Smallest Range Covering K Lists

📋

Smallest Range Covering K Lists

Edge cases:

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IDE

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

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

class Solution {
    public int[] smallestRangeBruteForce(List<List<Integer>> nums) {
        // Write your solution here
        return new int[0];
    }
}

#include <vector>
using namespace std;

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

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

Common Bugs to Avoid

Wrong: [1,3]Greedy approach picking minimum elements without ensuring coverage of all lists.✅ Use a min-heap to track current minimum element and advance pointers only when all lists are covered.

Wrong: [] or nullFailing to handle empty input or lists with zero elements gracefully.✅ Add input validation to return [] immediately if any list is empty or input is empty.

Wrong: [0,0]Incorrectly assuming zero range when lists do not overlap or missing to update range properly.✅ Update range only when all lists are covered and track min and max elements correctly.

Wrong: [4,10]Not minimizing the range properly, returning a valid but not minimal range.✅ Continuously update the range when a smaller range covering all lists is found.

✓

Test Cases

Focus on handling empty and minimal inputs correctly before tackling edge cases.

Review your approach for correctness on tricky inputs that can break greedy or naive solutions.

Optimize your solution to handle large inputs efficiently using heaps or two-pointer methods.

t1_01basic

Input{"nums":[[4,10,15,24,26],[0,9,12,20],[5,18,22,30]]}

Expected[20,24]

⏱ Performance - must finish in 2000ms

The range [20,24] includes 24 from the first list, 20 from the second list, and 22 from the third list, covering all lists with the smallest possible range.

💡 Consider merging all lists and tracking elements from each list.

💡 Use a min-heap to efficiently find the current minimum and track the maximum element in the current window.

💡 Slide the window to minimize the range while ensuring all lists are represented.

Why it failed: Incorrect output means the algorithm failed to track all lists in the range or did not minimize the range properly. Fix by ensuring the window always contains at least one element from each list and update the range only when all lists are covered.

✓ Correctly finds the smallest range covering all lists.

t1_02basic

Input{"nums":[[1,5,8],[4,12],[7,8,10]]}

Expected[4,8]

⏱ Performance - must finish in 2000ms

The range [4,8] includes 5 from the first list, 4 from the second list, and 7 or 8 from the third list, covering all lists with the smallest possible range.

💡 Track the current minimum and maximum elements from the pointers in each list.

💡 Use a heap to efficiently find the minimum element and update the range accordingly.

💡 Move the pointer of the list that contributed the minimum element to try to find a smaller range.

Why it failed: If output is wrong, likely the algorithm does not correctly update pointers or fails to maintain coverage of all lists. Fix by advancing the pointer of the list with the current minimum element and updating the range only when all lists are included.

✓ Correctly computes the minimal range covering all lists.

t2_01edge

Input{"nums":[]}

Expected[]

⏱ Performance - must finish in 2000ms

Empty input means no lists to cover, so no valid range exists; return empty list.

💡 Check if the input list is empty before processing.

💡 Return an empty list immediately if there are no lists.

💡 Add a guard clause to handle empty input to avoid errors.

Why it failed: Algorithm crashes or returns non-empty output on empty input. Fix by adding a check for empty input and returning [] immediately.

✓ Correctly handles empty input by returning an empty list.

t2_02edge

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

Expected[7,7]

⏱ Performance - must finish in 2000ms

Single list with one element means the smallest range is that element itself.

💡 For a single list, the smallest range is the element itself.

💡 Check if k=1 and return the single element as both start and end.

💡 Return [element, element] when only one list with one element is given.

Why it failed: Fails to handle single list or single element case, returning empty or incorrect range. Fix by returning [nums[0][0], nums[0][0]] when only one list with one element.

✓ Correctly returns the single element as the smallest range.

t2_03edge

Input{"nums":[[-10,-5,0],[0,5,10],[10,15,20]]}

Expected[0,10]

⏱ Performance - must finish in 2000ms

Range [0,10] covers 0 from first and second lists and 10 from third list, crossing zero with negative and positive numbers.

💡 Ensure algorithm handles negative and positive numbers correctly.

💡 Check that comparisons and range calculations handle negative values.

💡 Use correct min and max calculations without assumptions about sign.

Why it failed: Incorrect range due to mishandling negative numbers or sign assumptions. Fix by using standard min/max functions without sign bias.

✓ Correctly handles negative and positive numbers in range calculation.

t2_04edge

Input{"nums":[[100000],[100000],[100000]]}

Expected[100000,100000]

⏱ Performance - must finish in 2000ms

All lists contain the same large element; smallest range is that element itself.

