Arithmetic Slices II (All Subsequences)

🎯

Arithmetic Slices II (All Subsequences)

hardDPAmazonGoogle

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

💡 This problem is about counting all arithmetic subsequences, which is tricky because subsequences can be non-contiguous and overlap. Beginners often struggle to efficiently count these without enumerating all subsequences explicitly.

📋

Problem Statement

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

💡

Example

Input"[2, 4, 6, 8, 10]"

Output7

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 [2,6,10].

• All elements are the same → maximum arithmetic subsequences
• Array with length less than 3 → output 0
• Array with negative and positive numbers → correct difference handling
• Array with large values causing integer overflow → handle with long/int64

🔁

Why DP?

Why greedy fails:

Greedy fails because choosing local arithmetic subsequences does not guarantee counting all global subsequences. For example, picking only contiguous triples misses non-contiguous subsequences with the same difference.

DP state:

dp[i][diff] = number of arithmetic subsequences ending at index i with common difference diff, including subsequences of length 2 (pairs).

Recurrence:dp[i][diff] += dp[j][diff] + 1 for all j < i where diff = nums[i] - nums[j]

For each pair (j, i), extend all subsequences ending at j with difference diff by nums[i], and add the new pair (j, i) as a subsequence of length 2.

⚠️

Common Mistakes

Counting pairs as arithmetic subsequences

Overcounting results, since subsequences must have length ≥ 3

✅ Only add dp[j][diff] to total, not the '+1' which counts pairs

Integer overflow when calculating difference

Wrong counts or runtime errors

✅ Use long or 64-bit integer types for difference calculation

Not initializing dp maps correctly

Key errors or incorrect counts

✅ Initialize dp array with empty hash maps or default dictionaries

Adding all dp[i][diff] to total instead of only dp[j][diff]

Double counting subsequences

✅ Add only dp[j][diff] to total, since dp[i][diff] includes pairs

🧠

Brute Force (Enumerate All Subsequences)

💡 This approach exists to understand the problem's combinatorial explosion and why naive enumeration is infeasible.

Intuition

Try all subsequences of length ≥ 3 and check if they are arithmetic. Count those that satisfy the condition.

Algorithm

1. Generate all subsequences of length at least 3 using recursion or bitmask.
2. For each subsequence, check if it is arithmetic by verifying equal differences.
3. If arithmetic, increment the count.
4. Return the total count.

💡 The challenge is the huge number of subsequences, making this approach impractical for large inputs.

</>

Code

from itertools import combinations

def numberOfArithmeticSlices(nums):
    n = len(nums)
    count = 0
    for length in range(3, n+1):
        for comb in combinations(range(n), length):
            diff = nums[comb[1]] - nums[comb[0]]
            arithmetic = True
            for i in range(2, length):
                if nums[comb[i]] - nums[comb[i-1]] != diff:
                    arithmetic = False
                    break
            if arithmetic:
                count += 1
    return count

# Example usage
if __name__ == '__main__':
    print(numberOfArithmeticSlices([2,4,6,8,10]))  # Output: 7

Line Notes

from itertools import combinationsImport combinations to generate subsequences easily

for length in range(3, n+1)Only consider subsequences of length at least 3

for comb in combinations(range(n), length)Generate all index subsequences of given length

diff = nums[comb[1]] - nums[comb[0]]Calculate common difference for the subsequence

if nums[comb[i]] - nums[comb[i-1]] != diffCheck if difference is consistent for all pairs

import java.util.*;

public class Solution {
    public int numberOfArithmeticSlices(int[] nums) {
        int n = nums.length;
        int count = 0;
        for (int length = 3; length <= n; length++) {
            int[] indices = new int[length];
            count += backtrackCount(nums, indices, 0, 0, length);
        }
        return count;
    }

    private boolean isArithmetic(int[] nums, int[] indices) {
        int diff = nums[indices[1]] - nums[indices[0]];
        for (int i = 2; i < indices.length; i++) {
            if (nums[indices[i]] - nums[indices[i - 1]] != diff) return false;
        }
        return true;
    }

    private int backtrackCount(int[] nums, int[] indices, int start, int depth, int length) {
        if (depth == length) {
            return isArithmetic(nums, indices) ? 1 : 0;
        }
        int count = 0;
        for (int i = start; i < nums.length; i++) {
            indices[depth] = i;
            count += backtrackCount(nums, indices, i + 1, depth + 1, length);
        }
        return count;
    }

    public static void main(String[] args) {
        Solution sol = new Solution();
        System.out.println(sol.numberOfArithmeticSlices(new int[]{2,4,6,8,10})); // 7
    }
}

