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. You are given a list of words sorted lexicographically in an unknown language. You need to determine the order of characters in this alien language. Which algorithmic approach guarantees an optimal solution to find a valid character order?

easy

A. Topological sort using Kahn's algorithm with cycle detection and edge deduplication

B. Greedy approach comparing adjacent words without cycle detection

C. Dynamic programming to find longest common subsequence between words

D. Brute force pairwise comparison without graph construction

Solution

1. Step 1: Understand problem requires ordering characters based on precedence constraints

   Characters form a directed graph where edges represent order constraints derived from adjacent words.

2. Step 2: Identify algorithm that finds linear order respecting all edges and detects cycles

   Kahn's algorithm performs topological sort with cycle detection and can handle duplicate edges efficiently.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Topological sort with cycle detection is standard for ordering with partial constraints [OK]

Hint: Topological sort solves ordering with cycle detection [OK]

Common Mistakes:

• Assuming greedy comparison suffices
• Using DP unrelated to ordering
• Ignoring cycle detection

2. You are given two sorted arrays and need to find the median of their combined elements efficiently without fully merging them. Which approach guarantees the optimal time complexity for this problem?

easy

A. Merge both arrays completely and then find the median from the merged array.

B. Use binary search to partition the smaller array and find the correct partition in the other array.

C. Use a two-pointer approach to partially merge elements until the median position is reached.

D. Apply dynamic programming to store medians of subarrays and combine results.

Solution

1. Step 1: Understand the problem constraints

   The problem requires finding the median of two sorted arrays without fully merging them to achieve better than O(m+n) time.

2. Step 2: Identify the optimal approach

   Binary search on the smaller array to find a partition that balances elements on both sides guarantees O(log(min(m,n))) time, which is optimal.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Binary search on partition is the known optimal solution [OK]

Hint: Optimal median uses binary search on smaller array [OK]

Common Mistakes:

• Assuming full merge is necessary
• Using two pointers without binary search
• Confusing DP with median finding

3. You are given an undirected graph with n nodes. You need to find the shortest path length that visits every node at least once. Which algorithmic approach guarantees finding the optimal solution efficiently for this problem?

easy

A. Simple DFS traversal with backtracking to explore all paths

B. Greedy algorithm that always moves to the nearest unvisited node

C. Dynamic programming on subsets without BFS, enumerating all permutations

D. Bidirectional BFS with state compression using bitmasks to track visited nodes

Solution

1. Step 1: Understand problem constraints

   The problem requires visiting all nodes with the shortest path, which is a variant of the Traveling Salesman Problem (TSP) on an unweighted graph.

2. Step 2: Identify suitable algorithm

   Greedy and simple DFS approaches do not guarantee shortest paths and are inefficient. DP on subsets is correct but has O(n*2^n) complexity. Bidirectional BFS with bitmask state compression reduces complexity to O(n*2^(n/2)), making it optimal and efficient.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Bitmask BFS tracks visited nodes efficiently and bidirectional search halves the state space [OK]

Hint: Bitmask + bidirectional BFS optimizes shortest path visiting all nodes [OK]

Common Mistakes:

• Assuming greedy nearest neighbor always finds shortest path
• Using DFS without pruning leads to exponential blowup
• Confusing DP on subsets with simpler DP variants

4. Consider the following buggy implementation of the two-pointer approach for trapping rain water. Which line contains the subtle bug that can cause incorrect trapped water calculation?

medium

A. Line 7: if height[left] < height[right]:

B. Line 15: if height[right] >= right_max:

C. Line 17: water += right_max - height[right]

D. Line 9: if height[left] > left_max:

Solution

1. Step 1: Compare condition for updating left_max

   The original buggy code used '>' instead of '>=' which causes left_max not to update when height[left] equals left_max.

2. Step 2: Effect of bug

   When height[left] == left_max, water is incorrectly added (left_max - height[left] = 0), which is harmless but logically incorrect and can cause issues in some cases.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Using '>=' ensures correct max update and prevents unnecessary water addition [OK]

Hint: Use '>=' when updating max heights to avoid underestimation [OK]

Common Mistakes:

• Using '>' instead of '>=' for max updates
• Incorrect pointer movement
• Adding water without checking max heights

5. Suppose the problem is modified so that digits can be reused any number of times to form expressions (e.g., "1+1+2+3" from "123"). Which modification to the backtracking algorithm correctly handles this variant without infinite loops?

hard

A. Remove the index increment in recursive calls to allow reuse of digits indefinitely

B. Add a visited set to track used indices and prevent reusing digits more than once

C. Allow recursive calls to reuse digits by not incrementing index, but limit expression length to avoid infinite recursion

D. Modify recursion to restart from index 0 after each operator insertion, but track expression length to prevent infinite loops

Solution

1. Step 1: Understand reuse implications

   Allowing reuse means digits can be chosen multiple times, so index must be reset or not incremented to allow reuse.

2. Step 2: Prevent infinite recursion

   Resetting index to 0 after operator insertion allows reuse, but expression length must be tracked and limited to avoid infinite loops.

3. Step 3: Evaluate options

   Modify recursion to restart from index 0 after each operator insertion, but track expression length to prevent infinite loops correctly resets index and tracks length. Remove the index increment in recursive calls to allow reuse of digits indefinitely causes infinite loops by never incrementing index. Add a visited set to track used indices and prevent reusing digits more than once forbids reuse. Allow recursive calls to reuse digits by not incrementing index, but limit expression length to avoid infinite recursion risks infinite recursion without explicit length limit.

4. Final Answer:

   Option D -> Option D

5. Quick Check:

   Reset index with length limit prevents infinite recursion while allowing reuse [OK]

Hint: Reuse requires index reset plus recursion depth control [OK]

Common Mistakes:

• Not limiting recursion depth causing infinite loops
• Using visited set forbidding reuse