Longest Alternating Subarray


This Ruby function, alternating_subarray(a)
, checks the given array a
and finds the length of the longest subarray that alternates between two numbers. The algorithm works as follows:
Initialize the size
n
of the arraya
, the resultresult
which will hold the length of the longest alternating subarray, and a counterj
which will be used to iterate over the array.Loop over each element in the array
a
. For each element, incrementj
to point to the next element.In a nested loop, while
j
is within the bounds of the array and thej
th element ofa
equals thei
th element plus the remainder of the division(j  i) % 2
(which will be 0 or 1), do the following: Update
res
to be the maximum of its current value and the length of the current alternating subarray (j  i + 1
).  Increment
j
to move to the next element.
 Update
After the loops, if
res
is greater than 1 (which means that an alternating subarray of length greater than 1 was found), returnres
. Otherwise, return 1.
This function uses two pointers (i
and j
) to iterate over the array and find the longest subarray that alternates between two numbers. It returns the length of this subarray if it’s greater than 1, or 1 if no such subarray exists.
For instance, given the array [1, 2, 1, 2, 3, 4, 3]
, the function would return 4
, which is the length of the longest alternating subarray [1, 2, 1, 2]
.
An alternating array is one in which the difference between consecutive elements is alternately 1 and 1.
The pattern is:
 s1  s0 = 1, which means the second element is 1 greater than the first.
 s2  s1 = 1, which means the third element is 1 less than the second.
 s3  s2 = 1, which means the fourth element is 1 greater than the third.
So, the pattern alternates between increasing by 1 and decreasing by 1. In other words, the difference between the current element and the first element equals the position of the current element modulo 2. This is the expression “si  s0 = i % 2”.
To find an alternating subarray, you would look at each element, A[i], and find the longest subarray where A[j] == A[i] + (j  i) % 2. This condition is checking if the element A[j] fits the alternating pattern relative to A[i].
The length of the subarray is then j  i + 1, because j and i are indices and arrays are 0indexed. If i and j are the same, then the length of the subarray is 1. If i is 0 and j is 1, then the length of the subarray is 2, and so forth.
So the goal is to find the longest subarray where the pattern of alternating differences is maintained.
The expression “si  s0 = i % 2” comes from observing the pattern in an alternating array.
Let’s consider the first few elements of such an array:
 For i = 0, si = s0, so si  s0 = 0.
 For i = 1, si = s1 = s0 + 1, so si  s0 = 1.
 For i = 2, si = s2 = s1  1 = s0, so si  s0 = 0.
 For i = 3, si = s3 = s2 + 1 = s1 = s0 + 1, so si  s0 = 1.
 And so on…
In this pattern, you can see that si  s0 alternates between 0 and 1, which are also the results of i % 2. So that’s why we have the expression “si  s0 = i % 2”. It’s a way of expressing the alternating pattern using a simple mathematical equation.
An alternating array is one in which the difference between consecutive elements alternates between +1 and 1. For instance, [1,2,1,2,1,2] is an alternating array.
In the example given, [1,2,1,2,3,2,3], there are two alternating subarrays, [1,2,1,2] and [2,3,2,3]. These two subarrays overlap by two elements, [2,1].
In the solution approach, after finding the first alternating subarray [1,2,1,2] starting at A[0], the next step would be to check the subarray starting at A[1]. However, we don’t need to do that, because any alternating subarray starting at A[1] would be part of the longer alternating subarray we have already found, starting at A[0].
This same logic applies to all other starting positions within the first found alternating subarray. Any alternating subarray starting at A[2], A[3], or A[4] will also be part of the longer subarray [1,2,1,2].
Therefore, instead of starting the next search at A[1], A[2], A[3] or A[4], we can skip ahead to the position right after the end of the first found alternating subarray. In this example, that means starting the next search at the last 2 in [1,2,1,2,1,2,3].
This approach is a more efficient way to solve the problem, because it eliminates the need to check many starting positions that would only lead to shorter subarrays. It is a twopointers technique, where one pointer marks the start of the current subarray, and the other pointer moves ahead to find the end of the subarray.


Can I ask you what led you to believe it can be done by two pointers? Did you start by putting the math equation on paper? I’m trying to understand how you approach problem solving so that I can get better.
