Comprehensive Guide with Examples and Algorithms

In the world of algorithms and competitive programming, solving problems involving multiple lists is common. One such problem is finding the smallest range that includes at least one element from each of K sorted lists. This problem is not only an interesting challenge but also has practical applications in data merging, information retrieval, and more. In this comprehensive guide, we will explore the problem of finding the smallest range for K lists, discuss the strategies to solve it, provide step-by-step examples, and explain the algorithmic approach to derive the solution.

Problem Statement

Given K sorted lists of integers, the task is to find the smallest range that includes at least one number from each of the K lists.

Example:

Input: 
nums = [
  [4, 10, 15, 24, 26],
  [0, 9, 12, 20],
  [5, 18, 22, 30]
]

Output: [20, 24]

Explanation: 
The smallest range that includes at least one number from each of the K lists is [20, 24].

Constraints:

• Each list is sorted in non-decreasing order.
• There can be up to K lists, and each list can contain up to N elements.
• The goal is to return the smallest possible range [a, b] such that there is at least one element from each list within this range.

Understanding the Problem

The problem is about merging multiple sorted lists while maintaining a focus on identifying the smallest range that covers at least one element from each list. The key challenge is to ensure that all lists are represented within the range while minimizing the range’s size.

The problem can be visualized as having multiple markers (one on each list), where you try to move these markers to cover the smallest possible section of the number line that includes elements from all the lists.

Approach to Solve the Problem

There are multiple approaches to solve this problem, but one of the most efficient methods involves using a min-heap (or priority queue). This approach is not only intuitive but also ensures that the solution is found in an optimized manner.

Steps to Solve the Problem Using a Min-Heap:

1. Initialization:

• Use a min-heap to keep track of the minimum element among the K lists.
• Maintain a variable to track the maximum element in the current range (since all elements are in sorted order, the maximum is simply the last inserted element in the range).
• Start by inserting the first element from each list into the heap.

1. Heap Operations:

• Extract the minimum element from the heap (this represents the current start of the range).
• Calculate the current range as the difference between the maximum element and the extracted minimum element.
• If this current range is smaller than the previously found range, update the smallest range.
• Move the marker forward in the list from which the minimum element was extracted, and insert the next element from that list into the heap.
• Update the maximum element if the newly inserted element is larger.

1. Termination:

• The process continues until you can no longer extract a full set of elements from all K lists (i.e., one of the lists is exhausted).

1. Result:

• The smallest range found during this process is the answer.

Detailed Example

Let’s walk through the given example step by step.

Example:

Input: 
nums = [
  [4, 10, 15, 24, 26],
  [0, 9, 12, 20],
  [5, 18, 22, 30]
]

Output: [20, 24]

Step-by-Step Process:

1. Initialization:

• Insert the first element of each list into the min-heap: [4, 0, 5].
• The initial maximum value is 5 (among [4, 0, 5]).
• The current range is [0, 5].

1. First Iteration:

• Extract 0 from the heap. The range is [0, 5].
• Insert the next element from the second list, which is 9.
• The new heap is [4, 5, 9], and the new maximum value is 9.
• The current range is [4, 9], which is larger than [0, 5].

1. Second Iteration:

• Extract 4 from the heap. The range is [4, 9].
• Insert the next element from the first list, which is 10.
• The new heap is [5, 9, 10], and the new maximum value is 10.
• The current range is [5, 10], which is smaller than [4, 9].

1. Third Iteration:

• Extract 5 from the heap. The range is [5, 10].
• Insert the next element from the third list, which is 18.
• The new heap is [9, 10, 18], and the new maximum value is 18.
• The current range is [9, 18], which is larger than [5, 10].

1. Fourth Iteration:

• Extract 9 from the heap. The range is [9, 18].
• Insert the next element from the second list, which is 12.
• The new heap is [10, 12, 18], and the new maximum value is 18.
• The current range is [10, 18], which is larger than [9, 18].

1. Fifth Iteration:

• Extract 10 from the heap. The range is [10, 18].
• Insert the next element from the first list, which is 15.
• The new heap is [12, 15, 18], and the new maximum value is 18.
• The current range is [12, 18], which is larger than [10, 18].

1. Sixth Iteration:

• Extract 12 from the heap. The range is [12, 18].
• Insert the next element from the second list, which is 20.
• The new heap is [15, 18, 20], and the new maximum value is 20.
• The current range is [15, 20], which is larger than [12, 18].

