Longest Chain of Pairs

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Steps

setup

Sort pairs by their end time

The pairs are sorted by their second element (end time) to prepare for the greedy selection.

💡 Sorting by end time ensures we always consider the earliest finishing pair first, which is key to maximizing chain length.

Line:

pairs.sort(key=lambda x: x[1])

💡 Sorting sets the foundation for the greedy approach by ordering pairs to consider earliest finishing first.

setup

Initialize current_end and count

Set current_end to negative infinity and count to zero before starting to iterate pairs.

💡 current_end tracks the end of the last included pair; count tracks how many pairs are included so far.

Line:

current_end = float('-inf')
count = 0

💡 Initialization ensures the first pair can always be considered for inclusion.

compare

Check first pair [1,2]

Compare the start of the first pair (1) with current_end (-inf). Since 1 > -inf, include this pair.

💡 Including the first pair starts the chain and updates current_end to this pair's end.

Line:

if pair[0] > current_end:
    count += 1
    current_end = pair[1]

💡 The first pair always qualifies to start the chain.

compare

Check second pair [2,3]

Compare the start of the second pair (2) with current_end (2). Since 2 is not greater than 2, skip this pair.

💡 Pairs that start at or before current_end cannot be chained after the last included pair.

Line:

if pair[0] > current_end:
    count += 1
    current_end = pair[1]

💡 This shows the strict inequality needed to chain pairs.

compare

Check third pair [3,4]

Compare the start of the third pair (3) with current_end (2). Since 3 > 2, include this pair.

💡 Including this pair extends the chain beyond the previous pairs.

Line:

if pair[0] > current_end:
    count += 1
    current_end = pair[1]

💡 This step extends the chain length to 2.

reconstruct

Iteration complete, return count

All pairs have been processed. The count of included pairs is 2, which is the length of the longest chain.

💡 Returning the count gives the final answer to the problem.

Line:

return count

💡 The final count reflects the maximum number of pairs chained.

prune

Visual summary of chain decisions

Highlight which pairs were included (green) and which were skipped (red) to summarize the chain.

💡 Seeing included vs skipped pairs visually reinforces the greedy selection logic.

💡 The chain consists of pairs [1,2] and [3,4].

fill_cells

Confirm monotonic growth of count

Observe that the count never decreases as we move through pairs; it either stays the same or increases.

💡 Monotonic growth confirms that chain length only grows or remains stable as pairs are processed.

💡 This property is important for greedy correctness.

traverse

Review active element after last inclusion

The active element is the last included pair [3,4] with current_end 4 and count 2.

💡 Tracking the active element helps understand the chain's current boundary.

💡 The chain ends at 4, so no further pairs can be added.

reconstruct

Final answer confirmed

The algorithm returns 2 as the length of the longest chain of pairs.

💡 This final step confirms the output matches the expected result.

Line:

return count

💡 The longest chain length is 2 for the input pairs.

from typing import List

class Solution:
    def findLongestChain(self, pairs: List[List[int]]) -> int:
        pairs.sort(key=lambda x: x[1])  # STEP 1: Sort pairs by end time
        current_end = float('-inf')      # STEP 2: Initialize current_end
        count = 0                       # STEP 2: Initialize count
        for i, pair in enumerate(pairs):
            if pair[0] > current_end:   # STEP 3-5: Check if pair can be included
                count += 1             # STEP 3-5: Include pair
                current_end = pair[1]  # STEP 3-5: Update current_end
        return count                   # STEP 6,10: Return final count

if __name__ == '__main__':
    sol = Solution()
    print(sol.findLongestChain([[1,2],[2,3],[3,4]]))  # Output: 2

📊

Longest Chain of Pairs - Watch the Algorithm Execute, Step by Step

Watching each pair's evaluation and how the chain end updates helps you understand the greedy choice and why sorting by end time is crucial.

Initializing dp array

i\w	0	1	2
i=0	?	?	?

[1,2]

Initializing dp array

i\w	0	1	2
i=0	?	?	?

Item 0 - wt:1 val:2

i\w	0	1	2
i=0	1	?	?

