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Solution 13: Minimal Domain Covering

Approach Explanation

We maintain a window of size k containing one element from each list using a Min-Heap. We also track the maximum value currently in the heap. The current range is [heap[0].val, current_max].

Step-by-Step Logic

  1. For each list, maintain an index (initially 0).
  2. Store the first element of each list in a Min-Heap: (value, list_index, index_in_list).
  3. Track the max_val across all elements in the heap.
  4. While the heap size is k:
    • Calculate current range [heap_min, max_val]. Update if smaller.
    • Pop the smallest element.
    • If that list has more elements, push the next one and update max_val.
    • If one list is exhausted, we cannot form a range with all lists anymore.

Complexity

  • Time Complexity: O(N log K).
  • Space Complexity: O(K).

Code

import heapq

def smallest_range(nums):
    # Min-heap to find the minimum of the k pointers
    h = []
    max_val = -float('inf')
    
    # Init with first element of every list
    for i in range(len(nums)):
        heapq.heappush(h, (nums[i][0], i, 0))
        max_val = max(max_val, nums[i][0])
        
    ans = [-10**5, 10**5]
    
    while True:
        min_val, list_idx, val_idx = heapq.heappop(h)
        
        # Update range if current is smaller
        if max_val - min_val < ans[1] - ans[0]:
            ans = [min_val, max_val]
            
        # Move forward in the list we just popped from
        if val_idx + 1 < len(nums[list_idx]):
            nxt_val = nums[list_idx][val_idx + 1]
            heapq.heappush(h, (nxt_val, list_idx, val_idx + 1))
            max_val = max(max_val, nxt_val)
        else:
            # Cannot form any more ranges including all lists
            break
            
    return ans