Given a sequence of integers as an array, determine whether it is possible to obtain a strictly increasing sequence by removing no more than one element from the array.

Example

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.

You can remove 3 from the array to get the strictly increasing sequence [1, 2]. Alternately, you can remove 2 to get the strictly increasing sequence [1, 3].

My code:

def almostIncreasingSequence(sequence): c= 0 for i in range(len(sequence)-1): if sequence[i]>=sequence[i+1]: c +=1 return c<1

But it can't pass all tests.

input: [1, 3, 2]
Output:false
Expected Output:true
Input: [10, 1, 2, 3, 4, 5]
Output: false
Expected Output: true
Input: [0, -2, 5, 6]
Output: false
Expected Output: true
input: [1, 1]
Output: false
Expected Output: true
Input: [1, 2, 3, 4, 3, 6]
Output: false
Expected Output: true
Input: [1, 2, 3, 4, 99, 5, 6]
Output: false
Expected Output: true

3

25 Answers

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]))

8

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.

1

This is mine. Hope you find this helpful:

def almostIncreasingSequence(sequence): #Take out the edge cases if len(sequence) <= 2: return True #Set up a new function to see if it's increasing sequence def IncreasingSequence(test_sequence): if len(test_sequence) == 2: if test_sequence[0] < test_sequence[1]: return True else: for i in range(0, len(test_sequence)-1): if test_sequence[i] >= test_sequence[i+1]: return False else: pass return True for i in range (0, len(sequence) - 1): if sequence[i] >= sequence [i+1]: #Either remove the current one or the next one test_seq1 = sequence[:i] + sequence[i+1:] test_seq2 = sequence[:i+1] + sequence[i+2:] if IncreasingSequence(test_seq1) == True: return True elif IncreasingSequence(test_seq2) == True: return True else: return False

0

The reason why your modest algorithm fails here (apart from the missing '=' in return) is, it's just counting the elements which are greater than the next one and returning a result if that count is more than 1.

What's important in this is to look at the list after removing one element at a time from it, and confirm that it is still a sorted list.

My attempt at this is really short and works for all scenario. It fails the time constraint on the last hidden test set alone in the exercise.

As the problem name suggests, I directly wanted to compare the list to its sorted version, and handle the 'almost' case later - thus having the almostIncreasingSequence. i.e.:

if sequence==sorted(sequence): . .

But as the problem says:

determine whether it is possible to obtain a strictly increasing sequence by removing no more than one element from the array (at a time).

I started visualizing the list by removing an element at a time during iteration, and check if the rest of the list is a sorted version of itself. Thus bringing me to this:

for i in range(len(sequence)): temp=sequence.copy() del temp[i] if temp==sorted(temp): . .

It was here when I could see that if this condition is true for the full list, then we have what is required - an almostIncreasingSequence! So I completed my code this way:

def almostIncreasingSequence(sequence): t=0 for i in range(len(sequence)): temp=sequence.copy() del temp[i] if temp==sorted(temp): t+=1 return(True if t>0 else False)

This solution still fails on lists such as [1, 1, 1, 2, 3]. As @rory-daulton noted in his comments, we need to differentiate between a 'sorted' and an 'increasingSequence' in this problem. While the test [1, 1, 1, 2, 3] is sorted, its on an increasingSequence as demanded in the problem. To handle this, following is the final code with a one line condition added to check for consecutive same numbers:

def almostIncreasingSequence(sequence): t=0 for i in range(len(sequence)): temp=sequence.copy() del temp[i] if temp==sorted(temp) and not(any(i==j for i,j in zip(sorted(temp), sorted(temp)[1:]))): t+=1 return t>0

---

As this still fails the execution time limit on the last of the test (the list must be really big), I am still looking if there is a way to optimize this solution of mine.