1. Seventh Iteration:

• Extract 15 from the heap. The range is [15, 20].
• Insert the next element from the first list, which is 24.
• The new heap is [18, 20, 24], and the new maximum value is 24.
• The current range is [18, 24], which is larger than [15, 20].

1. Eighth Iteration:

• Extract 18 from the heap. The range is [18, 24].
• Insert the next element from the third list, which is 22.
• The new heap is [20, 22, 24], and the new maximum value is 24.
• The current range is [20, 24], which is smaller than [18, 24].

1. Ninth Iteration:
   • Extract 20 from the heap. The range is [20, 24].
   • There are no more elements in the second list to insert into the heap.
   • The process stops here.

The smallest range found during this process is [20, 24], which is the final answer.

Java Implementation

Now, let’s implement the above logic in Java.

import java.util.PriorityQueue;

public class SmallestRange {

    static class Element {
        int value;
        int listIndex;
        int elementIndex;

        Element(int value, int listIndex, int elementIndex) {
            this.value = value;
            this.listIndex = listIndex;
            this.elementIndex = elementIndex;
        }
    }

    public static int[] smallestRange(int[][] nums) {
        PriorityQueue<Element> minHeap = new PriorityQueue<>((a, b) -> a.value - b.value);
        int maxValue = Integer.MIN_VALUE;
        int rangeStart = 0;
        int rangeEnd = Integer.MAX_VALUE;

        // Initialize the heap and find the initial max value
        for (int i = 0; i < nums.length; i++) {
            minHeap.add(new Element(nums[i][0], i, 0));
            maxValue = Math.max(maxValue, nums[i][0]);
        }

        while (minHeap.size() == nums.length) {
            Element current = minHeap.poll();

            if (maxValue - current.value < rangeEnd - rangeStart) {
                rangeStart = current.value;
                rangeEnd = maxValue;
            }

            if (current.elementIndex + 1 < nums[current.listIndex].length) {
                int nextValue = nums[current.listIndex][current.elementIndex + 1];
                minHeap.add(new Element(nextValue, current.listIndex, current.elementIndex + 1));
                maxValue = Math.max(maxValue, nextValue);
            }
        }

        return new int[]{rangeStart, rangeEnd};
    }

    public static void main(String[] args)

 {
        int[][] nums = {
            {4, 10, 15, 24, 26},
            {0, 9, 12, 20},
            {5, 18, 22, 30}
        };

        int[] result = smallestRange(nums);
        System.out.println("Smallest Range: [" + result[0] + ", " + result[1] + "]");
    }
}

Explanation of the Code

• Element Class: This helper class stores the value of the element, the index of the list from which it came, and the index of the element within that list.
• Min-Heap: The priority queue is used to always extract the smallest element among the current candidates from each list.
• Max Value Tracking: maxValue tracks the maximum value among the current elements in the heap, which is used to calculate the current range.
• Range Update: The code updates the smallest range whenever a smaller range is found.
• Termination: The loop continues until the heap can no longer maintain one element from each list, meaning one list is exhausted.

Complexity Analysis

• Time Complexity: The time complexity of the algorithm is O(N * log K), where N is the total number of elements across all lists, and K is the number of lists. This is because each insertion and extraction operation in the heap takes O(log K) time, and we perform these operations N times.
• Space Complexity: The space complexity is O(K) for storing elements in the heap.

Edge Cases

1. Single List: If there’s only one list, the smallest range is simply the first and last elements of that list.
2. Lists of Unequal Lengths: The algorithm handles lists of different lengths by automatically stopping when the shortest list is exhausted.
3. Overlapping Ranges: The algorithm naturally handles overlapping ranges by always choosing the smallest possible range that includes all lists.

Practical Applications

This problem has real-world applications in scenarios like data aggregation from multiple sources, where the goal is to find the most consistent data range across different datasets. It is also applicable in scheduling problems, where the smallest time range covering all necessary events or tasks must be identified.

Finding the smallest range that covers at least one element from each of K sorted lists is a challenging yet intriguing problem. The use of a min-heap or priority queue is an effective way to solve this problem efficiently. By understanding and implementing this algorithm, you can tackle similar problems in competitive programming and real-world applications with confidence.

Further Exploration

For those interested in exploring further, consider modifying the problem constraints, such as finding the smallest range that covers exactly two elements from each list, or optimizing the algorithm for very large datasets. Additionally, exploring similar problems on platforms like LeetCode can deepen your understanding of these concepts.

References

• LeetCode – Smallest Range
• GeeksforGeeks – Smallest Range in K Lists
• Priority Queue in Java