632. Smallest Range Covering Elements from K Lists

Problem Description

You are provided k lists which contain sorted integers in non-decreasing order. Your task is to find the smallest range that contains at least one number from each of the k lists. To determine which range is the smallest, you must use the rules given below:

1. If range [a, b] has a smaller length than range [c, d] (meaning b - a < d - c), then [a, b] is considered smaller.
2. If the lengths are equal (meaning b - a == d - c), then range [a, b] is considered smaller if a < c.

The objective is to find the smallest such range [a, b] according to the criteria above.

Intuition

To arrive at the solution, consider the following approach:

• The brute force method would naturally involve checking all possible ranges, which would be prohibitively expensive as it might take O(N^2) time where N is the total number of elements in all lists combined.
• A better approach is to consider that since the lists are sorted, we can use a sliding window along with a pointer for each list to keep track of whether we have at least one number from each list within our current window.
• To implement this idea efficiently, we can use a min-heap to keep track of the smallest element outside our current range and a hash map or counter to keep track of the elements within our range.
• We line up all the elements from all the lists in sorted order with their corresponding list indices, then move the right boundary of our window to include elements, while keeping track of which lists are represented within the window using the counter.
• When all k lists are represented within the window, we try to minimize the range by moving the left boundary to the right while ensuring that at least one number from each list is still inside the window.
• Each time we have all k lists represented, we check if the current range is smaller than our best found range. If it is, we update our answer.
• This gives us an O(NlogN) solution because of the initial sort, and then for each of the N elements, we perform operations that can be considered in constant time on average.

The provided code follows this intuition and finds the smallest range efficiently.

Learn more about Greedy, Sorting, Sliding Window and Heap (Priority Queue) patterns.

Solution Approach

The code provided implements an efficient solution as follows:

1. Flatten and Sort: The algorithm starts by flattening all k sorted lists into a single list t and tagging each element with the index of the list it came from. This is done using list comprehension. Then, it sorts this combined list which will allow efficient traversal using a sliding window.

   1t = [(x, i) for i, v in enumerate(nums) for x in v]
   2t.sort()

2. Initialize Counter and Answer: A Counter is initialized to keep track of how many numbers from each list are within the current window. The starting answer range is set from -infinity to infinity to ensure any valid range found will be smaller than this initial range.

3. Sliding Window: The algorithm then uses a sliding window technique to find the smallest range. The variable j is the left boundary and b iterating over t is the right boundary.

   1cnt = Counter()
   2ans = [-inf, inf]
   3j = 0

4. Expand and Shrink Window: As we process each element b from the sorted list t, we increment the count of elements seen from its list (given by v). We try to expand the window until it includes at least one number from every list. We do this by checking if the length of the counter is equal to the number of lists.

   1for b, v in t:
   2    cnt[v] += 1
   3    while len(cnt) == len(nums):  # Shrink the window if possible
   4        # Rest of the code for shrinking the window

5. Update Smallest Range: If the current window includes at least one number from each list, we check if this window range is smaller than our current smallest range (ans). If it's smaller or starts earlier with the same size, we update the answer.

   1a = t[j][0]
   2x = b - a - (ans[1] - ans[0])
   3if x < 0 or (x == 0 and a < ans[0]):
   4    ans = [a, b]

6. Remove Left Boundary Elements: To potentially get a smaller range, we advance the left boundary of the window (j). We decrement the count of the left boundary's list number, and if the count falls to zero, we remove it from the counter, in effect narrowing our window until it no longer covers all k lists.

   1w = t[j][1]
   2cnt[w] -= 1
   3if cnt[w] == 0:
   4    cnt.pop(w)
   5j += 1

7. Return Answer: After processing all elements with this sliding window, the smallest range is stored in ans, which is returned as the result.

The implementation leverages the sorting to line up potential candidates for the range and uses a sliding window with a counter to efficiently find and attempt to minify the range while keeping track of the requirement to have at least one number from each of the k lists.

