What is the time complexity of Smallest Range Covering Elements from K Lists?

The optimal solution has a time complexity of O(n*log(k)).

What data structures are used to solve Smallest Range Covering Elements from K Lists?

This problem uses Array, Hash Table, Greedy, Sliding Window, Sorting, Heap (Priority Queue). Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Smallest Range Covering Elements from K Lists?

Insight 1: Since each list is sorted, we can maintain a pointer to the current element of each list. We need to find the smallest range formed by the elements pointed to by these pointers. Insight 2: Using a min-heap (priority queue) is essential for efficiently tracking the smallest element among the current elements of all lists.

Which companies ask LeetCode 632. Smallest Range Covering Elements from K Lists in interviews?

This problem is frequently asked at Databricks, Flipkart, Lyft, PhonePe, Pinterest and 1 more companies.

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Smallest Range Covering Elements from K Lists - LeetCode 632 Solution

Q: How to solve Smallest Range Covering Elements from K Lists?

The provided solutions use a min-heap to keep track of the smallest element among the current elements from each list. The algorithm proceeds as follows: 1. Initialize a min-heap with the first element from each list. Also, keep track of the current maximum value among these elements. 2. Iteratively: a. Extract the smallest element from the min-heap. This element is the current minimum of the range. b. If the list containing this smallest element has more elements, add the next element from that list to the min-heap and update the current maximum value if necessary. c. Update the best range found so far if the current range (current max - current min) is smaller than the best range or if the ranges have the same width, but the current range starts earlier. d. If any list is exhausted (no more elements to add), we've checked all possible ranges and can terminate.

Smallest Range Covering Elements from K Lists - Complete Solution Guide

Smallest Range Covering Elements from K Lists is LeetCode problem 632, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You have k lists of sorted integers in non-decreasing order . Find the smallest range that includes at least one number from each of the k lists. We define the range [a, b] is smaller than range [c, d] if b - a < d - c or a < c if b - a == d - c . Example 1: Input: nums = [[4,10,15,24,26],[0,9,12,20],[5,18,22,30]] Output: [20,24] Explanation: List 1: [4, 10, 15, 24,26], 24 is in range [20,24]. List 2: [0, 9, 12, 20], 20 is in range [20,24]. List 3: [5, 18, 22, 30], 22 is in range [20,24]. Exampl

Detailed Explanation

The problem asks us to find the smallest range (defined by its start and end points) that includes at least one number from each of the k sorted lists given as input. The 'smallest' range is determined by its width (end - start), and if two ranges have the same width, the one with the smaller start value is considered smaller. We are given k sorted lists of integers, and we need to return the start and end values of the smallest such range.

Solution Approach

The provided solutions use a min-heap to keep track of the smallest element among the current elements from each list. The algorithm proceeds as follows: 1. Initialize a min-heap with the first element from each list. Also, keep track of the current maximum value among these elements. 2. Iteratively: a. Extract the smallest element from the min-heap. This element is the current minimum of the range. b. If the list containing this smallest element has more elements, add the next element from that list to the min-heap and update the current maximum value if necessary. c. Update the best range found so far if the current range (current max - current min) is smaller than the best range or if the ranges have the same width, but the current range starts earlier. d. If any list is exhausted (no more elements to add), we've checked all possible ranges and can terminate.

Step-by-Step Algorithm

Step 1: Initialize a min-heap (priority queue) to store the first element from each list, along with its list index and element index within that list. The heap is ordered by element value.
Step 2: Initialize 'current_max' to the maximum of the initial elements in the min-heap.
Step 3: Initialize 'best_range_width' to infinity (or a very large number) and 'result_start' and 'result_end' to store the smallest range found so far.
Step 4: While the min-heap has elements:
Step 5: Extract the minimum element (val, list_idx, elem_idx) from the min-heap.
Step 6: If the extracted element is the last element in its list, break the loop because there are no more elements to consider from that list.
Step 7: Get the next element (next_val) from the same list (list_idx) by incrementing the element index (elem_idx).
Step 8: Push the next element (next_val, list_idx, next_elem_idx) into the min-heap.
Step 9: Update 'current_max' to the maximum of 'current_max' and 'next_val'.
Step 10: Find the new minimum value in the heap. This step is slightly different in the C++ code which is not the most efficient way to find the min, it could peek into the heap instead of re-building a temp heap.
Step 11: Calculate the current range width as 'current_max - min_val'.
Step 12: If 'current_range_width' is smaller than 'best_range_width', update 'best_range_width', 'result_start', and 'result_end'. If 'current_range_width' equals 'best_range_width', and 'min_val' is smaller than 'result_start', update 'result_start' and 'result_end'.
Step 13: After the loop finishes, return the array ['result_start', 'result_end'] which represents the smallest range.

