Arithmetic Slices II (All Subsequences)

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Problem

Imagine you are analyzing stock price trends and want to count all possible arithmetic patterns in the price changes, not just contiguous ones.

Given an integer array nums, return the number of all the arithmetic subsequences of nums. A sequence of numbers is arithmetic if it consists of at least three elements and if the difference between any two consecutive elements is the same. A subsequence of an array is a sequence that can be formed by removing some elements (possibly none) from the array without changing the order of the remaining elements.

1 ≤ nums.length ≤ 1000-2^31 ≤ nums[i] ≤ 2^31 - 1

Edge cases: All elements are the same → maximum arithmetic subsequencesArray with length less than 3 → output 0Array with negative and positive numbers → correct difference handling

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IDE

def numberOfArithmeticSlices(nums: list[int]) -> int:public int numberOfArithmeticSlices(int[] nums)int numberOfArithmeticSlices(vector<int>& nums)function numberOfArithmeticSlices(nums)

def numberOfArithmeticSlices(nums):
    # Write your solution here
    pass

class Solution {
    public int numberOfArithmeticSlices(int[] nums) {
        // Write your solution here
        return 0;
    }
}

#include <vector>
using namespace std;

int numberOfArithmeticSlices(vector<int>& nums) {
    // Write your solution here
    return 0;
}

function numberOfArithmeticSlices(nums) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: 0Returning 0 for all inputs due to missing subsequence counting logic or incorrect base case handling.✅ Add DP logic to count subsequences with difference maps and handle arrays with length >= 3.

Wrong: Incorrect small number (e.g., 1 or 2) for canonical exampleCounting only contiguous arithmetic slices or longest subsequence instead of all subsequences.✅ Use DP with hash maps to count all subsequences sharing the same difference, not just contiguous ones.

Wrong: Wrong large number or overflowed valueInteger overflow in difference calculation causing incorrect map keys or counts.✅ Use 64-bit integers or long type for difference calculations and map keys.

Wrong: Timeout or no output on large inputUsing brute force enumeration instead of O(n^2) DP approach.✅ Implement DP with hash maps to achieve O(n^2) time complexity.

✓

Test Cases

Focus on handling small and boundary inputs correctly before tackling edge cases.

Understand common pitfalls like greedy traps and integer overflow.

Optimize your solution to O(n^2) DP to handle large inputs efficiently.

t1_01basic

Input{"nums":[2,4,6,8,10]}

Expected7

⏱ Performance - must finish in 2000ms

The arithmetic subsequences are: [2,4,6], [4,6,8], [6,8,10], [2,4,6,8], [4,6,8,10], [2,4,6,8,10], and [2,4,10].

💡 Think about counting subsequences with the same difference using dynamic programming.

💡 Use a map at each index to store counts of subsequences ending there with a given difference.

💡 For each pair (i,j), update dp[i][diff] += dp[j][diff] + 1 to count new subsequences.

Why it failed: Incorrect count likely due to missing subsequences or incorrect difference handling. Fix by properly updating dp maps for each pair and including subsequences of length >= 3 only.

✓ Correctly counts all arithmetic subsequences using DP with difference maps.

t1_02basic

Input{"nums":[7,7,7,7]}

Expected5

⏱ Performance - must finish in 2000ms

All subsequences of length >=3 with difference 0: [7,7,7] (3 times), [7,7,7,7] (1 time), total 5 arithmetic subsequences.

💡 Consider that difference zero subsequences can be many when elements are identical.

💡 Count subsequences with difference 0 carefully using DP maps.

💡 Remember to count only subsequences of length at least 3.

Why it failed: Failing to count difference zero subsequences or counting subsequences of length < 3. Fix by including only subsequences with length >= 3 and correctly summing counts for difference 0.

✓ Handles identical elements and counts all valid arithmetic subsequences.

t2_01edge

Input{"nums":[]}

Expected0

⏱ Performance - must finish in 2000ms

Empty array has no subsequences of length >= 3.

💡 Check how your code handles empty input arrays.

💡 Ensure no subsequences are counted when array length < 3.

💡 Return 0 immediately if nums length < 3.

Why it failed: Code crashes or returns non-zero for empty input. Fix by adding a guard clause for nums length < 3 returning 0.

✓ Correctly returns 0 for empty input.

t2_02edge

Input{"nums":[5]}

Expected0

⏱ Performance - must finish in 2000ms

Single element array cannot form arithmetic subsequences of length >= 3.

