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Smallest Range Covering Elements

Certification: Advanced Level Accuracy: 100% Submissions: 1 Points: 15

Write a Java program to find the smallest range that includes at least one number from each of k sorted lists. The range [a, b] is smaller than range [c, d] if (b - a) < (d - c), or if (b - a) == (d - c) and a < c. You have k sorted lists of integers, and you need to find the smallest range that includes at least one number from each of the k lists.

Example 1
    
  
  • Input: nums = [[4,10,15,24,26], [0,9,12,20], [5,18,22,30]]
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  • Output: [20, 24]
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  • Explanation: 
    List 1: [4,10,15,24,26]
    List 2: [0,9,12,20]
    List 3: [5,18,22,30]
    The range [20,24] contains 24 from list 1, 20 from list 2, and 22 from list 3.
    No smaller range can be found that contains at least one element from each list.
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Example 2
    
  
  • Input: nums = [[1,2,3], [1,2,3], [1,2,3]]
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  • Output: [1, 1]
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  • Explanation: 
    Each list contains elements 1, 2, and 3.
    The range [1,1] contains the element 1 from each list.
    This is the smallest possible range as its size is 0.
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Constraints
    
  
  • nums.length == k, where k is between 1 and 3500
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  • 1 ≤ nums[i].length ≤ 50 for all 0 ≤ i < k
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  • -10^5 ≤ nums[i][j] ≤ 10^5 for all 0 ≤ i < k and 0 ≤ j < nums[i].length
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  • nums[i] is sorted in non-decreasing order
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  • Time Complexity: O(n * log(k)) where n is the total number of elements across all lists
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  • Space Complexity: O(k)
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Priority QueueCapgeminiWalmart
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Hint
Solution

Solution Hints

    
  
  • Use a min-heap (priority queue) to track the smallest element from each list
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  • Keep a current maximum value across all elements in the heap
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  • Initialize the heap with the first element from each list
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  • Repeatedly pop the smallest element, update the range if needed, and add the next element from the same list
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  • Track the range with the minimum difference
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  • Continue until any one of the lists is exhausted
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  • The problem is similar to merging k sorted lists, but with a focus on finding the minimum range
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Steps to solve by this approach:

 Step 1: Create a min-heap (priority queue) to track the smallest element from each list along with its list index and position.

 Step 2: Initialize the heap with the first element from each list, and keep track of the maximum value.

 Step 3: While the heap contains elements from all k lists, extract the minimum element from the heap.

 Step 4: Calculate the current range as (max - current min value) and update the smallest range if needed.

 Step 5: Add the next element from the same list as the extracted element to the heap, if available.

 Step 6: Update the maximum value if the newly added element is larger.

 Step 7: Continue until any list is exhausted, then return the smallest range found.

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