Arithmetic Slices II - Subsequence

Hard

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ArraysDynamic Programming

Given an integer array nums, return the number of all the arithmetic subsequences of nums.

A sequence of numbers is called arithmetic if it consists of at least three elements and if the difference between any two consecutive elements is the same.

For example, [1, 3, 5, 7, 9], [7, 7, 7, 7], and [3, -1, -5, -9] are arithmetic sequences.
For example, [1, 1, 2, 5, 7] is not an arithmetic sequence.

A subsequence of an array is a sequence that can be formed by removing some elements (possibly none) of the array.

For example, [2,5,10] is a subsequence of [1,2,1,2,4,1,5,10].

The test cases are generated so that the answer fits in 32-bit integer.

Example 1:

Input: nums = [2,4,6,8,10]
Output: 7
Explanation: All arithmetic subsequence slices are:
[2,4,6]
[4,6,8]
[6,8,10]
[2,4,6,8]
[4,6,8,10]
[2,4,6,8,10]
[2,6,10]

Example 2:

Input: nums = [7,7,7,7,7]
Output: 16
Explanation: Any subsequence of this array is arithmetic.

Constraints:

1 <= nums.length <= 1000
-2³¹ <= nums[i] <= 2³¹ - 1

Arithmetic Slices II - Subsequence

Hard

Google

2 views

Topics:

ArraysDynamic Programming

Given an integer array nums, return the number of all the arithmetic subsequences of nums.

A sequence of numbers is called arithmetic if it consists of at least three elements and if the difference between any two consecutive elements is the same.

For example, [1, 3, 5, 7, 9], [7, 7, 7, 7], and [3, -1, -5, -9] are arithmetic sequences.
For example, [1, 1, 2, 5, 7] is not an arithmetic sequence.

A subsequence of an array is a sequence that can be formed by removing some elements (possibly none) of the array.

For example, [2,5,10] is a subsequence of [1,2,1,2,4,1,5,10].

The test cases are generated so that the answer fits in 32-bit integer.

Example 1:

Input: nums = [2,4,6,8,10]
Output: 7
Explanation: All arithmetic subsequence slices are:
[2,4,6]
[4,6,8]
[6,8,10]
[2,4,6,8]
[4,6,8,10]
[2,4,6,8,10]
[2,6,10]

Example 2:

Input: nums = [7,7,7,7,7]
Output: 16
Explanation: Any subsequence of this array is arithmetic.

Constraints:

1 <= nums.length <= 1000
-2³¹ <= nums[i] <= 2³¹ - 1

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values within the input array, and what data type are we using (e.g., 32-bit integer, 64-bit integer)?
Can the input array `nums` be empty or null?
If no arithmetic subsequence exists, what should I return?
Are duplicate numbers allowed in the input array, and how should they be handled when forming subsequences?
Is the order of elements in the subsequence important, or can I choose any valid subsequence?

Brute Force Solution

Approach

The brute force approach solves this problem by checking every possible combination of numbers from the given list. It essentially tries out all possible subsequences to see if they form an arithmetic progression.

Here's how the algorithm would work step-by-step:

Start by considering every possible combination of numbers from the input list. This means looking at every possible group of numbers, no matter how big or small.
For each of these groups, check if the numbers form a valid arithmetic sequence. This means verifying if the difference between consecutive numbers is the same throughout the entire group.
If a group does form a valid arithmetic sequence, count it as a valid solution.
After going through all possible combinations, add up the number of valid arithmetic sequence solutions you've found. That's the final answer.