0

Here's my simple solution

def almostIncreasingSequence(sequence): removed_one = False prev_maxval = None maxval = None for s in sequence: if not maxval or s > maxval: prev_maxval = maxval maxval = s elif not prev_maxval or s > prev_maxval: if removed_one: return False removed_one = True maxval = s else: if removed_one: return False removed_one = True return True

1

I'm still working on mine. Wrote it like this but I can't pass the last 3 hidden tests.

def almostIncreasingSequence(sequence):
boolMe = 0
checkRep = 0
for x in range(0, len(sequence)-1): if sequence[x]>sequence[x+1]: boolMe = boolMe + 1 if (x!=0) & (x!=(len(sequence)-2)): if sequence[x-1]>sequence[x+2]: boolMe = boolMe + 1 if sequence.count(sequence[x])>1: checkRep = checkRep + 1 if (boolMe > 1) | (checkRep > 2): return False
return True

1

There are two possibilities whenever you hit the condition of the sequence[i-1]>=sequence[i]

• Delete index i-1
• Delete index i

So my idea was to create copy and delete the indexes and check if they are sorted and then at the end you can do the or and return if the ans is attainable. Complexity will be O(N2)[because of del] and space O(N)

def almostIncreasingSequence(sequence): c0,c1=1,1 n=len(sequence) l1=[] l2=[] for i in sequence: l1.append(i) l2.append(i) for i in range(1,n): if sequence[i-1]>=sequence[i]: del l1[i] del l2[i-1] break for i in range(1,n-1): if l1[i-1]>=l1[i]: c0=0 break for i in range(1,n-1): if l2[i-1]>=l2[i]: c1=0 break return bool(c0 or c1)

This is accepted solution.

2

This was a pretty cool exercise.

I did it like this:

def almostIncreasingSequence(list): removedIdx = [] #Indexes that need to be removed for idx, item in enumerate(list): tmp = [] #Indexes between current index and 0 that break the increasing order for i in range(idx-1, -1, -1): if list[idx]<=list[i]: #Add index to tmp if number breaks order tmp.append(i) if len(tmp)>1: #If more than one of the former numbers breaks order removedIdx.append(idx) #Add current index to removedIdx else: if len(tmp)>0: #If only one of the former numbers breaks order removedIdx.append(tmp[0]) #Add it to removedIdx return len(set(removedIdx))<=1
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]))
print(almostIncreasingSequence([1, 1, 1, 2, 3]))

With Python3, I started with something like this...

def almostIncreasingSequence(sequence): for i, x in enumerate(sequence): ret = False s = sequence[:i]+sequence[i+1:] for j, y in enumerate(s[1:]): if s[j+1] <= s[j]: ret = True break if ret: break if not ret: return True return False

But kept timing out on Check #29.

I kicked myself when I realized that this works, too, but still times out on #29. I have no idea how to speed it up.

def almostIncreasingSequence(sequence): for i, x in enumerate(sequence): s = sequence[:i] s.extend(sequence[i+1:]) if s == sorted(set(s)): return True return False

Well, here's also my solution, I think it's a little bit cleaner than other solutions proposed here so I'll just bring it below.

What it does is it basically checks for an index in which i-th value is larger than (i+1)-th value, if it finds such an index, checks whether removing any of those two makes the list into an increasing sequence.

def almostIncreasingSequence(sequence): def is_increasing(lst): for idx in range(len(lst)-1): if lst[idx] >= lst[idx + 1]: return False return True for idx in range(len(sequence) - 1): if sequence[idx] >= sequence[idx + 1]: fixable = is_increasing([*sequence[:idx], *sequence[idx+1:]]) or is_increasing([*sequence[:idx+1], *sequence[idx+2:]]) if not fixable: return False return True

2

def almostIncreasingSequence(sequence): if len(sequence) == 1: return True decreasing = 0 for i in range(1,len(sequence)): if sequence[i] <= sequence[i-1]: decreasing +=1 if decreasing > 1: return False if sequence[i] <= sequence[i-2] and i-2 >=0: if i != len(sequence)-1 and sequence[i+1] <= sequence[i-1]: return False return True

1

C++ Answer with looping array only once

bool almostIncreasingSequence(std::vector<int> a)
{ int n=a.size(), p=-1, c=0; for (int i=1;i<n;i++) if (a[i-1]>=a[i]) p=i, c++; if (c>1) return 0; if (c==0) return 1; if (p==n-1 || p==1) return 1; if (a[p-1] < a[p+1]) return 1; if (a[p-2] < a[p]) return 1; return 0;
}

I worked on this problem using JavaScript, the idea is to find the breakpoints where the sequence is not strictly increasing, from [0, ..., x - 1] and [x, ..., n.length - 1]. If there are more than 1 breakpoint, then return false.

