413. Arithmetic Slices

Problem Description

The goal of the given problem is to find the number of contiguous subarrays within a given array of integers, where each subarray forms an arithmetic sequence. An arithmetic sequence is defined as a sequence of at least three numbers where the difference between consecutive elements is the same throughout.

For example, if the difference between the first and second elements is 2, then every subsequent pair of consecutive elements must also have a difference of 2 for the array to be considered arithmetic.

Our task is to count how many such arithmetic subarrays exist in the nums array.

Intuition

The intuition behind the solution comes from the realization that we can count each possible arithmetic subarray as we iterate through the array nums and keep a running count of the number of times we encounter consecutive pairs with the same difference.

If we find consecutive pairs with the same difference (d), this means we've found an additional element that can extend the arithmetic subarray formed so far. Each time we find a new pair with the same difference, it means we've formed an additional arithmetic subarray including the new element.

In detail, if we have a sequence of n consecutive elements that have the same difference d between them, the number of arithmetic subarrays that can be formed from these elements is equal to the sum of the first n-2 positive integers. This is because an arithmetic subarray needs at least three elements.

So, to reach the solution, we keep variables for the current pair difference (d) and a counter (cnt) for consecutive pairs with the same difference found so far. We initialize d to a value outside any possible difference we could get from elements in nums to make sure we don't wrongly increment cnt at the start.

As we iterate through each pair of consecutive elements using the pairwise function, if the difference is the same as d, we increment the cnt because we've found an additional element that extends the arithmetic subarrays we've counted so far. We add the current cnt value to the answer (ans) as this is the number of new arithmetic subarrays formed by including this new element.

If the difference is not the same, it means we've reached the end of the current sequence of elements forming arithmetic subarrays, so we set cnt back to zero and set d to the current pair's difference.

By the end of the iteration, ans will contain the total number of arithmetic subarrays within nums.

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Solution Approach

The given solution in Python takes advantage of the pairwise utility which creates an iterator that returns consecutive pairs of elements from the input nums. In other words, if nums is [a, b, c, d, e], pairwise(nums) would yield (a, b), (b, c), (c, d), and (d, e) consecutively.

Here's the step-by-step implementation of the solution:

Initialize Variables: Before looping through the nums, we initialize the ans variable to 0. This variable will accumulate the total count of arithmetic subarrays. We also initialize a cnt variable to 0; it keeps track of consecutive pairs with the same difference. The d variable, which holds the current common difference between pairs, is set to a number outside the valid range (3000) to handle the edge case at the beginning of the array.
Loop through pairwise(nums): We iterate over each pair (a, b) in pairwise(nums). Here, a and b represent consecutive elements in nums.
Check the Difference and Update Counter: We compare the difference b - a with the current d. If they're equal, increment cnt by 1, because this extends the current arithmetic subarray sequence by one element. If they're different, update d to the new difference b - a and reset cnt to 0, because we're starting to count a new set of arithmetic subarrays.
Update the Answer: Add the current cnt to ans in each iteration. By adding cnt, we're accounting for all the new arithmetic subarrays that end at the current element b. The reason adding cnt works is that for each extension of an arithmetic subarray by one element, we introduce exactly cnt new subarrays where cnt is the count of previous consecutive elements that were part of such subarrays.
Return Result: After the loop completes, ans represents the total number of arithmetic subarrays in nums, which is then returned as the final answer.

This algorithm is efficient, as it needs only a single pass through the array nums, making it an O(n) time complexity solution, where n is the length of nums. The space complexity is O(1) since it uses a constant amount of extra space.

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Example Walkthrough

Let's demonstrate the solution approach with a small example. Consider the array nums = [1, 3, 5, 7, 9, 15].

Initialize Variables: Set ans = 0, cnt = 0, and d = 3000.
Loop through pairwise(nums): We'll be looking at the following pairs as we loop: (1, 3), (3, 5), (5, 7), (7, 9), and (9, 15).
Check the Difference and Update Counter:
- For the first pair (1, 3), the difference b - a is 2. Since d is initialized to 3000, d does not equal b - a. So, update d to 2 and keep cnt at 0.
- Moving to the next pair (3, 5), the difference is 2 which matches the current d. Increment cnt by 1. Now, cnt = 1 and ans = 1.
- For (5, 7), the difference is still 2. Increment cnt again, cnt = 2 and ans = 1 + 2 = 3.
- The pair (7, 9) also has the same difference, so cnt = 3 and ans = 3 + 3 = 6.
- The final pair (9, 15) has a different difference of 6. This ends the current sequence of arithmetic subarrays, so reset cnt to 0 and update d to 6.
Update the Answer: We've updated ans throughout the steps as we went along.
Return Result: The value of ans after evaluating all pairs is 6. This means there are 6 contiguous subarrays within nums that form an arithmetic sequence.

To conclude, given nums = [1, 3, 5, 7, 9, 15], the arithmetic subarrays are [1, 3, 5], [3, 5, 7], [5, 7, 9], [1, 3, 5, 7], [3, 5, 7, 9], and [1, 3, 5, 7, 9], resulting in a count of 6.

