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 an array representing the heights of adjacent bars in a histogram. The task is to find the largest rectangular area that can be formed within the bounds of the histogram bars. Which algorithmic approach guarantees an optimal O(n) time solution for this problem?

easy

A. Greedy approach selecting the tallest bar and expanding outwards

B. Brute force nested loops checking all possible rectangles

C. Dynamic programming with memoization of subproblems

D. Single pass stack-based approach that tracks indices of bars

Solution

1. Step 1: Understand problem constraints and requirements

   The problem requires finding the largest rectangle in a histogram efficiently, which involves checking contiguous bars and their heights.

2. Step 2: Identify optimal approach using stack

   The single pass stack-based approach uses a stack to keep track of indices of bars in ascending order, allowing calculation of maximal rectangle areas in O(n) time by popping bars when a smaller height is encountered.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Stack approach is known optimal for largest rectangle in histogram [OK]

Hint: Stack-based approach achieves O(n) time [OK]

Common Mistakes:

• Assuming brute force is efficient enough
• Confusing this with a DP problem
• Thinking greedy expansion from tallest bar suffices

2. Consider the following optimized sliding window code for Minimum Window Substring:

def min_window_optimized(s: str, t: str) -> str:
    if not s or not t:
        return ""
    from collections import defaultdict
    t_count = [0] * 128
    window_count = [0] * 128
    for c in t:
        t_count[ord(c)] += 1
    have, need = 0, sum(1 for x in t_count if x > 0)
    res, res_len = [-1, -1], float('inf')
    left = 0
    for right in range(len(s)):
        c = s[right]
        window_count[ord(c)] += 1
        if t_count[ord(c)] > 0 and window_count[ord(c)] == t_count[ord(c)]:
            have += 1
        while have == need:
            if (right - left + 1) < res_len:
                res = [left, right]
                res_len = right - left + 1
            window_count[ord(s[left])] -= 1
            if t_count[ord(s[left])] > 0 and window_count[ord(s[left])] < t_count[ord(s[left])]:
                have -= 1
            left += 1
    left, right = res
    return s[left:right+1] if res_len != float('inf') else ""

What is the returned substring for s = "ADOBECODEBANC" and t = "ABC"?

easy

A. "ADOBEC"

B. "CODEBA"

C. "BANC"

D. "DOBECODEBA"

Solution

1. Step 1: Trace window expansion

   Expand right pointer until window contains all chars 'A', 'B', 'C'. The first valid window is "ADOBEC" but not minimal.

2. Step 2: Shrink window to find minimal substring

   By moving left pointer, the minimal window found is "BANC" which contains all required chars with minimal length.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   "BANC" is the known minimal window substring for this input [OK]

Hint: Minimal window substring for this input is "BANC" [OK]

Common Mistakes:

• Returning first valid window instead of minimal
• Off-by-one errors in substring indices

3. Consider the following buggy code snippet for the optimal median of two sorted arrays. Which line contains the subtle bug that can cause incorrect median calculation or runtime error?

medium

A. Line 10: max_left_nums1 = float('-inf') if i == 0 else nums1[i - 1]

B. Line 7: i = (low + high) // 2

C. Line 17: elif max_left_nums1 > min_right_nums2:

D. Line 2: if len(nums1) < len(nums2): nums1, nums2 = nums2, nums1

Solution

1. Step 1: Check array swap condition

   The code swaps arrays only if len(nums1) < len(nums2), which is incorrect. It should swap if nums1 is larger to ensure binary search on smaller array.

2. Step 2: Consequences of wrong swap

   Without correct swap, binary search range can be large, causing inefficiency or index errors.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Swap condition reversed causes subtle bugs [OK]

Hint: Swap arrays so binary search is on smaller one [OK]

Common Mistakes:

• Swapping arrays incorrectly or not at all
• Off-by-one errors in partition indices
• Incorrect median calculation for odd/even lengths

4. The following code attempts to find the minimum window substring but contains a subtle bug:

def min_window_buggy(s: str, t: str) -> str:
    if not s or not t:
        return ""
    t_count = [0] * 128
    window_count = [0] * 128
    for c in t:
        t_count[ord(c)] += 1
    have, need = 0, sum(1 for x in t_count if x > 0)
    res, res_len = [-1, -1], float('inf')
    left = 0
    for right in range(len(s)):
        c = s[right]
        window_count[ord(c)] += 1
        if t_count[ord(c)] > 0 and window_count[ord(c)] == t_count[ord(c)]:
            have += 1
        while have == need:
            if (right - left + 1) < res_len:
                res = [left, right]
                res_len = right - left + 1
            window_count[ord(s[left])] -= 1
            # BUG: Missing decrement of 'have' when window_count falls below t_count
            left += 1
    left, right = res
    return s[left:right+1] if res_len != float('inf') else ""

Which line contains the bug?

medium

A. Line where window_count[ord(s[left])] is decremented

B. Line where have is incremented when window_count matches t_count

C. Line where left pointer is incremented inside the while loop

D. Line where t_count array is initialized from string t

Solution

1. Step 1: Identify decrement of window_count

   When shrinking the window, window_count for s[left] is decremented, but the code does not check if this causes window_count to fall below t_count.

2. Step 2: Missing decrement of 'have'

   Without decrementing 'have' when window_count falls below required count, the while loop condition remains true incorrectly, causing wrong results.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Bug is missing update of 'have' after decrementing window_count [OK]

Hint: Must update 'have' when window_count falls below required count [OK]

Common Mistakes:

• Forgetting to decrement 'have' when shrinking window
• Incorrectly updating pointers without adjusting counts

5. Suppose the Burst Balloons problem is modified so that you can burst the same balloon multiple times (balloons regenerate after bursting). Which of the following changes to the algorithm correctly adapts the solution to this variant?

hard

A. Modify the DP to consider infinite bursting by changing the recurrence to allow repeated indices and use memoization with state tracking of burst counts

B. Switch to a greedy approach that always bursts the balloon with the highest product value repeatedly

C. Use the same bottom-up DP but allow dp states to revisit the same balloon multiple times without changes

D. Use a top-down DFS without memoization to explore all sequences with repeated bursts

Solution

1. Step 1: Understand the new problem variant

   Allowing multiple bursts of the same balloon means the state must track how many times each balloon has been burst or if it is still available.

2. Step 2: Adapt DP to handle repeated bursting

   The original DP assumes each balloon bursts once. To handle repeats, the recurrence must be modified to track burst counts or states, and memoization must store these states to avoid recomputation.

3. Step 3: Evaluate options

   Use the same bottom-up DP but allow dp states to revisit the same balloon multiple times without changes ignores state changes and is incorrect. Switch to a greedy approach that always bursts the balloon with the highest product value repeatedly is greedy and suboptimal. Use a top-down DFS without memoization to explore all sequences with repeated bursts is exponential without memoization. Modify the DP to consider infinite bursting by changing the recurrence to allow repeated indices and use memoization with state tracking correctly modifies DP with state tracking and memoization.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   State tracking and memoization needed for repeated bursts [OK]

Hint: Repeated bursting requires state tracking in DP [OK]

Common Mistakes:

• Using original DP unchanged
• Greedy approach
• Ignoring memoization