Once I am able to locate the break point, just check the combination both [0, ..., x - 1, x + 1, ..., n.length - 1] and [0, ..., x - 2, x, ..., n.length - 1], to check whether breakpoint existed in those 2 arrays. It can be simplified to check 2 pairs of points also.

Here is my solution:

function almostIncreasingSequence(sequence) { const breakpoints = findBreak(sequence); if (breakpoints.length === 0) return true; if (breakpoints.length > 1) return false; const removedSeq = sequence .slice(0, breakpoints[0]) .concat(sequence.slice(breakpoints[0] + 1, sequence.length)); const removedSeq2 = sequence .slice(0, breakpoints[0] - 1) .concat(sequence.slice(breakpoints[0], sequence.length)); return ( findBreak(removedSeq).length === 0 || findBreak(removedSeq2).length === 0 );
}
const findBreak = (sequence) => { const breakpoints = []; for (let i = 1; i < sequence.length; i++) { if (sequence[i] <= sequence[i - 1]) { breakpoints.push(i); } } return breakpoints;
};

Here is a solution in Java.

boolean almostIncreasingSequence(int[] sequence) { int count = 0; for(int i=1; i< sequence.length; i++){ if(sequence[i] <= sequence[i-1]){ count++; if( i > 1 && i < sequence.length -1 && sequence[i] <= sequence[i-2] && sequence[i+1] <= sequence[i-1] ) { count++; } } } return count <= 1;
}

boolean almostIncreasingSequence(int[] sequence) { int length = sequence.length; if(length ==1) return true; if(length ==2 && sequence[1] > sequence[0]) return true; int count = 0; int index = 0; boolean iter = true; while(iter){ index = checkSequence(sequence,index); if(index != -1){ count++; index++; if(index >= length-1){ iter=false; }else if(index-1 !=0){ if(sequence[index-1] <= sequence[index]){ iter=false; count++; }else if(((sequence[index] <= sequence[index-2])) && ((sequence[index+1] <= sequence[index-1]))){ iter=false; count++; } } }else{ iter = false; } } if(count > 1) return false; return true;
} int checkSequence(int[] sequence, int index){ for(; index < sequence.length-1; index++){ if(sequence[index+1] <= sequence[index]){ return index; } } return -1;
}

1

Below is the Python3 code that I used and it worked fine:

def almostIncreasingSequence(sequence):
flag = False
if(len(sequence) < 3): return True
if(sequence == sorted(sequence)): if(len(sequence)==len(set(sequence))): return True
bigFlag = True
for i in range(len(sequence)): if(bigFlag and i < len(sequence)-1 and sequence[i] < sequence[i+1]): bigFlag = True continue tempSeq = sequence[:i] + sequence[i+1:] if(tempSeq == sorted(tempSeq)): if(len(tempSeq)==len(set(tempSeq))): flag = True break bigFlag = False
return flag

0

This works on most cases except has problems with performance.

def almostIncreasingSequence(sequence): if len(sequence)==2: return sequence==sorted(list(sequence)) else: for i in range(0,len(sequence)): newsequence=sequence[:i]+sequence[i+1:] if (newsequence==sorted(list(newsequence))) and len(newsequence)==len(set(newsequence)): return True break else: result=False return result

1

This is my Solution,

def almostIncreasingSequence(sequence):

def hasIncreasingOrder(slicedSquence, lengthOfArray): count =0 output = True while(count < (lengthOfArray-1)) : if slicedSquence[count] >= slicedSquence[count+1] : output = False break count = count +1 return output
count = 0
seqOutput = False
lengthOfArray = len(sequence)
while count < lengthOfArray: newArray = sequence[:count] + sequence[count+1:] if hasIncreasingOrder(newArray, lengthOfArray-1): seqOutput = True break count = count+1
return seqOutput

This one works well.

