Russian Doll Envelopes

📋

Russian Doll Envelopes

Edge cases:

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IDE

def maxEnvelopes(envelopes: List[List[int]]) -> int:public int maxEnvelopes(int[][] envelopes)int maxEnvelopes(vector<vector<int>>& envelopes)function maxEnvelopes(envelopes)

def maxEnvelopes(envelopes: List[List[int]]) -> int:
    # Write your solution here
    pass

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

#include <vector>
using namespace std;

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

function maxEnvelopes(envelopes) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: 2Using LIS on widths only or ignoring height dimension.✅ Sort envelopes by width ascending and height descending, then apply LIS on heights with strict increasing condition.

Wrong: 4Greedy approach picking envelopes by width only, ignoring height constraints.✅ Use LIS on heights after sorting widths ascending and heights descending to ensure strict nesting.

Wrong: 0Not handling empty input or returning uninitialized dp max value.✅ Add base case to return 0 if envelopes list is empty.

Wrong: 3Allowing non-strict increasing heights in LIS, counting envelopes with equal heights.✅ Enforce strict inequality in LIS comparison: next height must be strictly greater than previous.

✓

Test Cases

Focus on handling empty and minimal inputs correctly.

Understand sorting order and LIS strictness to avoid common pitfalls.

Implement LIS with binary search to handle large inputs efficiently.

t1_01basic

Input{"envelopes":[[5,4],[6,4],[6,7],[2,3]]}

Expected3

⏱ Performance - must finish in 2000ms

The sequence [2,3] fits into [5,4], which fits into [6,7], forming a chain of length 3.

💡 Sort envelopes by width ascending and height ascending before processing.

💡 Use Longest Increasing Subsequence (LIS) on heights after sorting widths.

💡 Sort by width ascending and height descending to handle equal widths, then find LIS on heights.

Why it failed: Incorrect output means LIS logic or sorting order is wrong. Fix by sorting widths ascending and heights descending, then apply LIS on heights.

✓ Correctly finds the longest chain of envelopes nested.

t1_02basic

Input{"envelopes":[[1,1],[2,2],[3,3],[4,4]]}

Expected4

⏱ Performance - must finish in 2000ms

All envelopes can nest inside each other in increasing order of size, chain length is 4.

💡 Check if sorting by width and height ascending produces a strictly increasing sequence.

💡 Apply LIS on heights after sorting to find maximum nesting.

💡 Since all envelopes increase strictly, LIS length equals number of envelopes.

Why it failed: If output less than 4, LIS or sorting is incorrect. Ensure strict increasing condition and correct LIS implementation.

✓ Correctly identifies full nesting chain of length 4.

t2_01edge

Input{"envelopes":[]}

Expected0

⏱ Performance - must finish in 2000ms

Empty input means no envelopes to nest, so result is 0.

💡 Consider the case when input list is empty.

💡 Return 0 immediately if no envelopes are given.

💡 Add a base case check for empty input before processing.

Why it failed: Returning non-zero means missing empty input check. Fix by returning 0 if envelopes list is empty.

✓ Correctly handles empty input by returning 0.

t2_02edge

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

Expected1

⏱ Performance - must finish in 2000ms

Only one envelope means maximum nesting is 1.

💡 Check behavior when only one envelope is present.

💡 LIS on single element should return 1.

💡 Ensure code handles single-element input without errors.

Why it failed: If output is not 1, LIS initialization or base case is wrong. Fix by initializing dp or LIS length to 1 for single element.

✓ Correctly returns 1 for single envelope input.

t2_03edge

Input{"envelopes":[[3,3],[3,3],[3,3]]}

Expected1

⏱ Performance - must finish in 2000ms

All envelopes have identical width and height, so only one can be chosen.

💡 Check how equal widths and heights affect nesting.

💡 Sorting by width ascending and height descending helps handle duplicates.

💡 LIS on heights with duplicates should count only one envelope in chain.

Why it failed: If output > 1, LIS or sorting does not handle equal envelopes correctly. Fix by sorting heights descending for equal widths and enforcing strict increasing condition.

✓ Correctly returns 1 when all envelopes are identical.

t2_04edge

Input{"envelopes":[[5,5],[4,4],[3,3],[2,2],[1,1]]}

Expected1

⏱ Performance - must finish in 2000ms

Envelopes sorted in descending order; no nesting possible beyond one envelope.

💡 Check sorting order and LIS logic for descending inputs.

💡 LIS requires strictly increasing sequences; descending order yields length 1.

💡 Ensure sorting is ascending before LIS.

Why it failed: If output > 1, sorting or LIS logic is incorrect. Fix by sorting ascending and enforcing strict LIS.

✓ Correctly returns 1 for descending order envelopes.

t3_01corner

Input{"envelopes":[[1,10],[2,9],[3,8],[4,7],[5,6]]}

Expected1

⏱ Performance - must finish in 2000ms

Widths increase but heights decrease, so no two envelopes can nest; max chain length is 1.

💡 Check if sorting by width ascending and height descending is done correctly.

💡 LIS on heights must be strictly increasing; here heights decrease so LIS length is 1.

💡 Fix LIS to require strictly increasing heights, not non-decreasing.

Why it failed: If output > 1, LIS allows non-strict increasing heights. Fix by enforcing strict inequality in LIS comparison.

✓ Correctly identifies no nesting possible beyond single envelope.

t3_02corner

Input{"envelopes":[[2,3],[5,4],[6,4],[6,7]]}

Expected3

⏱ Performance - must finish in 2000ms

Tests correct handling of envelopes with equal widths but different heights; max chain length is 3.

💡 Sort by width ascending and height descending to handle equal widths.

💡 Apply LIS on heights to find max nesting chain.