Key Insights

Insight 1: Since each list is sorted, we can maintain a pointer to the current element of each list. We need to find the smallest range formed by the elements pointed to by these pointers.
Insight 2: Using a min-heap (priority queue) is essential for efficiently tracking the smallest element among the current elements of all lists.
Insight 3: The key idea is to maintain a 'sliding window' of numbers, where the width of the window is the range we're currently considering. We iteratively expand the window by replacing the smallest element in the window with the next element from the list it belongs to.

Complexity Analysis

Time Complexity: O(n*log(k))

Space Complexity: O(k)

Topics

This problem involves: Array, Hash Table, Greedy, Sliding Window, Sorting, Heap (Priority Queue).

Companies

Asked at: Databricks, Flipkart, Lyft, PhonePe, Pinterest, WinZO.

Smallest Range Covering Elements from K Lists | LeetCode 632

Hard

Array Hash Table Greedy Sliding Window Sorting Heap (Priority Queue)

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

Solution Code

Available in:

import java.util.List;
import java.util.PriorityQueue;
import java.lang.Integer;

class Solution {
    public int[] smallestRange(List<List<Integer>> nums) {
        PriorityQueue<int[]> minHeap = new PriorityQueue<>((a, b) -> a[0] - b[0]);
        int currentMax = Integer.MIN_VALUE;

        for (int i = 0; i < nums.size(); i++) {
            int val = nums.get(i).get(0);
            minHeap.offer(new int[]{val, i, 0});
            currentMax = Math.max(currentMax, val);
        }

        int minVal = minHeap.peek()[0];
        int bestRangeWidth = currentMax - minVal;
        int resultStart = minVal;
        int resultEnd = currentMax;

        while (true) {
            int[] top = minHeap.poll();
            int val = top[0];
            int listIdx = top[1];
            int elemIdx = top[2];

            if (elemIdx + 1 == nums.get(listIdx).size()) {
                break;
            }

            int nextElemIdx = elemIdx + 1;
            int nextVal = nums.get(listIdx).get(nextElemIdx);
            minHeap.offer(new int[]{nextVal, listIdx, nextElemIdx});

            currentMax = Math.max(currentMax, nextVal);
            minVal = minHeap.peek()[0];

            int currentRangeWidth = currentMax - minVal;

            if (currentRangeWidth < bestRangeWidth) {
                bestRangeWidth = currentRangeWidth;
                resultStart = minVal;
                resultEnd = currentMax;
            } else if (currentRangeWidth == bestRangeWidth && minVal < resultStart) {
                resultStart = minVal;
                resultEnd = currentMax;
            }
        }

        return new int[]{resultStart, resultEnd};
    }
}

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Problem Info

Description

Problem 632: Smallest Range Covering Elements from K Lists - Detailed explanation and solution approaches.

Topics

Array Hash Table Greedy Sliding Window Sorting Heap (Priority Queue)