Code Implementation

def arithmetic_slices_ii_subsequence_brute_force(numbers):
    total_arithmetic_subsequences = 0
    list_length = len(numbers)

    # Iterate through all possible subsequences
    for i in range(1 << list_length):
        subsequence = []
        for j in range(list_length):
            # Check if the j-th element is included in the subsequence
            if (i >> j) & 1:
                subsequence.append(numbers[j])

        # A subsequence needs at least 3 elements to be an arithmetic sequence
        if len(subsequence) >= 3:

            # Check if the subsequence is an arithmetic sequence
            is_arithmetic = True
            difference = subsequence[1] - subsequence[0]

            for k in range(2, len(subsequence)): 
                if subsequence[k] - subsequence[k - 1] != difference:
                    is_arithmetic = False
                    break

            # Increment count if subsequence is an arithmetic sequence
            if is_arithmetic:
                total_arithmetic_subsequences += 1

    return total_arithmetic_subsequences

Big(O) Analysis

Time Complexity

O(2^n * n) – The brute force approach involves generating all possible subsequences of the input array of size n. There are 2^n possible subsequences. For each subsequence, we need to check if it forms an arithmetic progression. Checking if a subsequence of length k is an arithmetic progression takes O(k) time, which is at most O(n) in the worst case. Therefore, the overall time complexity is O(2^n * n).

Space Complexity

O(1) – The brute force approach, as described, iterates through all possible subsequences but doesn't explicitly create auxiliary data structures to store these subsequences. It checks each subsequence for being an arithmetic progression on the fly. Therefore, the auxiliary space required is limited to a few variables for iteration and comparison, independent of the input size N (the number of elements in the input list). This implies a constant auxiliary space complexity.

Optimal Solution

Approach

This problem asks us to find special number sequences within a larger list. The smart way to do this involves remembering the number sequences we've already seen to avoid repeating work, and building upon existing sequences.

Here's how the algorithm would work step-by-step:

Think about each number in the list one by one.
For each number, check every number that comes before it.
When looking at two numbers, calculate their difference.
See if you've already found number sequences ending with the earlier number that have this same difference.
If you have, it means you can extend those sequences by adding the current number to them. If not, start a new sequence.
Keep track of the number of sequences of each length you find. This is what we need to calculate in the end.
Adding the sequences instead of recounting them, ensures we only have to calculate the sequences ending at each position once. This avoids checking every possibility, making the solution much faster.

Code Implementation

def number_of_arithmetic_slices(sequence):
    sequence_length = len(sequence)
    total_arithmetic_slices = 0

    # Store count of sequences ending at each index with a given difference.
    dp = [{} for _ in range(sequence_length)]

    for i in range(1, sequence_length):
        for j in range(i):
            difference = sequence[i] - sequence[j]

            # Sequences of length at least 3 are arithmetic slices.
            previous_count = dp[j].get(difference, 0)

            # Extend arithmetic subsequence if possible.
            if difference in dp[i]:
                dp[i][difference] += previous_count + 1
            else:
                dp[i][difference] = previous_count + 1

            # Accumulate counts from subsequences of length >= 3
            total_arithmetic_slices += previous_count

    return total_arithmetic_slices

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through each of the n numbers in the input array. For each number, it iterates through all the numbers preceding it. This pair checking is the dominant operation. Therefore, the total operations approximate n * (n-1) / 2 which simplifies to O(n²).

Space Complexity

O(N^2) – The solution involves iterating through each number in the list and checking every number that comes before it. For each pair of numbers, a hash map (dictionary) is potentially stored, to track the number of arithmetic subsequences ending at each index with a specific difference. In the worst case, each pair (i, j) where j < i can have a unique difference and a corresponding entry in the hash map at index i. Therefore, the space complexity is O(N^2) because each of the N elements can store up to N entries in its hash map. Each hash map stores the number of arithmetic sequences ending at the current index.

Edge Cases

Case	How to Handle
Empty input array	Return 0 since an empty array cannot have any arithmetic subsequences.
Array with less than 3 elements	Return 0 since an arithmetic subsequence requires at least 3 elements.
Integer overflow when calculating the difference	Use long data type to store the difference to avoid potential overflow issues.
Array with all identical values	Handle this case by correctly counting all subsequences of length 3 or more.
Array with a large number of elements (scalability)	The solution should use dynamic programming with hashmaps to avoid exponential time complexity.
Input array contains negative numbers	The difference calculation should handle negative differences correctly.
Input array contains 0	The difference calculation should handle 0 differences correctly, possibly resulting in sequences with repeated numbers.
The same arithmetic subsequence with elements at different indexes.	Use hashmaps to count the number of arithmetic subsequences ending at each index, to differentiate the index and allow accurate calculation.