💡 Check handling of large values near constraints.

💡 Ensure no integer overflow or incorrect comparisons.

💡 Return [element, element] when all lists share the same element.

Why it failed: Fails due to overflow or incorrect handling of large values. Fix by using appropriate data types and comparisons.

✓ Correctly handles large values without overflow.

t3_01corner

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

Expected[1,1]

⏱ Performance - must finish in 2000ms

All lists contain the same elements; smallest range is [1,1] covering all lists.

💡 Check if all lists have overlapping elements.

💡 If all lists share a common element, the smallest range is that element.

💡 Return [common_element, common_element] when found.

Why it failed: Fails to detect common elements across all lists, returning larger range. Fix by checking intersection or updating range when all lists share an element.

✓ Correctly identifies smallest range when all lists share elements.

t3_02corner

Input{"nums":[[1,5,9],[2,6,10],[3,7,11]]}

Expected[3,7]

⏱ Performance - must finish in 2000ms

Greedy approach picking minimums fails; correct smallest range is [3,7] covering all lists.

💡 Beware of greedy approaches that pick minimums without considering all lists.

💡 Use a sliding window or heap to track coverage of all lists simultaneously.

💡 Advance pointers carefully to maintain coverage and minimize range.

Why it failed: Greedy approach picks [1,3] or [1,2] which misses coverage or is not minimal. Fix by using a min-heap and sliding window to cover all lists.

✓ Correctly avoids greedy trap and finds minimal range.

t3_03corner

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

Expected[3,9]

⏱ Performance - must finish in 2000ms

Tests correct handling of disjoint lists and advancing pointers to find minimal range.

💡 Check that algorithm advances pointer of list with current minimum element.

💡 Ensure range updates only when all lists are covered.

💡 Use a min-heap to efficiently find minimum and update range.

Why it failed: Fails to advance pointers correctly or update range, returning incorrect or larger range. Fix by advancing pointer of list with minimum element and updating range only when all lists covered.

✓ Correctly handles disjoint lists and finds minimal range.

t4_01performance

Input{"nums":[[0,2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36,38,40,42,44,46,48,50,52,54,56,58,60,62,64,66,68,70,72,74,76,78,80,82,84,86,88,90,92,94,96,98],[1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39,41,43,45,47,49,51,53,55,57,59,61,63,65,67,69,71,73,75,77,79,81,83,85,87,89,91,93,95,97,99],[50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99],[-50,-49,-48,-47,-46,-45,-44,-43,-42,-41,-40,-39,-38,-37,-36,-35,-34,-33,-32,-31,-30,-29,-28,-27,-26,-25,-24,-23,-22,-21,-20,-19,-18,-17,-16,-15,-14,-13,-12,-11,-10,-9,-8,-7,-6,-5,-4,-3,-2,-1],[25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74],[10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59],[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49],[20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69],[30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79],[40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89]]}

Expectednull

⏱ Performance - must finish in 2000ms

n=10 lists with up to 50 elements each (total ~500 elements). O(n log n) or O(n log k) solution must complete within 2s. Brute force O(m^k) will TLE.

💡 Brute force enumerating all combinations is exponential and will time out.

💡 Use a min-heap to merge k lists and track ranges efficiently.

💡 Implement sliding window or two pointers on merged array for optimal performance.

Why it failed: TLE occurs due to exponential brute force complexity O(m^k). Fix by implementing O(n log k) heap-based approach or O(n log n) two-pointer approach.

✓ Algorithm runs within time limits using efficient heap or two-pointer approach.

Practice

1. Consider the following code snippet for computing the maximum path sum in a binary tree. Given the tree with root value 1, left child 2, and right child 3, what is the final returned maximum path sum?

easy

A. 3

B. 6

C. 5

D. 4

Solution

1. Step 1: Trace helper on left child (2):

   left and right children are null, so left=0, right=0, max_sum updated to max(-inf, 2+0+0)=2, returns 2+max(0,0)=2.

2. Step 2: Trace helper on right child (3):

   Similarly, max_sum updated to max(2, 3+0+0)=3, returns 3+max(0,0)=3.

3. Step 3: Trace helper on root (1):

   left=2, right=3, max_sum updated to max(3, 1+2+3)=6, returns 1+max(2,3)=4.