Example Walkthrough

Let's illustrate the solution approach using a small example. Suppose we have k = 3 lists of sorted integers:

1List 1: [4, 10, 15]
2List 2: [1, 13, 15]
3List 3: [5, 10, 20]

To find the smallest range which contains at least one number from each list following the steps of the solution approach:

1. Flatten and Sort: We start by combining and tagging each element with its respective list index:

   1Combined: [(4, 0), (10, 0), (15, 0), (1, 1), (13, 1), (15, 1), (5, 2), (10, 2), (20, 2)]

   After sorting we get:

   1Sorted:    [(1, 1), (4, 0), (5, 2), (10, 0), (10, 2), (13, 1), (15, 0), (15, 1), (20, 2)]

2. Initialize Counter and Answer: We create a counter and set the starting answer range:

   1Counter: {}
   2Answer Range: [-inf, inf]

3. Sliding Window: Initialize the sliding window boundaries:

   1Left Boundary (j): 0
   2Right Boundary (to be determined in loop): -

4. Expand and Shrink Window: We iterate through the sorted elements and expand the window by increasing the count for each tag:

   1Iteration 1 (Element: (1, 1)):
   2Counter: {1: 1}
   3Window: [1, -]
   4
   5Iteration 2 (Element: (4, 0)):
   6Counter: {1: 1, 0: 1}
   7Window: [1, 4]
   8
   9Iteration 3 (Element: (5, 2)):
   10Counter: {1: 1, 0: 1, 2: 1}
   11Window: [1, 5]  // Now we have all 3 lists represented

   Upon finding all lists are represented, we can potentially start to shrink the window from the left to find the smallest range.

5. Update Smallest Range: Since this is the first window that includes numbers from all lists, we update our answer range:

   1Answer Range: [1, 5]

   We will check the range each time we have a new window containing elements from all lists and update if a smaller range is found.

6. Remove Left Boundary Elements: Now, we will attempt to shrink our window:

   1Moving Left Boundary from 1 to 4:
   2Counter after decrement: {1: 0, 0: 1, 2: 1}
   3Since the count of number from list 1 is now 0, we remove it, and our window is no longer valid.
   4
   5We continue this sliding window approach for the rest of the elements.

   This process will continue until we've exhausted the sorted list.

7. Return Answer: Eventually, after sliding the window over the entire sorted list, we get our smallest range, which is [1, 5] in this example.

The final output is the answer range. In this example, the smallest range containing at least one number from each of the k lists would be [1, 5]. This method ensures we have covered all the lists while keeping the range as small as possible.

Solution Implementation

Time and Space Complexity

The given Python code finds the smallest numerical range that includes at least one number from each of the k lists. The approach is a type of sliding window algorithm with two pointers.

Time Complexity

Let's break down the time complexity:

1. Creating the merged list (t): This involves iterating over each list in nums and each element within those lists. If n is the total number of elements across all k lists, creating the merged list t is O(n).

2. Sorting the merged list (t.sort()): The sorting step has a time complexity of O(n log n) since it is sorting n elements.

3. Sliding window over t: The sliding window loop processes each element in t once. Incrementing the second pointer j and checking and updating the counter results in traversing all n elements once. Thus, this part has a time complexity of O(n).

Combining these steps, the dominant term is the sort, and thus the overall time complexity of the code is O(n log n).

Space Complexity

Now, let's consider space complexity:

1. Merged list (t): The merged list t contains n tuples where n is the total number of elements across all k lists, so space complexity for t is O(n).

2. Counter (cnt): The counter keeps track of the presence of elements from each list within the current window. In the worst case, it will have k keys where k is the number of lists in nums. Since the space for cnt is dependent on the number of lists and not the number of elements, it's O(k).

3. Range list (ans): The range list has a constant size of 2, hence O(1).

Thus, the overall space complexity of the algorithm is dominated by the space taken by the merged list t, which is O(n) where n is the total number of elements across all the lists.

Learn more about how to find time and space complexity quickly using problem constraints.