Line Notes

for (int length = 3; length <= n; length++)Iterate over subsequence lengths starting from 3

private boolean isArithmeticHelper to check if a subsequence is arithmetic

for (int i = start; i < nums.length; i++)Generate all subsequences recursively

count += backtrackCount(...)Accumulate count of valid arithmetic subsequences

#include <iostream>
#include <vector>
using namespace std;

bool isArithmetic(const vector<int>& nums, const vector<int>& indices) {
    int diff = nums[indices[1]] - nums[indices[0]];
    for (int i = 2; i < indices.size(); i++) {
        if (nums[indices[i]] - nums[indices[i - 1]] != diff) return false;
    }
    return true;
}

int backtrackCount(const vector<int>& nums, vector<int>& indices, int start, int depth, int length) {
    if (depth == length) {
        return isArithmetic(nums, indices) ? 1 : 0;
    }
    int count = 0;
    for (int i = start; i < nums.size(); i++) {
        indices[depth] = i;
        count += backtrackCount(nums, indices, i + 1, depth + 1, length);
    }
    return count;
}

int numberOfArithmeticSlices(const vector<int>& nums) {
    int n = nums.size();
    int count = 0;
    for (int length = 3; length <= n; length++) {
        vector<int> indices(length);
        count += backtrackCount(nums, indices, 0, 0, length);
    }
    return count;
}

int main() {
    vector<int> nums = {2,4,6,8,10};
    cout << numberOfArithmeticSlices(nums) << endl; // 7
    return 0;
}

Line Notes

for (int length = 3; length <= n; length++)Loop over subsequence lengths starting at 3

bool isArithmeticCheck if given subsequence indices form arithmetic sequence

for (int i = start; i < nums.size(); i++)Generate subsequences recursively

count += backtrackCount(...)Sum counts of valid arithmetic subsequences

function numberOfArithmeticSlices(nums) {
    const n = nums.length;
    let count = 0;
    function isArithmetic(indices) {
        const diff = nums[indices[1]] - nums[indices[0]];
        for (let i = 2; i < indices.length; i++) {
            if (nums[indices[i]] - nums[indices[i - 1]] !== diff) return false;
        }
        return true;
    }
    function backtrack(indices, start, depth, length) {
        if (depth === length) {
            if (isArithmetic(indices)) count++;
            return;
        }
        for (let i = start; i < n; i++) {
            indices[depth] = i;
            backtrack(indices, i + 1, depth + 1, length);
        }
    }
    for (let length = 3; length <= n; length++) {
        backtrack([], 0, 0, length);
    }
    return count;
}

// Example usage
console.log(numberOfArithmeticSlices([2,4,6,8,10])); // 7

Line Notes

for (let length = 3; length <= n; length++)Iterate over subsequence lengths starting at 3

function isArithmetic(indices)Check if subsequence is arithmetic

for (let i = start; i < n; i++)Generate subsequences recursively

if (isArithmetic(indices)) count++Increment count if subsequence is arithmetic

⏱

Complexity

TimeO(n * 2^n)

SpaceO(n)

Enumerating all subsequences is exponential in n, and checking each subsequence takes linear time.

💡 For n=20, this means over a million subsequences, which is too large to compute in reasonable time.

Interview Verdict: TLE

This approach is too slow for large inputs but helps understand the problem's combinatorial nature.

🧠

DP with Hash Maps (Count Subsequences by Difference)

💡 This approach uses DP with hash maps to count arithmetic subsequences efficiently by difference, avoiding explicit enumeration.

Intuition

For each pair (j, i), calculate difference diff = nums[i] - nums[j]. The number of subsequences ending at i with diff is increased by those ending at j with diff plus the new pair (j, i).

Algorithm

1. Initialize an array of hash maps dp where dp[i] maps difference to count of subsequences ending at i.
2. Iterate over pairs (j, i) with j < i.
3. Calculate diff = nums[i] - nums[j].
4. Update dp[i][diff] by adding dp[j][diff] + 1 (the new pair).
5. Accumulate dp[j][diff] to the result since subsequences of length ≥ 3 are counted from dp[j][diff].
6. Return the total count.