💡 Check base case for arrays with length less than 3.

💡 No subsequences of length >= 3 possible with single element.

💡 Return 0 if nums length < 3.

Why it failed: Incorrectly counts subsequences for single element input. Fix by returning 0 when nums length < 3.

✓ Correctly returns 0 for single element input.

t2_03edge

Input{"nums":[1,3,5]}

Expected1

⏱ Performance - must finish in 2000ms

Only one arithmetic subsequence of length 3: [1,3,5].

💡 Check that your code counts subsequences exactly of length 3 and above.

💡 Verify difference calculation and subsequence counting.

💡 Ensure dp updates include subsequences starting from length 2 to build longer ones.

Why it failed: Fails to count subsequences exactly at boundary length 3. Fix by correctly initializing and updating dp counts for subsequences of length >= 3.

✓ Correctly counts subsequences at the exact boundary length.

t3_01corner

Input{"nums":[1,2,4,7,11]}

Expected3

⏱ Performance - must finish in 2000ms

Arithmetic subsequences: [1,4,7], [4,7,11], and [1,4,7,11] with differences 3.

💡 Beware of greedy approaches that pick only longest subsequences.

💡 Count all valid subsequences, not just longest or contiguous ones.

💡 Use DP with maps to count all subsequences sharing the same difference.

Why it failed: Greedy approach counts only longest subsequence, missing others. Fix by counting all subsequences using DP maps, not just longest.

✓ Counts all arithmetic subsequences, not just longest.

t3_02corner

Input{"nums":[1,3,5,7,9,11]}

Expected16

⏱ Performance - must finish in 2000ms

All arithmetic subsequences with difference 2, including all subsequences of length >= 3.

💡 Check if your code mistakenly treats this as unbounded knapsack or similar problem.

💡 Remember subsequences must maintain order and difference consistency.

💡 Use DP with difference maps to count subsequences correctly.

Why it failed: Confusing subsequences with combinations or permutations leads to wrong count. Fix by using DP maps to count subsequences preserving order and difference.

✓ Correctly counts all arithmetic subsequences preserving order.

t3_03corner

Input{"nums":[1,1000000000,2000000000]}

Expected1

⏱ Performance - must finish in 2000ms

One arithmetic subsequence: [1, 1000000000, 2000000000] with large differences.

💡 Check for integer overflow in difference calculations.

💡 Use 64-bit integers or long type for differences.

💡 Ensure difference calculations do not overflow and are stored correctly.

Why it failed: Overflow in difference calculation causes wrong counts. Fix by using 64-bit integer types for difference variables.

✓ Handles large integer differences without overflow.

t4_01performance

Input{"nums":[1000,999,998,997,996,995,994,993,992,991,990,989,988,987,986,985,984,983,982,981,980,979,978,977,976,975,974,973,972,971,970,969,968,967,966,965,964,963,962,961,960,959,958,957,956,955,954,953,952,951,950,949,948,947,946,945,944,943,942,941,940,939,938,937,936,935,934,933,932,931,930,929,928,927,926,925,924,923,922,921,920,919,918,917,916,915,914,913,912,911,910,909,908,907,906,905,904,903,902,901]}

Expectednull

⏱ Performance - must finish in 2000ms

Input size n=100 with descending sequence to test O(n^2) DP performance within 2 seconds.

💡 Brute force O(n*2^n) will time out for n=100.

💡 Use DP with hash maps to achieve O(n^2) time complexity.

💡 Optimize difference storage and avoid unnecessary recomputations.

Why it failed: Brute force or exponential solutions time out on n=100 input. Fix by implementing O(n^2) DP with hash maps.

✓ Solution runs within time limit using O(n^2) DP approach.

Practice

1. Given the following code snippet and input person_caps = [[5, 100], [1, 5, 100], [1, 2]], what is the final returned value of count_ways_space_optimized(person_caps)?

easy

A. 3

B. 6

C. 4

D. 5

Solution

1. Step 1: Trace dp array updates for each cap

   Initially dp[0]=1. Caps 1,2,5,100 appear. Assigning caps to persons updates dp masks. For example, cap 5 can assign to person 0 or 1, cap 1 to person 1 or 2, etc.

