How do you solve Smallest Range Covering Elements from K Lists on LeetCode?

Smallest Range Covering Elements from K Lists (LeetCode #632) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log k) time complexity and O(k) space complexity. Key insight: Using a min-heap allows efficient tracking of the smallest element across multiple lists.

What is the brute force solution for Smallest Range Covering Elements from K Lists?

The brute force approach for Smallest Range Covering Elements from K Lists is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible range that can be formed using the elements from the lists. This means we will generate ranges for every combination of elements from the lists and check if they cover all lists. The optimal Optimal Solution approach improves this to O(n log k).

What are the interview tips for Smallest Range Covering Elements from K Lists?

Explain your thought process clearly while coding. Consider edge cases, such as lists with only one element. Practice writing clean and efficient code with proper variable names.

What are common mistakes when solving Smallest Range Covering Elements from K Lists?

Not updating the maximum value correctly after popping from the heap. Forgetting to check if there are more elements in the list after popping.

#632

Smallest Range Covering Elements from K Lists

Hard

ArrayHash TableGreedySliding WindowSortingHeap (Priority Queue)HeapSliding Window

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log k)
Space	O(1)	O(k)

💡

Intuition

Time O(n log k)Space O(k)

The optimal solution uses a min-heap to efficiently track the smallest elements from each list while maintaining the largest element seen so far. This allows us to dynamically adjust the range as we iterate through the lists.

⚙️

Algorithm

5 steps

1Step 1: Initialize a min-heap and add the first element of each list along with its list index.
2Step 2: Track the maximum value from the elements in the heap.
3Step 3: Continuously extract the minimum element from the heap, calculate the current range, and update the best range if the current range is smaller.
4Step 4: If the extracted element has a next element in its list, push that next element into the heap and update the maximum value if necessary.
5Step 5: Repeat until we can no longer extract elements from the heap.

solution.py22 lines

1# Full working Python code
2import heapq
3
4def smallestRange(nums):
5    min_heap = []
6    current_max = float('-inf')
7    for i in range(len(nums)):
8        heapq.heappush(min_heap, (nums[i][0], i, 0))
9        current_max = max(current_max, nums[i][0])
10    best_range = [float('-inf'), float('inf')]
11
12    while min_heap:
13        current_min, list_index, element_index = heapq.heappop(min_heap)
14        if current_max - current_min < best_range[1] - best_range[0]:
15            best_range = [current_min, current_max]
16        if element_index + 1 < len(nums[list_index]):
17            next_element = nums[list_index][element_index + 1]
18            heapq.heappush(min_heap, (next_element, list_index, element_index + 1))
19            current_max = max(current_max, next_element)
20        else:
21            break
22    return best_range

ℹ

Complexity note: The time complexity is O(n log k) because we are using a min-heap to manage the elements from k lists, where n is the total number of elements and k is the number of lists.

1Using a min-heap allows efficient tracking of the smallest element across multiple lists.
2Maintaining the maximum element seen so far helps in calculating the range quickly.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.