💡 The key is to count subsequences ending at each index by difference, building on previous counts.

Recurrence:dp[i][diff] = dp[i][diff] + dp[j][diff] + 1

</>

Code

from collections import defaultdict

def numberOfArithmeticSlices(nums):
    n = len(nums)
    dp = [defaultdict(int) for _ in range(n)]
    total = 0
    for i in range(n):
        for j in range(i):
            diff = nums[i] - nums[j]
            count_at_j = dp[j][diff]
            dp[i][diff] += count_at_j + 1
            total += count_at_j
    return total

# Example usage
if __name__ == '__main__':
    print(numberOfArithmeticSlices([2,4,6,8,10]))  # Output: 7

Line Notes

dp = [defaultdict(int) for _ in range(n)]Initialize dp array with hash maps for difference counts

for i in range(n)Iterate over each element as subsequence end

for j in range(i)Check all previous elements to form pairs

diff = nums[i] - nums[j]Calculate difference between pair elements

dp[i][diff] += count_at_j + 1Add existing subsequences plus new pair

total += count_at_jAdd count of subsequences extended (length ≥ 3)

import java.util.*;

public class Solution {
    public int numberOfArithmeticSlices(int[] nums) {
        int n = nums.length;
        List<Map<Long, Integer>> dp = new ArrayList<>();
        for (int i = 0; i < n; i++) dp.add(new HashMap<>());
        int total = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < i; j++) {
                long diff = (long)nums[i] - (long)nums[j];
                int count_at_j = dp.get(j).getOrDefault(diff, 0);
                int count_at_i = dp.get(i).getOrDefault(diff, 0);
                dp.get(i).put(diff, count_at_i + count_at_j + 1);
                total += count_at_j;
            }
        }
        return total;
    }

    public static void main(String[] args) {
        Solution sol = new Solution();
        System.out.println(sol.numberOfArithmeticSlices(new int[]{2,4,6,8,10})); // 7
    }
}

Line Notes

List<Map<Long, Integer>> dp = new ArrayList<>()Initialize list of maps to store counts by difference

for (int i = 0; i < n; i++)Iterate over array elements as subsequence ends

long diff = (long)nums[i] - (long)nums[j]Use long to avoid integer overflow in difference

dp.get(i).put(diff, count_at_i + count_at_j + 1)Update dp for current index and difference

total += count_at_jAdd counts of subsequences extended to total

#include <iostream>
#include <vector>
#include <unordered_map>
using namespace std;

int numberOfArithmeticSlices(vector<int>& nums) {
    int n = nums.size();
    vector<unordered_map<long long, int>> dp(n);
    int total = 0;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < i; j++) {
            long long diff = (long long)nums[i] - (long long)nums[j];
            int count_at_j = dp[j].count(diff) ? dp[j][diff] : 0;
            dp[i][diff] += count_at_j + 1;
            total += count_at_j;
        }
    }
    return total;
}

int main() {
    vector<int> nums = {2,4,6,8,10};
    cout << numberOfArithmeticSlices(nums) << endl; // 7
    return 0;
}

Line Notes

vector<unordered_map<long long, int>> dp(n)DP array of hash maps for difference counts

for (int i = 0; i < n; i++)Outer loop for subsequence ends

long long diff = (long long)nums[i] - (long long)nums[j]Calculate difference safely with long long

dp[i][diff] += count_at_j + 1Update dp with extended subsequences and new pair

total += count_at_jAdd count of subsequences extended to total

function numberOfArithmeticSlices(nums) {
    const n = nums.length;
    const dp = Array.from({length: n}, () => new Map());
    let total = 0;
    for (let i = 0; i < n; i++) {
        for (let j = 0; j < i; j++) {
            const diff = nums[i] - nums[j];
            const count_at_j = dp[j].get(diff) || 0;
            dp[i].set(diff, (dp[i].get(diff) || 0) + count_at_j + 1);
            total += count_at_j;
        }
    }
    return total;
}

// Example usage
console.log(numberOfArithmeticSlices([2,4,6,8,10])); // 7

Line Notes

const dp = Array.from({length: n}, () => new Map())Initialize dp array of maps for difference counts

for (let i = 0; i < n; i++)Iterate over array elements as subsequence ends

const diff = nums[i] - nums[j]Calculate difference between elements

dp[i].set(diff, (dp[i].get(diff) || 0) + count_at_j + 1)Update dp with extended subsequences and new pair

total += count_at_jAdd counts of subsequences extended to total

⏱

Complexity

TimeO(n^2)

SpaceO(n^2)

We iterate over all pairs and store counts in hash maps, which can be up to O(n^2) in worst case.