💡 Ensure LIS uses strict increasing condition on heights.

Why it failed: If output less than 3, sorting or LIS logic for equal widths is incorrect. Fix by sorting heights descending for equal widths and strict LIS.

✓ Correctly handles equal widths with different heights.

t3_03corner

Input{"envelopes":[[1,1],[1,2],[1,3],[2,2],[2,3],[3,3]]}

Expected3

⏱ Performance - must finish in 2000ms

Tests greedy trap: picking envelopes by width only fails; correct LIS on heights needed.

💡 Greedy by width alone fails; sort by width ascending and height descending.

💡 Use LIS on heights to find max nesting chain.

💡 Fix greedy approach by applying LIS with binary search on heights.

Why it failed: If output is 4 or more, greedy approach is used incorrectly. Fix by applying LIS on heights after sorting widths ascending and heights descending.

✓ Correctly avoids greedy trap and finds max nesting length.

t4_01performance

Input{"envelopes":[[1,999999999],[2,999999998],[3,999999997],[4,999999996],[5,999999995],[6,999999994],[7,999999993],[8,999999992],[9,999999991],[10,999999990],[11,999999989],[12,999999988],[13,999999987],[14,999999986],[15,999999985],[16,999999984],[17,999999983],[18,999999982],[19,999999981],[20,999999980],[21,999999979],[22,999999978],[23,999999977],[24,999999976],[25,999999975],[26,999999974],[27,999999973],[28,999999972],[29,999999971],[30,999999970],[31,999999969],[32,999999968],[33,999999967],[34,999999966],[35,999999965],[36,999999964],[37,999999963],[38,999999962],[39,999999961],[40,999999960],[41,999999959],[42,999999958],[43,999999957],[44,999999956],[45,999999955],[46,999999954],[47,999999953],[48,999999952],[49,999999951],[50,999999950],[51,999999949],[52,999999948],[53,999999947],[54,999999946],[55,999999945],[56,999999944],[57,999999943],[58,999999942],[59,999999941],[60,999999940],[61,999999939],[62,999999938],[63,999999937],[64,999999936],[65,999999935],[66,999999934],[67,999999933],[68,999999932],[69,999999931],[70,999999930],[71,999999929],[72,999999928],[73,999999927],[74,999999926],[75,999999925],[76,999999924],[77,999999923],[78,999999922],[79,999999921],[80,999999920],[81,999999919],[82,999999918],[83,999999917],[84,999999916],[85,999999915],[86,999999914],[87,999999913],[88,999999912],[89,999999911],[90,999999910],[91,999999909],[92,999999908],[93,999999907],[94,999999906],[95,999999905],[96,999999904],[97,999999903],[98,999999902],[99,999999901],[100,999999900]]}

Expectednull

⏱ Performance - must finish in 2000ms

n=100, O(n log n) solution must complete within 2 seconds; brute force O(n^2) likely to TLE.

💡 Use sorting and LIS with binary search for O(n log n) complexity.

💡 Avoid nested loops for LIS; use patience sorting or binary search.

💡 Optimize sorting and LIS steps to meet time constraints.

Why it failed: TLE indicates O(n^2) or brute force approach used. Fix by implementing LIS with binary search for O(n log n) time.

✓ Efficient O(n log n) solution confirmed by passing performance test.

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. Consider the following code snippet implementing the topological sort based DP for longest increasing path. Given the input matrix = [[9,9,4],[6,6,8],[2,1,1]], what is the final returned value?

easy

A. 3

B. 5

C. 4

D. 6

Solution

1. Step 1: Trace dp updates for smallest cell (1,2) with value 1

   Initialize dp with 1s. Process cells in ascending order: start from value 1 cells, update neighbors with larger values increasing dp accordingly.

2. Step 2: Identify longest path length after processing all cells

   Longest increasing path is 4, e.g., 1 -> 2 -> 6 -> 9.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Longest path length matches known example output [OK]

Hint: Trace dp updates in ascending order [OK]

Common Mistakes:

• Off-by-one in dp updates
• Confusing max path length with number of edges
• Ignoring boundary checks

4. Consider the following Fenwick Tree based code snippet for counting the number of longest increasing subsequences. Given the input array nums = [1, 3, 2], what is the final returned count of LIS?

easy

A. 2

B. 3

C. 1

D. 4

Solution

1. Step 1: Trace Fenwick tree updates for nums=[1,3,2]

   For 1: query(0) -> (0,0), count=1, update(1,(1,1)) For 3: query(2) -> max_len=1,count=1, update(3,(2,1)) For 2: query(1) -> max_len=1,count=1, update(2,(2,1))

2. Step 2: Final query on full range

   query(3) returns max_len=2, total_count=2 (from 3 and 2 updates), so total LIS count is 2.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Two LIS: [1,2] and [1,3] [OK]

Hint: Count increments when length ties in Fenwick tree [OK]

Common Mistakes:

• Off-by-one in Fenwick indexing
• Forgetting to set count=1 when zero
• Mixing length and count in update

5. What is the time complexity of the optimal greedy solution that sorts pairs by their end values and then iterates through them once to find the longest chain?

medium

A. O(n log n)

B. O(n²)

C. O(n)

D. O(n log n + n²)

Solution

1. Step 1: Identify sorting cost

   Sorting pairs by end values takes O(n log n) time.

2. Step 2: Identify iteration cost

   Iterating through sorted pairs once is O(n).

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Dominant cost is sorting -> O(n log n) [OK]

Hint: Sorting dominates, iteration is linear [OK]

Common Mistakes:

• Assuming iteration is O(n²) due to nested loops
• Confusing DP approach complexity with greedy
• Ignoring sorting cost