bool almostIncreasingSequence(std::vector<int> sequence) {
/*
if(is_sorted(sequence.begin(), sequence.end())){ return true; }
*/
int max = INT_MIN;
int secondMax = INT_MIN;
int count = 0;
int i = 0;
while(i < sequence.size()){ if(sequence[i] > max){ secondMax = max; max = sequence[i];
}else if(sequence[i] > secondMax){ max = sequence[i]; count++; cout<<"count after increase = "<<count<<endl; }else {count++; cout<<"ELSE count++ = "<<count<<endl;} i++;
}
return count <= 1;
}

def almostIncreasingSequence(sequence): if len(sequence) == 1: return False if len(sequence) == 2: return True c = 0 c1 = 0 for i in range(1,len(sequence)): if sequence[i-1] >= sequence[i]: c += 1 if i != 0 and i+1 < len(sequence): if sequence[i-1] >= sequence[i+1]: c1 += 1 if c > 1 or c1 > 1: return False return c1 == 1 or c == 1

this is mine and it runs fine. I just remove suggested elements and see if new list is strictly increasing

To check if a list is strictly increasing. I check if there are any duplicates first. I then check if the sorted list is the same as the original list

import numpy as np
def IncreasingSequence(sequence): temp=sequence.copy() temp.sort() if (len(sequence) != len(set(sequence))): return False if (sequence==temp): return True return False
def almostIncreasingSequence(sequence): for i in range(len(sequence)-1): if sequence[i] >= sequence[i+1]: sequence_temp=sequence.copy() sequence_temp.pop(i) # print(sequence_temp) # print(IncreasingSequence(sequence_temp)) if (IncreasingSequence(sequence_temp)): return True # Might be the neighbor that is worth removing sequence_temp=sequence.copy() sequence_temp.pop(i+1) if (IncreasingSequence(sequence_temp)): return True return False

I spent a whole day trying to make it as short as possible, but no luck. But here is my accepted answer in CodeSignal.

def almostIncreasingSequence(sequence): if len(sequence)<=2: return True def isstepdown(subsequence): return [a>=b for a,b in zip(subsequence, subsequence[1:])] stepdowns = isstepdown(sequence) n_stepdown = sum(stepdowns) if n_stepdown>1: return False else: sequence2 = sequence.copy() sequence.pop(stepdowns.index(True)) stepdowns_temp = isstepdown(sequence) n_stepdown = sum(stepdowns_temp) sequence2.pop(stepdowns.index(True)+1) stepdowns_temp = isstepdown(sequence2) n_stepdown += sum(stepdowns_temp) if n_stepdown<=1: return True else: return False

Here is another solution. Passes all the tests. Just one method. One pass through the list.

def almostIncreasingSequence(sequence):

count = 0
if len(sequence) < 3: return True
for i in range(1, len(sequence) - 2): #to test only inner elements if (sequence[i] >= sequence[i+1]): count += 1 if count == 2: # the second time this occurs return False #check if skipping one of these items solves the problem if sequence[i-1] >= sequence[i+1] and sequence[i] >= sequence[i+2]: return False i += 1
#handle the first element
if sequence[0] >= sequence[1]: count += 1 if count == 2: return False
#handle the last element
if sequence[-2] >= sequence[-1] and count == 1: return False
return True 

Works on CodeSignal test cases

def almostIncreasingSequence(sequence): s = sequence # for ease

prevMax = s[0] # stores previous max value to which current element has to be compared
found = False
maxI = 0 #index of prevMax
for i in range(1, len(s)): if s[i] <= prevMax: if found: return False else: found = True if maxI > 0 : #checks if current item is smaller thant the prevMax and the value before that if s[i] <= s[maxI] and s[i] > s[maxI - 1]: prevMax = s[i] maxI = i else: # checks if the current and next element are smaller than the prevMax value if (i+1) < len(s) and s[i+1] <= s[maxI]: prevMax = s[i] maxI = i else: prevMax = s[i] maxI = i
return True

Here is the solution for javascript, take those principles and convert them to the desired lang. It makes two array copies (sequence1, sequence2) and checks several things for both:

1. sequence1 - deletes one or more next array items where the next item is smaller than the former item. Then it counts number of deletions (iterator1).

2. sequence2 - deletes one or more former array items where the former item is bigger than the next item. Then it counts number of deletions (iterator2).

