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 Python code that merges k sorted arrays using divide and conquer. Given the input arrays [[1,4,5],[1,3,4],[2,6]], what is the value of merged_arrays after the first iteration of the while loop inside merge_k_sorted_arrays?

easy

A. [[1,1,3,4,4,5,2,6]]

B. [[1,1,3,4,4,5],[2,6]]

C. [[1,3,4,5],[1,2,4,6]]

D. [[1,1,3,4,4,5],[2,6]]

Solution

1. Step 1: Trace merging pairs in first iteration

   Arrays are [[1,4,5],[1,3,4],[2,6]]. Merge first two: merge_two_arrays([1,4,5],[1,3,4]) -> [1,1,3,4,4,5]. The last array [2,6] remains as is because no pair.

2. Step 2: Form merged_arrays after first iteration

   merged_arrays = [[1,1,3,4,4,5],[2,6]]

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   First pair merged correctly, last array unchanged -> matches [[1,1,3,4,4,5],[2,6]] [OK]

Hint: Merge pairs left to right, last array unchanged if odd count [OK]

Common Mistakes:

• Merging last array incorrectly
• Off-by-one in loop
• Appending arrays instead of merging

2. Suppose the problem is modified so that elements from nums2 can be reused multiple times in pairs (i.e., pairs can reuse the same element from nums2 multiple times). Which modification to the min-heap approach correctly handles this variant?

hard

A. Push all pairs (nums1[i] + nums2[j]) for all i, j into the heap initially and pop k times.

B. After popping a pair (i, j), push (i, j+1) and also push (i+1, j) if within bounds, allowing reuse of nums2 elements.

C. After popping a pair (i, j), push (i, j) again to allow reuse of nums2[j].

D. Modify the heap to store sums with counts of usage and only push new pairs when usage count is less than k.

Solution

1. Step 1: Understand reuse requirement

   Allowing reuse of nums2 elements means after using nums2[j], we can also try nums2[j+1] and nums2[j] again with next nums1 element.

2. Step 2: Modify heap push logic

   After popping (i, j), pushing (i, j+1) continues along nums2, and pushing (i+1, j) allows pairing next nums1 element with the same nums2[j], enabling reuse.

3. Step 3: Reject other options

   Push all pairs (nums1[i] + nums2[j]) for all i, j into the heap initially and pop k times. is brute force and inefficient. After popping a pair (i, j), push (i, j) again to allow reuse of nums2[j]. causes infinite loops by pushing same pair repeatedly. Modify the heap to store sums with counts of usage and only push new pairs when usage count is less than k. is unnecessarily complex and not standard.

4. Final Answer:

   Option B -> Option B

5. Quick Check:

   After popping a pair (i, j), push (i, j+1) and also push (i+1, j) if within bounds, allowing reuse of nums2 elements. correctly expands pairs to allow reuse without infinite loops [OK]

Hint: Push next nums1 with same nums2 to allow reuse [OK]

Common Mistakes:

• Pushing same pair repeatedly causing infinite loops
• Brute forcing all pairs initially
• Overcomplicating usage counts

3. Suppose the array elements can be reused multiple times (duplicates allowed and repeated). Which modification to the k-sorted array sorting algorithm is necessary to correctly handle this scenario?

hard

A. No modification needed; the existing min-heap approach works correctly with duplicates.

B. Use a max-heap instead of a min-heap to handle duplicates correctly.

C. Add a frequency map to track counts and adjust heap operations accordingly.

D. Sort the array fully using a standard sort to handle duplicates reliably.

Solution

1. Step 1: Understand duplicates in k-sorted array

   Duplicates do not affect the heap property; min-heap handles equal elements correctly.

2. Step 2: Confirm no algorithmic change needed

   Heap operations remain valid; duplicates are naturally ordered by heapq.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Min-heap supports duplicates without extra data structures [OK]

Hint: Min-heap naturally supports duplicates without changes [OK]

Common Mistakes:

• Thinking max-heap is needed for duplicates
• Adding unnecessary frequency maps
• Defaulting to full sort unnecessarily

4. If the problem changes so that super ugly numbers can be reused multiple times (i.e., duplicates allowed in the sequence), which modification to the min-heap approach is correct?

hard

A. Use a BFS approach with a queue instead of a heap to allow duplicates naturally.

B. Remove the duplicate check when appending values to ugly and push new heap elements as usual.

C. Keep the duplicate check but also maintain a global set to avoid pushing duplicates into the heap.

D. Modify the heap push to multiply primes with the last appended ugly number instead of indexed ugly numbers.

Solution

1. Step 1: Understand the reuse variant

   Allowing duplicates means we no longer need to skip repeated values; all generated multiples are valid.

2. Step 2: Adjust algorithm accordingly

   Removing the duplicate check when appending allows duplicates; heap push logic remains unchanged.

3. Step 3: Verify correctness

   Heap still merges sequences correctly; duplicates appear naturally without filtering.

4. Final Answer:

   Option B -> Option B

5. Quick Check:

   Removing duplicate check aligns with allowing reuse [OK]

Hint: Allow duplicates -> remove duplicate filtering, keep heap logic [OK]

Common Mistakes:

• Keeping duplicate check breaks reuse requirement
• Adding global sets wastes space unnecessarily
• Changing multiplication logic breaks sequence order

5. Suppose the problem changes: now you want to find the nth ugly number but you can reuse the same ugly number multiple times in generating new numbers (i.e., duplicates allowed). Which modification to the min-heap approach correctly handles this variant without losing correctness or efficiency?

hard

A. Remove the 'seen' set entirely and push all generated numbers into the heap, allowing duplicates.

B. Keep the 'seen' set but reset it after every iteration to allow reuse of numbers in the next iteration.

C. Use a counter map to track how many times each number is generated and only push if count is below a threshold.

D. Keep the 'seen' set as is to avoid duplicates, since duplicates do not affect correctness or efficiency.

Solution

1. Step 1: Understand the variant allowing duplicates

   Allowing reuse means duplicates are valid and should be generated and processed multiple times.

2. Step 2: Modify the algorithm accordingly

   Removing the 'seen' set allows duplicates to be pushed and popped, correctly reflecting the new problem constraints.

3. Step 3: Consider efficiency

   Though duplicates increase heap size, this is necessary to correctly generate all numbers under the new rules.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Removing 'seen' allows duplicates as required [OK]

Hint: Allow duplicates by removing 'seen' when reuse is allowed [OK]

Common Mistakes:

• Keeping 'seen' and missing duplicates
• Resetting 'seen' incorrectly
• Using counters unnecessarily