💡 For n=1000, this means about 1 million operations, which is feasible in interviews.

Interview Verdict: Accepted

This is the standard efficient solution for this problem.

🧠

DP with Long for Difference and Tabulation

💡 This approach is similar to Approach 2 but emphasizes safe handling of large integer differences and uses explicit tabulation.

Intuition

Use a DP table of hash maps indexed by i storing counts by difference, carefully using long integers to avoid overflow.

Algorithm

1. Create dp array of hash maps to store counts of subsequences by difference.
2. Iterate i from 0 to n-1, and for each i, iterate j from 0 to i-1.
3. Calculate difference as long to avoid overflow.
4. Update dp[i][diff] by adding dp[j][diff] + 1.
5. Accumulate dp[j][diff] to result as these represent subsequences extended.
6. Return total count.

💡 The main difference is careful use of long integers and explicit tabulation order.

Recurrence:dp[i][diff] = dp[i][diff] + dp[j][diff] + 1

</>

Code

from collections import defaultdict

def numberOfArithmeticSlices(nums):
    n = len(nums)
    dp = [defaultdict(int) for _ in range(n)]
    total = 0
    for i in range(n):
        for j in range(i):
            diff = nums[i] - nums[j]
            count_at_j = dp[j][diff]
            dp[i][diff] += count_at_j + 1
            total += count_at_j
    return total

if __name__ == '__main__':
    print(numberOfArithmeticSlices([2,4,6,8,10]))  # 7

Line Notes

dp = [defaultdict(int) for _ in range(n)]Initialize dp array with maps for difference counts

for i in range(n)Process elements in order to build subsequences

for j in range(i)Check all previous elements to extend subsequences

diff = nums[i] - nums[j]Calculate difference safely

dp[i][diff] += count_at_j + 1Add counts of subsequences extended plus new pair

total += count_at_jAdd counts of subsequences of length ≥ 3

import java.util.*;

public class Solution {
    public int numberOfArithmeticSlices(int[] nums) {
        int n = nums.length;
        List<Map<Long, Integer>> dp = new ArrayList<>();
        for (int i = 0; i < n; i++) dp.add(new HashMap<>());
        int total = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < i; j++) {
                long diff = (long)nums[i] - (long)nums[j];
                int count_at_j = dp.get(j).getOrDefault(diff, 0);
                dp.get(i).put(diff, dp.get(i).getOrDefault(diff, 0) + count_at_j + 1);
                total += count_at_j;
            }
        }
        return total;
    }

    public static void main(String[] args) {
        Solution sol = new Solution();
        System.out.println(sol.numberOfArithmeticSlices(new int[]{2,4,6,8,10})); // 7
    }
}

Line Notes

List<Map<Long, Integer>> dp = new ArrayList<>()Initialize dp list of maps for difference counts

for (int i = 0; i < n; i++)Process elements in increasing order

long diff = (long)nums[i] - (long)nums[j]Calculate difference with long to avoid overflow

dp.get(i).put(diff, dp.get(i).getOrDefault(diff, 0) + count_at_j + 1)Update dp with extended subsequences and new pair

total += count_at_jAdd counts of subsequences extended to total

#include <iostream>
#include <vector>
#include <unordered_map>
using namespace std;

int numberOfArithmeticSlices(vector<int>& nums) {
    int n = nums.size();
    vector<unordered_map<long long, int>> dp(n);
    int total = 0;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < i; j++) {
            long long diff = (long long)nums[i] - (long long)nums[j];
            int count_at_j = dp[j].count(diff) ? dp[j][diff] : 0;
            dp[i][diff] += count_at_j + 1;
            total += count_at_j;
        }
    }
    return total;
}

int main() {
    vector<int> nums = {2,4,6,8,10};
    cout << numberOfArithmeticSlices(nums) << endl; // 7
    return 0;
}