Included

Item 1 - wt:1 val:2

i\w	0	1	2
i=0	1	1	?

Skipped

Item 2 - wt:1 val:4

i\w	0	1	2
i=0	1	1	2

Included

Item 2 - wt:1 val:4

i\w	0	1	2
i=0	1	1	2

Answer

Item 2 - wt:1 val:4

i\w	0	1	2
i=0	1	1	2

Included

Item 2 - wt:1 val:4

i\w	0	1	2
i=0	1	1	2

1

Item 2 - wt:1 val:4

i\w	0	1	2
i=0	1	1	2

Active

Item 2 - wt:1 val:4

i\w	0	1	2
i=0	1	1	2

Answer

Key Takeaways

✓ Sorting pairs by their end time is essential for the greedy approach to work correctly.

Without sorting, the algorithm might include pairs in a suboptimal order, missing longer chains.

✓ The chain length count only increases when a pair's start is strictly greater than the current chain end.

This strict inequality ensures pairs do not overlap or touch, maintaining chain validity.

✓ Pairs that cannot be chained are skipped, but the count remains stable, showing monotonic growth.

This helps visualize how the algorithm prunes invalid pairs without losing progress.

Practice

1. You are given a binary tree where each node contains a non-negative integer value representing money. You want to maximize the amount of money you can rob without robbing two directly connected nodes (parent and child). Which algorithmic approach guarantees an optimal solution for this problem?

easy

A. Dynamic Programming on trees using post-order traversal to consider robbing or skipping each node

B. Greedy algorithm that always robs the node with the highest value first

C. Simple recursion without memoization that tries all combinations

D. Breadth-first search to select nodes level by level

Solution

1. Step 1: Understand problem constraints

   The problem requires maximizing sum of node values without robbing adjacent nodes, which is a classic tree DP problem.

2. Step 2: Identify suitable algorithm

   Greedy and BFS fail because they don't consider subtree dependencies; simple recursion is exponential. DP on trees with post-order traversal correctly aggregates rob/not_rob states bottom-up.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   DP on trees handles adjacency constraints optimally [OK]

Hint: Tree DP with post-order traversal handles adjacency constraints [OK]

Common Mistakes:

• Thinking greedy or BFS can solve adjacency constraints optimally

2. 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

3. Examine the following code snippet for longest increasing path. Which line contains a subtle bug that can cause incorrect results or runtime errors?

medium

A. Line returning max value from dp

B. Line initializing dp with 1s

C. Line sorting cells by their matrix values

D. Line checking if matrix[x][y] > matrix[i][j] without boundary check

Solution

1. Step 1: Identify missing boundary check

   The code accesses matrix[x][y] without verifying x,y are within matrix bounds, risking IndexError.

2. Step 2: Confirm other lines are correct

   dp initialization, sorting, and return statements are correct and standard.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Boundary checks are essential before accessing neighbors [OK]

Hint: Always check boundaries before neighbor access [OK]

Common Mistakes:

• Forgetting boundary checks causing crashes
• Misplaced memoization causing TLE
• Incorrect dp initialization

4. 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

5. Suppose the problem is modified so that you can revisit cells multiple times (cycles allowed) and the path must be strictly increasing. Which approach correctly handles this variant without infinite loops?

hard

A. Use DFS with memoization and cycle detection to avoid infinite loops

B. Use topological sort based DP after detecting and removing cycles

C. Use BFS ignoring value order to find longest path

D. Use DFS without memoization to explore all paths

Solution

1. Step 1: Understand cycle implications

   Allowing revisits with strict increasing values can create cycles if not carefully handled, risking infinite loops.

2. Step 2: Identify safe approach

   DFS with memoization plus cycle detection prevents infinite recursion and correctly computes longest paths even with cycles.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Memoization + cycle detection ensures correctness and termination [OK]

Hint: Cycle detection needed when revisits allowed [OK]

Common Mistakes:

• Using naive DFS causing infinite loops
• Ignoring cycles in topological sort
• Using BFS without value order