3. checks separately for both arrays sequence1 and sequence2 if they have duplicated item which will set separate result flags to false (bigger1, bigger2)

4. check to see whether any of the two arrays (sequence1, sequence2) pass the criteria to be valid. If yes return true for that array and end program. Criteria for checking is: number of deletions of items from the array in question cannot be larger than 1, and there should be no duplicated items (bigger1, bigger2). If there are duplicates it means that two number are the same thus one is not bigger than another one.

However, for some reason testing page almostincreasingSequence at Codesignal (test 12) returns the array [1, 1] to be true which is not correct because the criteria they have been describing says that the next number has to be strictly bigger than the former one, not equal to it. I have made the code according to their written criteria (next number has to be bigger, not equal to).

Code:

function solution(sequence) { let iterator1 = 0; let iterator2 = 0; let bigger1 = true; let bigger2 = true; let sequence1 = sequence.slice(); // copy of the orig array let sequence2 = sequence.slice(); for (let i = 0; i < sequence1.length; i++) { if (typeof sequence1[i+1] !== 'undefined' && sequence1[i+1] < sequence1[i]) { sequence1.splice(i+1, 1); // delete number iterator1++; // count number of deletions i = -1; continue; // restart for loop from i = 0 } } for (let i = 0; i < sequence2.length; i++) { if (typeof sequence2[i+1] !== 'undefined' && sequence2[i] > sequence2[i+1]) { sequence2.splice(i, 1); iterator2++; i = -1; continue; } } for (let i = 0; i < sequence1.length; i++) { for (let k = i + 1; k < sequence1.length; k++) { if (sequence1[i] == sequence1[k]) { // if two numbers are equal.. bigger1 = false; // one is not bigger than another - false } } } for (let i = 0; i < sequence2.length; i++) { for (let k = i + 1; k < sequence2.length; k++) { if (sequence2[i] == sequence2[k]) { bigger2 = false; } } } if (iterator1 < 2 && bigger1 == true) { return true; } else if (iterator2 < 2 && bigger2 == true) { return true; } else { return false; }
}
/* Sample tests */
let arr = [1, 2, 1, 2]; // false
//let arr = [3, 6, 5, 8, 10, 20, 15]; // false
//let arr = [1, 3, 2, 1]; // false
//let arr = [1, 3, 2]; // true
//let arr = [10, 1, 2, 3, 4, 5]; // true
//let arr = [1, 2, 1, 3, 2]; // false
//let arr = [1, 1, 2, 3, 4, 4]; // false
//let arr = [1, 4, 10, 4, 2]; // false
//let arr = [1, 1, 1, 2, 3]; // false
//let arr = [0, -2, 5, 6]; // true
//let arr = [1, 2, 3, 4, 5, 3, 5, 6]; // false
//let arr = [40, 50, 60, 10, 20, 30]; // false
//let arr = [1, 1]; // false
//let arr = [1, 2, 5, 3, 5]; // true
//let arr = [1, 2, 5, 5, 5]; // false
//let arr = [10, 1, 2, 3, 4, 5, 6, 1]; // false
//let arr = [1, 2, 3, 4, 3, 6]; // true (Test 16)
//let arr = [1, 2, 3, 4, 99, 5, 6]; // true
//let arr = [123, -17, -5, 1, 2, 3, 12, 43, 45]; // true
//let arr = [3, 5, 67, 98, 3]; // true
solution(arr);

But if you want code to pass criteria despite of their instructions, thus to allow numbers to be equal like [1, 1] = true and not strictly next number bigger than former one, then solution would be this:

function solution(sequence) { let iterator1 = 0; let iterator2 = 0; let sequence1 = sequence.slice(); let sequence2 = sequence.slice(); for (let i = 0; i < sequence1.length; i++) { if (typeof sequence1[i+1] !== 'undefined' && sequence1[i+1] <= sequence1[i]) { sequence1.splice(i+1, 1); iterator1++; i = -1; continue; } } for (let i = 0; i < sequence2.length; i++) { if (typeof sequence2[i+1] !== 'undefined' && sequence2[i] >= sequence2[i+1]) { sequence2.splice(i, 1); iterator2++; i = -1; continue; } } if (iterator1 < 2) { return true; } else if (iterator2 < 2) { return true; } else { return false; }
}