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Problem Statement
Given an integer array nums
, return the number of arithmetic subarrays
of nums
.
If any integer array consists of at least
three
elements and if the difference
between any two consecutive elements
is the same
, it is called arithmetic
.
A subarray
is a contiguous subsequence
of the array.
Examples
Example 1:
- Input:
[1, 3, 5, 7, 9, 10, 11]
- Expected Output:
7
- Justification: The segments that form arithmetic sequences are
[1, 3, 5]
,[3, 5, 7]
,[5, 7, 9]
,[1, 3, 5, 7]
,[3, 5, 7, 9]
,[1, 3, 5, 7, 9]
, and[9, 10, 11]
. Each of these segments has a constant difference of 2 between consecutive elements.
Example 2:
- Input:
[7, 7, 7, 7]
- Expected Output:
3
- Justification: The qualifying segments are
[7, 7, 7]
,[7, 7, 7, 7]
(considering the first three and then all four elements), and the last three elements[7, 7, 7]
. Each segment has a constant difference of 0 between its elements.
Example 3:
- Input:
[1, 2, 4, 6, 8]
- Expected Output:
3
- Justification: Only three segments
[2,4,6]
,[4, 6, 8]
and[2, 4, 6, 8]
form arithmetic sequences with a constant difference of 2.
Solution
To solve this problem, we will iterate through the array while keeping track of the current arithmetic sequence's length. The approach hinges on understanding that for any arithmetic sequence with more than three elements, it inherently contains smaller arithmetic sequences. For instance, a sequence of four elements contains three smaller sequences. Thus, we calculate the total count of sequences as we identify them.
The rationale behind this approach is its efficiency and simplicity; it avoids unnecessary nested iterations, focusing instead on leveraging the arithmetic property of contiguous subarrays. This method is expected to work well because it directly counts sequences without explicitly enumerating all possible subarrays, thus optimizing both time and space complexity.
Step-by-Step Algorithm
-
Initialization: Start by checking if the input array
nums
has fewer than 3 elements. If yes, return 0 immediately since at least three elements are needed to form an arithmetic slice. -
Setup Counters: Initialize two counters:
count
to store the total number of arithmetic slices found, andconsecutive
to track the length of the current consecutive arithmetic slice sequence. -
Iterate Through Array: Begin iterating through the array starting from the third element (index 2), as the first two elements are used for initial comparison.
-
Check Arithmetic Sequence: For each element at index
i
, check if the difference between the current element and the previous one (nums[i] - nums[i-1]
) is the same as the difference between the previous element and the one before it (nums[i-1] - nums[i-2]
). This checks if the current set of three elements forms part of an arithmetic sequence. -
Update Counters: If the current set of elements continues an arithmetic sequence, increment the
consecutive
counter by 1. Then, add the value ofconsecutive
tocount
. This step accounts for the number of new arithmetic slices that include the current element. -
Reset
consecutive
if Needed: If the current set of elements does not continue an arithmetic sequence, reset theconsecutive
counter to 0. This prepares it for the next sequence of arithmetic slices. -
Complete Iteration and Return: Continue steps 4 through 6 for each element in the array. After completing the iteration, return the total
count
of arithmetic slices found.
Algorithm Walkthrough
Let's consider the Input: nums = [1, 3, 5, 7, 9, 10, 11]
-
Initialization: The input array has more than 3 elements, so we proceed.
-
Setup Counters: Initialize
count = 0
andconsecutive = 0
. -
First Iteration (i = 2, Element = 5):
- Calculate differences:
5 - 3 = 2
and3 - 1 = 2
. They are equal. - Increment
consecutive
to 1. - Add
consecutive
tocount
(count = 1
).
- Calculate differences:
-
Second Iteration (i = 3, Element = 7):
- Differences:
7 - 5 = 2
and5 - 3 = 2
. Still equal. - Increment
consecutive
to 2. - Add
consecutive
tocount
(count = 3
).
- Differences:
-
Third Iteration (i = 4, Element = 9):
- Differences:
9 - 7 = 2
and7 - 5 = 2
. Equal again. - Increment
consecutive
to 3. - Add
consecutive
tocount
(count = 6
).
- Differences:
-
Fourth Iteration (i = 5, Element = 10):
- Differences:
10 - 9 = 1
and9 - 7 = 2
. Not equal. - Reset
consecutive
to 0.
- Differences:
-
Fifth Iteration (i = 6, Element = 11):
- Differences:
11 - 10 = 1
and10 - 9 = 1
. Equal. - Increment
consecutive
to 1. - Add
consecutive
tocount
(count = 7
).
- Differences:
-
Completion: After completing the iteration, we find that the total
count
of arithmetic slices is 7.
Code
Complexity Analysis
Time Complexity
The time complexity of the algorithm is O(n), where n is the number of elements in the input array nums
. This is because the solution involves a single pass through the array to count arithmetic slices, with constant time operations performed for each element.
Space Complexity
The space complexity for the algorithm is O(1). This is because the amount of extra space used does not depend on the size of the input array.
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