Companies

DatabricksFlipkartLyftPhonePePinterestWinZO

Stats

Difficulty:Hard

Languages:4

Companies:6

Problem #:632

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

13 steps

1
Step 1: Initialize a min-heap (priority queue) to store the first element from each list, along with its list index and element index within that list. The heap is ordered by element value.
2
Step 2: Initialize 'current_max' to the maximum of the initial elements in the min-heap.
3
Step 3: Initialize 'best_range_width' to infinity (or a very large number) and 'result_start' and 'result_end' to store the smallest range found so far.
4
Step 4: While the min-heap has elements:
5
Step 5: Extract the minimum element (val, list_idx, elem_idx) from the min-heap.
6
Step 6: If the extracted element is the last element in its list, break the loop because there are no more elements to consider from that list.
7
Step 7: Get the next element (next_val) from the same list (list_idx) by incrementing the element index (elem_idx).
8
Step 8: Push the next element (next_val, list_idx, next_elem_idx) into the min-heap.
9
Step 9: Update 'current_max' to the maximum of 'current_max' and 'next_val'.
10
Step 10: Find the new minimum value in the heap. This step is slightly different in the C++ code which is not the most efficient way to find the min, it could peek into the heap instead of re-building a temp heap.
11
Step 11: Calculate the current range width as 'current_max - min_val'.
12
Step 12: If 'current_range_width' is smaller than 'best_range_width', update 'best_range_width', 'result_start', and 'result_end'. If 'current_range_width' equals 'best_range_width', and 'min_val' is smaller than 'result_start', update 'result_start' and 'result_end'.
13
Step 13: After the loop finishes, return the array ['result_start', 'result_end'] which represents the smallest range.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n*log(k))

💾

Space Complexity

O(k)

📝

Why?

The time complexity is derived from the fact that we potentially process each element across all lists once (n total elements). Each time we process an element, we perform heap operations (push and pop), which cost O(log k) time. Therefore, the overall time complexity is O(n log k). The space complexity is determined by the size of the min-heap, which holds at most one element from each of the k lists. This results in a space complexity of O(k).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Merge and Sort

Merge all the lists into a single sorted array, and then use a sliding window approach. Maintain a counter for each of the k lists to track how many elements from each list are in the current window. Expand the window until all k lists are represented, then shrink it from the left until one of the lists is no longer represented. Repeat.

Time: O(n log n), where n is the total number of elements.Space: O(n), to store the merged sorted array.

Trade-offs: This approach can be simpler to understand but has a higher time complexity than the heap-based approach. This would be better for smaller input sizes when the log k factor dominates or when memory is not a concern.

✨

Key Insights

1Insight 1: Since each list is sorted, we can maintain a pointer to the current element of each list. We need to find the smallest range formed by the elements pointed to by these pointers.
2Insight 2: Using a min-heap (priority queue) is essential for efficiently tracking the smallest element among the current elements of all lists.
3Insight 3: The key idea is to maintain a 'sliding window' of numbers, where the width of the window is the range we're currently considering. We iteratively expand the window by replacing the smallest element in the window with the next element from the list it belongs to.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Greedy Algorithm

Making the best choice at each step without looking ahead.

Sliding Window

A moving "window" over data to process subsets efficiently.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty input lists. The provided solutions do not explicitly handle empty input lists. If any of the input lists are empty, the algorithm might encounter an error or produce incorrect results. It would be beneficial to add a check at the beginning to handle this case.
Edge case 2: Lists with very large numbers. The solutions work fine as long as the integer range is within the limits of int (or long in Java), no specific handling is needed unless there are calculations that might lead to overflow.
Edge case 3: Duplicate numbers within or across the lists. The solution handles the presence of duplicate numbers correctly. Duplicates don't affect the correctness or complexity of the algorithm.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where one of the lists is exhausted. If the algorithm doesn't check when a list is exhausted, it can lead to an infinite loop or incorrect results.

✗Common mistake 2: Incorrectly updating the 'current_max' value. The 'current_max' value must be updated whenever a new element is added to the min-heap.

✗Common mistake 3: Not considering the secondary condition when the ranges are equal. Always compare the start points of range when range width is same.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Data aggregation from multiple sorted sources (e.g., log files from different servers) to find a time window where all sources have entries. The time is the integer, the logs from each server are each lists, you need to find the minimum time range when all the log sources have entries.

🏦Real-world scenario 2: Database query optimization. Finding the smallest range of values that satisfy certain conditions across multiple indexes (which can be viewed as sorted lists).

🎮Real-world scenario 3: Supply chain optimization. Imagine you have multiple suppliers (lists) each with scheduled delivery times. Find the smallest window where you can get a delivery from each supplier.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 1, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 14

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Smallest Range Covering Elements from K Lists"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?