The realization that this problem could be solved using a twopointer technique mainly comes from understanding the nature of alternating sequences and the constraints in the problem statement.
Understanding Alternating Sequences: Alternating sequences have a clear pattern: a value alternates between increasing and decreasing by 1. This predictability makes it possible to use pointers to track the start and end of these sequences.
Problem Constraints: The problem requires us to find the maximum length of the alternating sequence. Therefore, once we detect an alternating sequence, we need to keep moving our end pointer to track how long it is, updating our result if it’s the longest we’ve found so far. Once the sequence stops alternating, we can move our start pointer to the end of this sequence and start looking for the next one. This is the twopointer technique in action.
The decision to use two pointers does not always come from a mathematical equation, but rather from a strategic understanding of the problem. In this case, we recognized that we needed to track the start and end of sequences and that moving two pointers would allow us to do this efficiently.
Practicing problem solving is a great way to develop this strategic understanding. As you solve more problems, you start to recognize common patterns and techniques that can be applied.
Another approach is to learn and study common problemsolving techniques and algorithms. Twopointer technique, binary search, dynamic programming, etc., are common strategies that are useful in many situations. Once you’re familiar with these strategies, you’ll start recognizing situations where they can be applied.
In this case, seeing that we needed to keep track of the longest sequence of alternating numbers suggested that two pointers could be a useful approach.
In example 1 why subarray [2,3] is not a valid subarray
In the problem statement, an alternating subarray is defined as a subarray where each element alternates between increasing by 1 and decreasing by 1 compared to the previous element. In other words, if we denote the elements of the subarray as s0, s1, s2, …, sm, it should follow this pattern:
 s1  s0 = 1 (increasing by 1)
 s2  s1 = 1 (decreasing by 1)
 s3  s2 = 1 (increasing by 1)
 s4  s3 = 1 (decreasing by 1)
 and so on…
For the example 1 where nums = [2,3,4,3,4]
, if we consider the subarray [2,3], it only consists of two elements. According to the alternating rule, the second element should be one more than the first element, which is the case here (3 is 1 more than 2). However, we can’t say whether it’s truly an alternating subarray because it does not have any more elements to check for the alternating increase and decrease pattern.
The subarray [3,4] is considered an alternating subarray because it’s part of a larger alternating subarray [3,4,3,4]. The element after 4 is 3, which is 1 less than 4, confirming the alternating pattern.
In general, a subarray of only two elements can be considered as the start of an alternating subarray, but it’s not “alternating” in itself because it doesn’t demonstrate the full increase/decrease pattern. That’s why [2,3] is not considered as a valid alternating subarray in this context.
Input : [14,30,29,49,3,23,44,21,26,52] Output: 2 Expected: 1
How does this make sense? 30,29 are alternating so the output should be 2?
In this problem, a subarray needs to consist of more than two elements to be considered “alternating”. This is explicitly stated in the problem statement: “A subarray s of length m is called alternating if m is greater than 1.”
While 30,29 is technically alternating according to the definition given, the problem statement asks for the maximum length of all alternating subarrays present in nums or 1 if no such subarray exists.
So, if you only find twoelement subarrays that are alternating, like 30,29, they don’t satisfy the problem condition, and the correct output is 1.
The output you’ve mentioned is expected based on the problem’s conditions: it’s looking for alternating subarrays of more than two elements. If there are none, it returns 1.
m greater than 1, is it not 2?
Yes, when the problem statement mentions “m is greater than 1”, it implies m could be 2 or more. Therefore, a subarray with 2 elements is valid and should be considered alternating if it satisfies the remaining conditions specified in the problem statement.
Apologies for the confusion caused in my previous response. So, if the only alternating subarray you have in the entire nums array is of length 2, like [30,29], then the maximum length of alternating subarray would indeed be 2, not 1.
The problem in the example you provided ([14,30,29,49,3,23,44,21,26,52]) is that we can’t find an alternating subarray with more than one element, that’s why the output is 1. In case of [30,29], it doesn’t meet the condition for being an alternating subarray: s1  s0 = 1, which is not true for 30 and 29, so they can’t form an alternating subarray.