2. Step 2: Calculate final dp[(1<<3)-1] = dp[7]

   After processing all caps, dp[7] counts all unique assignments. Manual tracing shows dp[7] = 4.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Manual bitmask DP tracing confirms 4 unique assignments [OK]

Hint: Trace dp updates backward to count unique assignments [OK]

Common Mistakes:

• Off-by-one in mask indexing
• Forgetting to iterate masks backward

2. You are given a binary tree and need to find the length of the longest path between any two nodes in the tree. This path may or may not pass through the root. Which algorithmic approach guarantees an optimal solution for this problem?

easy

A. A brute force approach that computes the height of every subtree multiple times and checks all paths.

B. A bottom-up depth-first search (DFS) that computes heights and updates the diameter in a single traversal.

C. A greedy approach that always chooses the longest child path at each node without recursion.

D. A dynamic programming approach on arrays that stores intermediate results for subproblems unrelated to tree structure.

Solution

1. Step 1: Understand the problem requires longest path between any two nodes, not necessarily through root

   The problem is to find the diameter, which is the longest path between any two nodes in the tree, possibly passing through any node.

2. Step 2: Identify the approach that computes height and diameter simultaneously in one DFS traversal

   The bottom-up DFS approach computes the height of subtrees and updates the diameter during the same recursion, ensuring O(n) time complexity.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Bottom-up DFS avoids repeated height computations and covers all paths [OK]

Hint: Diameter requires single DFS updating max path sum [OK]

Common Mistakes:

• Assuming diameter passes only through root
• Using greedy without recursion
• Repeated height computations causing inefficiency

3. You are given a list of pairs of integers where each pair represents a start and end point. You want to find the longest sequence of pairs such that for every consecutive pair (a, b) and (c, d) in the sequence, b < c. Which algorithmic approach guarantees finding the optimal longest sequence efficiently?

easy

A. Dynamic Programming with sorting pairs by their start values and checking all subsequent pairs

B. Greedy algorithm sorting pairs by their end values and selecting pairs that start after the last selected pair ends

C. Pure brute force recursion checking all possible sequences without sorting

D. Sorting pairs by their start values and greedily selecting pairs with the earliest start

Solution

1. Step 1: Understand the problem constraints

   The problem requires selecting pairs so that each pair starts after the previous one ends, maximizing the chain length.

2. Step 2: Identify the optimal approach

   Sorting pairs by their end values allows a greedy selection of the earliest finishing pairs, ensuring maximum room for subsequent pairs.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Greedy by end time is a classic interval scheduling approach [OK]

Hint: Sort by end time for optimal chain [OK]

Common Mistakes:

• Sorting by start time and applying greedy fails on some inputs
• Using brute force recursion without pruning is inefficient
• Assuming DP must sort by start time

4. What is the time complexity of the optimized LIS algorithm using binary search on an input array of length n?

medium

A. O(n log n)

B. O(n)

C. O(n²)

D. O(log n)

Solution

1. Step 1: Identify outer loop complexity

   The algorithm iterates over all n elements once, so at least O(n).

2. Step 2: Identify inner operation complexity

   For each element, binary search on tails array takes O(log n) time.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   n iterations x log n binary search -> O(n log n) [OK]

Hint: Binary search inside loop gives n log n [OK]

Common Mistakes:

• Assuming O(n²) because of nested loops
• Ignoring binary search complexity
• Confusing with brute force exponential

5. Suppose the problem is modified so that caps can be assigned to multiple people (i.e., caps are no longer unique). Which of the following algorithmic changes correctly adapts the solution to count the number of ways to assign caps without uniqueness constraints?

hard

A. Remove bitmask tracking and use simple multiplication of counts per person

B. Keep bitmask DP but allow assigning the same cap multiple times by not checking mask bits

C. Use a DP over persons only, multiplying the number of caps each person can wear

D. Use a nested loop over persons and caps without bitmask, summing ways for each person independently

Solution

1. Step 1: Understand the uniqueness constraint removal

   Without uniqueness, caps can be reused freely, so no need to track assigned caps with bitmasks.

2. Step 2: Simplify counting approach

   Counting ways reduces to multiplying the number of caps each person can wear, as assignments are independent.

3. Step 3: Why other options fail

   Bitmask DP is unnecessary and incorrect if caps can be reused; summing independently without multiplication misses combinations.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Removing bitmask and multiplying counts per person matches problem relaxation [OK]

Hint: No uniqueness -> no bitmask needed, multiply counts [OK]

Common Mistakes:

• Trying to keep bitmask DP unnecessarily
• Summing instead of multiplying counts