never found a non-increasing pair, so the list is strictly increasing, so return True .

Answer 3

To solve this problem in linear time we have to first pre-process the given array.

Suppose the array is [1,2,3,4,5,2,6,7] , now we will make two additional arrays.
increasingPrefix which will be a boolean array, in this array i-th element is True iff the prefix till i is an strictly increasing sequence.
Similarly, we will make another array increasingSuffix which will also be a boolean array, here i-th element is True iff suffix till i is a strictly increasing sequence.

So

array = [1, 2, 3, 4, 5, 2, 6, 7]

increasingPrefix = [True, True, True, True, True, False, False, False]

increasingSuffix = [False, False, False, False, False, True, True, True]

Now to check if the given array can become strictly increasing by removing no more than 1 element,
we can simply traverse the given array and for each i-th element we will check if
increasingPrefix[i-1]==True and increasingSuffix[i+1]==True and array[i+1]>array[i-1] .
If above condition holds True in any i-th element than ans is True because we can remove i-th element and the remaining array will be strictly increasing.

Code for the above idea

def fun(arr):
    n=len(arr)
    
    # empty and 1 length sequences are strictly increasing
    if n<=1:
        return True
    increasingPrefix=[False]*n
    increasingSuffix=[False]*n
    increasingPrefix[0]=True
    increasingSuffix[n-1]=True

    # Calculate increasingPrefix values as described above
    for i in range(1,n):
        if arr[i]>arr[i-1]:
            increasingPrefix[i]=True
        else:
            for j in range(i,n):
                increasingPrefix[j]=False
            break

    # calculated increasingSuffix values as described above
    for i in range(n-2,-1,-1):
        if arr[i]<arr[i+1]:
            increasingSuffix[i]=True
        else:
            for j in range(i,-1,-1):
                increasingSuffix[j]=False
            break

    # traverse the given array and check for the conditions
    ans=False
    for i in range(n):
        if i==0:
            if increasingSuffix[i+1]:
                ans=True
                break
        elif i==n-1:
            if increasingPrefix[i-1]:
                ans=True
                break
        else:
            if increasingPrefix[i-1] and increasingSuffix[i+1] and arr[i+1]>arr[i-1]:
                ans=True
                break
 return ans