4. Final Answer:

   Option B -> Option B

5. Quick Check:

   Max path is 2 -> 1 -> 3 with sum 6 [OK]

Hint: Sum includes node plus max gains from both children [OK]

Common Mistakes:

• Returning sum of one child only
• Off-by-one in max_sum update
• Ignoring zero cutoff for negative gains

2. Consider the following code snippet implementing the optimal solution for Word Search II. Given the board = [['o','a','a','n'], ['e','t','a','e'], ['i','h','k','r'], ['i','f','l','v']] and words = ['oath', 'pea', 'eat', 'rain'], what is the content of the result list res after the function completes?

easy

A. ['oath', 'pea', 'eat', 'rain']

B. ['pea', 'eat']

C. ['oath', 'eat', 'rain']

D. ['oath', 'eat']

Solution

1. Step 1: Trace found words on board

   'oath' can be formed starting at (0,0) -> o-a-t-h; 'eat' can be formed starting at (1,1) -> e-a-t. 'pea' and 'rain' cannot be formed.

2. Step 2: Confirm duplicates avoided

   Code sets found word to None to avoid duplicates, so only unique found words remain.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Only 'oath' and 'eat' are found on the board [OK]

Hint: Only words fully traceable on board appear [OK]

Common Mistakes:

• Including words not on board
• Duplicates in result
• Missing found words

3. The following code attempts to solve the Russian Doll Envelopes problem optimally. Identify the line containing the subtle bug that causes incorrect results on some inputs.

medium

A. Line 4: Sorting envelopes with key (x[0], x[1]) ascending both

B. Line 6: Extracting heights from envelopes

C. Line 8: Using bisect_left to find insertion index

D. Line 11: Replacing tails element at found index

Solution

1. Step 1: Examine sorting key

   Sorting by (width asc, height asc) allows envelopes with same width and increasing height, which breaks strict nesting rules.

2. Step 2: Correct sorting key

   Sorting by (width asc, height desc) prevents envelopes with equal width but increasing height from being counted incorrectly.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Changing sorting to (x[0], -x[1]) fixes the bug [OK]

Hint: Sort height descending for equal widths to avoid counting duplicates [OK]

Common Mistakes:

• Sorting both width and height ascending
• Using <= instead of < in comparisons
• Not handling equal widths properly

4. Suppose the Alien Dictionary problem is extended so that words can be repeated multiple times in the input list, possibly introducing duplicate edges. Which modification to the optimal Kahn's algorithm approach correctly handles this scenario without corrupting the in-degree counts?

hard

A. Use a set to track edges and only add an edge and increment in-degree if it is not already present

B. Reset in-degree counts after building the graph to avoid duplicates

C. Remove the edge deduplication check and increment in-degree for every edge found

D. Use DFS instead of BFS to avoid counting duplicate edges

Solution

1. Step 1: Identify problem with duplicate edges

   Repeated words can cause multiple identical edges, inflating in-degree counts incorrectly.

2. Step 2: Use edge deduplication to prevent multiple increments

   Track edges in a set and only add/increment if edge is new, preserving correct in-degree.

3. Step 3: Confirm BFS with correct in-degree works properly

   Maintains accurate topological order and cycle detection despite duplicates.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Edge deduplication is essential for correctness with repeated words [OK]

Hint: Avoid incrementing in-degree multiple times for same edge [OK]

Common Mistakes:

• Ignoring duplicates
• Resetting in-degree incorrectly
• Switching to DFS unnecessarily

5. Suppose the problem is modified so that the same letter cell on the board can be used multiple times within the same word search path (i.e., reuse of cells is allowed). Which modification to the optimal backtracking with Trie approach correctly handles this variant without causing infinite loops?

hard

A. Use a separate visited set to track cells per path and allow revisiting after backtracking

B. Remove the in-place marking of visited cells and rely solely on Trie pruning

C. Keep the in-place marking but reset visited cells only after exploring all neighbors

D. Modify the Trie to store counts of allowed cell reuse and decrement on each visit

Solution

1. Step 1: Understand reuse impact on visited marking

   Allowing reuse means cells can be revisited, so marking cells as visited must be carefully managed to avoid infinite loops.

2. Step 2: Identify correct marking strategy

   Using a separate visited set per path allows revisiting cells after backtracking, preventing infinite loops while supporting reuse.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Separate visited set per path safely handles cell reuse [OK]

Hint: Track visited cells per path to allow safe reuse [OK]

Common Mistakes:

• Removing marking causes infinite loops
• Using global visited set without reset
• Modifying Trie for cell reuse incorrectly