Solution Implementation

1from itertools import pairwise
2
3class Solution:
4    def numberOfArithmeticSlices(self, nums: List[int]) -> int:
5        total_slices = 0  # Initialize the count of arithmetic slices
6        current_sequence_length = 0  # Tracks the length of the current arithmetic sequence
7      
8        # Set initial difference to a large number outside the possible range
9        previous_difference = 3000
10
11        # Iterate pairwise over the array to check differences between consecutive elements
12        for current_num, next_num in pairwise(nums):
13            current_difference = next_num - current_num
14          
15            # If the current difference equals the previous one, we extend the arithmetic sequence
16            if current_difference == previous_difference:
17                current_sequence_length += 1
18            else:
19                # If not, we start a new arithmetic sequence
20                previous_difference = current_difference
21                current_sequence_length = 0
22
23            # The number of arithmetic slices ending at the current position can be added to the total
24            total_slices += current_sequence_length
25
26        return total_slices  # Return the total count of arithmetic slices
27

1class Solution {
2    public int numberOfArithmeticSlices(int[] nums) {
3        int arithmeticSliceCount = 0;  // To store the number of arithmetic slices found.
4        int currentSliceLength = 0;    // To keep track of the current sequence length.
5        int difference = 3000;         // Initialize with a value outside the problem constraints.
6
7        // Iterate through the given array to find arithmetic slices.
8        for (int i = 0; i < nums.length - 1; ++i) {
9            // Check if the current pair of elements continue the arithmetic sequence.
10            if (nums[i + 1] - nums[i] == difference) {
11                // If they do, increment the count of arithmetic slices by increasing the current length.
12                ++currentSliceLength;
13            } else {
14                // If not, update the common difference to the new pair's difference.
15                difference = nums[i + 1] - nums[i];
16                // Reset the current sequence length because a new arithmetic sequence starts.
17                currentSliceLength = 0;
18            }
19            // Add the number of arithmetic slices ending at the current position to the total count.
20            arithmeticSliceCount += currentSliceLength;
21        }
22        // Return the total count of arithmetic slices in the array.
23        return arithmeticSliceCount;
24    }
25}
26

1class Solution {
2public:
3    int numberOfArithmeticSlices(vector<int>& nums) {
4        int totalSlices = 0; // This will hold the total number of arithmetic slices 
5        int currentStreak = 0; // This keeps track of the current streak of arithmetic slices
6        int previousDifference = 3000; // Initialize with a difference not likely to appear in the sequence
7      
8        for (int i = 0; i < nums.size() - 1; ++i) { // Loop through the vector, but not including the last element
9            int currentDifference = nums[i + 1] - nums[i]; // Calculate the difference between two consecutive elements
10          
11            // Check if the current difference is the same as the previous difference
12            if (currentDifference == previousDifference) {
13                ++currentStreak; // If so, increment the current streak of consecutive slices
14            } else {
15                previousDifference = currentDifference; // Otherwise, update the difference
16                currentStreak = 0; // And reset the current streak since a new difference has started
17            }
18          
19            totalSlices += currentStreak; // Add the current streak to the total slices count
20                                           // This works because an arithmetic slice of length n contributes
21                                           // (n - 1) + (n - 2) + ... + 1 slices.
22        }
23      
24        return totalSlices; // Return the total number of arithmetic slices found
25    }
26};
27

1function numberOfArithmeticSlices(nums: number[]): number {
2    let totalSlices = 0; // Initialize count of arithmetic slices
3    let currentStreak = 0; // Count of consecutive arithmetic pairs within a slice
4    let previousDifference = 3000; // Large initial difference to ensure the first pair forms a new slice
5  
6    // Iterate through the array of numbers to detect arithmetic slices
7    for (let i = 0; i < nums.length - 1; ++i) {
8        const currentNum = nums[i]; // The current element in nums
9        const nextNum = nums[i + 1]; // The next element in nums
10        const currentDifference = nextNum - currentNum; // Difference between the current and next element
11      
12        // Check if the current pair continues the streak of the same differences
13        if (currentDifference === previousDifference) {
14            // If so, increment the streak count since it extends an arithmetic slice
15            ++currentStreak;
16        } else {
17            // If not, reset the streak count and update the difference tracker
18            previousDifference = currentDifference;
19            currentStreak = 0;
20        }
21      
22        // Accumulate the count of arithmetic slices
23        // Each additional number in a streak adds to existing slices
24        totalSlices += currentStreak;
25    }
26  
27    // Return the total count of arithmetic slices found in the array
28    return totalSlices;
29}
30

Time and Space Complexity

The given Python function numberOfArithmeticSlices computes the number of continuous arithmetic slices (subarrays) in an array. The computation iterates over the array once, using a pairwise comparison of elements to determine if a current pair continues an arithmetic sequence or starts a new one.

Time Complexity:

The time complexity of the function is O(n), where n is the length of the input list nums. This is because the function makes a single pass through the list using pairwise(nums) to compare consecutive elements.

The inner operations of the for loop (checking conditions, updating the count cnt, updating the difference d, and incrementing ans) are all constant time operations, which means they do not depend on the size of the input list. Thus, they do not change the overall linear time complexity.

Space Complexity:

The space complexity of the function is O(1) as only a fixed number of extra variables are used (ans, cnt, d). The space used does not scale with the input size, so it remains constant regardless of the length of nums.

Learn more about how to find time and space complexity quickly using problem constraints.

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