Line Notes

vector<unordered_map<long long, int>> dp(n)DP array of maps for difference counts

for (int i = 0; i < n; i++)Process elements in order

long long diff = (long long)nums[i] - (long long)nums[j]Calculate difference safely

dp[i][diff] += count_at_j + 1Update dp with extended subsequences and new pair

total += count_at_jAdd counts of subsequences extended to total

function numberOfArithmeticSlices(nums) {
    const n = nums.length;
    const dp = Array.from({length: n}, () => new Map());
    let total = 0;
    for (let i = 0; i < n; i++) {
        for (let j = 0; j < i; j++) {
            const diff = nums[i] - nums[j];
            const count_at_j = dp[j].get(diff) || 0;
            dp[i].set(diff, (dp[i].get(diff) || 0) + count_at_j + 1);
            total += count_at_j;
        }
    }
    return total;
}

console.log(numberOfArithmeticSlices([2,4,6,8,10])); // 7

Line Notes

const dp = Array.from({length: n}, () => new Map())Initialize dp array of maps

for (let i = 0; i < n; i++)Process elements in order

const diff = nums[i] - nums[j]Calculate difference

dp[i].set(diff, (dp[i].get(diff) || 0) + count_at_j + 1)Update dp with extended subsequences and new pair

total += count_at_jAdd counts of subsequences extended to total

⏱

Complexity

TimeO(n^2)

SpaceO(n^2)

Similar to Approach 2, but with explicit attention to integer overflow.

💡 This approach is safer for large inputs with big integer values.

Interview Verdict: Accepted

This is a robust implementation suitable for all inputs.

📊

All Approaches - One-Glance Tradeoffs

💡 Approach 2 (DP with hash maps) is the best to code in interviews due to efficiency and clarity.

Approach	Time	Space	Stack Risk	Reconstruct	Use In Interview
1. Brute Force	O(n * 2^n)	O(n)	Yes (deep recursion)	No	Mention only - never code
2. DP with Hash Maps	O(n^2)	O(n^2)	No	Yes	Code this approach
3. DP with Long and Tabulation	O(n^2)	O(n^2)	No	Yes	Mention as robust implementation

💼

Interview Strategy

💡 Use this guide to understand the problem deeply, starting from brute force to optimized DP, and practice coding all approaches before interviews.

How to Present

Clarify problem constraints and definitions.Explain brute force approach and its inefficiency.Introduce DP with hash maps and explain the recurrence.Discuss handling of integer overflow and implementation details.Code the DP solution and test with examples.

Time Allocation

Clarify: 3min → Approach: 7min → Code: 10min → Test: 5min. Total ~25min

What the Interviewer Tests

Understanding of subsequences vs subarrays, DP state design, handling large differences, and coding hash map based DP efficiently.

Common Follow-ups

• What if subsequences must be contiguous? → Use simpler DP for arithmetic slices I.
• How to reconstruct one arithmetic subsequence? → Store predecessors and backtrack.

💡 These follow-ups test your understanding of variations and ability to extend solutions.

🔍

Pattern Recognition

When to Use

1) Problem asks for counting subsequences, 2) Arithmetic progression or equal difference, 3) Subsequence length ≥ 3, 4) Large input requiring efficient counting.

Signature Phrases

arithmetic subsequencesdifference between consecutive elementscount all subsequences

NOT This Pattern When

Problems counting contiguous subarrays or sequences without difference constraints.

Similar Problems

Arithmetic Slices I - counts contiguous arithmetic slicesLongest Arithmetic Subsequence - finds length of longest arithmetic subsequenceNumber of Arithmetic Triplets - counts triplets with fixed difference

Practice

1. You are given a 2D grid of integers and need to find the length of the longest strictly increasing path moving only up, down, left, or right. Which algorithmic approach guarantees an optimal solution for this problem?

easy

A. Greedy approach that always moves to the neighbor with the largest value

B. Topological sort based dynamic programming processing cells in ascending order of their values

C. Simple DFS from each cell without memoization

D. Breadth-first search layer by layer from the smallest values

Solution

1. Step 1: Understand problem constraints

   The problem requires finding the longest strictly increasing path in a matrix, which can be modeled as a directed graph where edges go from smaller to larger values.

2. Step 2: Identify suitable algorithm

   Topological sort based DP processes nodes in ascending order, ensuring all dependencies are resolved before updating dp values, guaranteeing optimal substructure and avoiding recomputation.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Topological sort ensures correct order for DP updates [OK]

Hint: Longest increasing path -> topological sort DP [OK]

Common Mistakes:

• Assuming greedy always works
• Using DFS without memoization causing TLE
• Using BFS layer by layer ignoring value order

2. What is the time complexity of the iterative postorder traversal solution for the Binary Tree Maximum Path Sum problem, assuming the tree has n nodes?

medium

A. O(n) for traversal but O(n) extra space for the stack and gains dictionary

B. O(n log n) due to repeated stack operations and subtree recomputations

C. O(n^2) because for each node, the algorithm recomputes gains from all descendants

D. O(n) because each node is visited once and gains are computed in constant time per node

Solution

1. Step 1: Identify traversal cost

   The iterative postorder traversal visits each node exactly once, pushing and popping it from the stack once.

2. Step 2: Analyze per-node computation

   At each node, gains are computed using stored values from children in O(1) time, no recomputation of subtrees.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Linear traversal with constant work per node -> O(n) time [OK]

Hint: Each node processed once, no repeated subtree work [OK]

Common Mistakes:

• Confusing recursion stack with repeated work
• Assuming recomputation of subtree gains
• Ignoring constant time per node

3. Examine the following buggy code snippet for the Binary Tree Maximum Path Sum problem. Identify the line that contains the subtle bug causing incorrect results on trees with negative values.

class Solution:
    def maxPathSum(self, root):
        self.max_sum = float('-inf')
        def max_gain(node):
            if not node:
                return 0
            left_gain = max_gain(node.left)
            right_gain = max_gain(node.right)
            price_newpath = node.val + left_gain + right_gain
            self.max_sum = max(self.max_sum, price_newpath)
            return node.val + left_gain + right_gain
        max_gain(root)
        return self.max_sum

medium

A. Line returning node.val + left_gain + right_gain

B. Line computing price_newpath = node.val + left_gain + right_gain

C. Line updating self.max_sum = max(self.max_sum, price_newpath)

D. Line returning 0 for null nodes

Solution

1. Step 1: Understand the return value's role

   The return value represents the maximum gain from the current node to its parent, which must be a path extending in one direction only.

2. Step 2: Identify the bug

   Returning node.val + left_gain + right_gain sums both children, which is invalid for the parent's gain calculation and leads to incorrect results.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Correct return should be node.val + max(left_gain, right_gain, 0) [OK]

Hint: Return max gain from one child, not sum of both [OK]

Common Mistakes:

• Returning sum of both children
• Not ignoring negative gains
• Updating max_sum only with one child path

4. Examine the following buggy code snippet for shortest path visiting all nodes. Which line contains the subtle bug that can cause incorrect results or infinite loops?

medium

A. Line setting base cases: dp[i][1 << i] = 0

B. Line updating next_mask: next_mask = mask ^ (1 << v)

C. Line with dp initialization: dp = [[float('inf')] * (1 << n) for _ in range(n)]

D. Line updating dp[v][next_mask]: dp[v][next_mask] = dp[u][mask] + 1

Solution

1. Step 1: Identify how visited nodes are tracked

   Visited nodes are represented by a bitmask. To add a visited node, bitwise OR (|) must be used.

2. Step 2: Spot the incorrect bitmask operation

   The code uses XOR (^) instead of OR (|) to update next_mask, which can incorrectly unset bits and cause wrong states or infinite loops.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Using XOR breaks the visited set semantics [OK]

Hint: Visited mask update must use OR, not XOR [OK]

Common Mistakes:

• Using XOR instead of OR for mask update
• Forgetting to initialize dp base cases
• Not checking dp[u][mask] == inf before continuing

5. Suppose the problem is modified so that the binary tree nodes can have negative values, and the diameter is defined as the longest path in terms of number of edges (unchanged). Which modification to the optimal bottom-up DFS algorithm is necessary to correctly compute the diameter?

hard

A. Change the diameter update to consider the sum of node values along the path instead of edge count.

B. Modify the height function to return 0 for negative-valued nodes to avoid decreasing path length.

C. No change needed; the diameter depends only on edges, so negative values do not affect the algorithm.

D. Use a top-down approach to track maximum path sums including negative values.

Solution

1. Step 1: Understand diameter definition depends on edges, not node values

   The diameter is the longest path in terms of edges, so node values (positive or negative) do not affect path length.

2. Step 2: Confirm algorithm only counts edges via heights, so no modification needed

   The existing height and diameter update logic counts edges, so negative values do not require changes.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Diameter counts edges, independent of node values [OK]

Hint: Diameter counts edges, not node values [OK]

Common Mistakes:

• Confusing diameter with max path sum
• Modifying height to exclude negative nodes unnecessarily