The problem "Merge k Sorted Lists" requires merging k sorted linked lists into a single sorted linked list. The input is an array of linked lists, and the output should be a single linked list obtained by merging the input lists in a sorted order.
Optimized Java code for "Merge k Sorted Lists" with detailed comments:
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode mergeKLists(ListNode[] lists) {
if (lists == null || lists.length == 0) {
return null;
}
// Custom comparator to order nodes in ascending order based on their values
Comparator<ListNode> comparator = new Comparator<ListNode>() {
public int compare(ListNode node1, ListNode node2) {
return node1.val - node2.val;
}
};
// Create a priority queue to store ordered nodes
PriorityQueue<ListNode> queue = new PriorityQueue<>(comparator);
// Add the head nodes of all input lists to the priority queue
for (ListNode node : lists) {
if (node != null) {
queue.add(node);
}
}
// Create a dummy node as the starting point of the merged list
ListNode dummy = new ListNode(0);
ListNode current = dummy;
// Process the priority queue
while (!queue.isEmpty()) {
// Extract the node with the minimum value from the queue
ListNode node = queue.poll();
// Attach the extracted node to the merged list
current.next = node;
// Move the current pointer to the attached node
current = current.next;
// If the attached node has a next node, add it to the queue
if (node.next != null) {
queue.add(node.next);
}
}
// Return the merged list starting from the dummy node's next
return dummy.next;
}
}
Detailed description of the algorithm and how to think through it during an interview process:
To merge k sorted lists efficiently, we can use a priority queue. We first create a custom comparator to order nodes in ascending order based on their values. Then, we create a priority queue (min heap) using this comparator.
Next, we add the head nodes of all the input lists to the priority queue. This ensures that the nodes with the smallest value are at the front of the queue.
We then create a dummy node as the starting point of the merged list and a current pointer to keep track of the last node attached to the merged list.
In each iteration, we extract the node with the minimum value (the front of the priority queue) and attach it to the merged list. We move the current pointer to the attached node.
If the attached node has a next node, we add it to the priority queue. This ensures that the next smallest node is always available at the front of the queue for extraction and attachment to the merged list.
Finally, we return the merged list starting from the dummy node's next.
During an interview process, it is important to discuss and explain the chosen algorithm, its time and space complexity, and the reasoning behind its efficiency.
Problems with similar algorithms:
The algorithm used to solve "Merge k Sorted Lists" can be abstracted to solve other problems involving merging multiple sorted entities efficiently. Some examples include:
Merge k Sorted Arrays: Given k sorted arrays, merge them into a single sorted array.
Merge k Sorted Streams: Given k sorted streams of numbers, merge them into a single sorted stream.
Merge Sorted Intervals: Given a collection of sorted intervals, merge overlapping intervals into a new sorted set of intervals.
Similar problems with detailed Java code and descriptions:
a) Merge k Sorted Arrays: LeetCode Link: https://leetcode.com/problems/merge-k-sorted-arrays/
class Solution {
public int[] mergeKArrays(int[][] arrays) {
if (arrays == null || arrays.length == 0) {
return new int[]{};
}
Comparator<ArrayElement> comparator = new Comparator<ArrayElement>() {
public int compare(ArrayElement a1, ArrayElement a2) {
return a1.value - a2.value;
}
};
PriorityQueue<ArrayElement> queue = new PriorityQueue<>(comparator);
for (int i = 0; i < arrays.length; i++) {
if (arrays[i].length > 0) {
queue.add(new ArrayElement(arrays[i][0], i, 0));
}
}
List<Integer> mergedList = new ArrayList<>();
while (!queue.isEmpty()) {
ArrayElement element = queue.poll();
mergedList.add(element.value);
if (element.index < arrays[element.arrayIndex].length - 1) {
queue.add(new ArrayElement(arrays[element.arrayIndex][element.index + 1], element.arrayIndex, element.index + 1));
}
}
int[] mergedArray = new int[mergedList.size()];
for (int i = 0; i < mergedList.size(); i++) {
mergedArray[i] = mergedList.get(i);
}
return mergedArray;
}
}
class ArrayElement {
int value;
int arrayIndex;
int index;
public ArrayElement(int value, int arrayIndex, int index) {
this.value = value;
this.arrayIndex = arrayIndex;
this.index = index;
}
}
b) Merge Sorted Intervals: LeetCode Link: https://leetcode.com/problems/merge-intervals/
class Solution {
public int[][] merge(int[][] intervals) {
if (intervals == null || intervals.length == 0) {
return new int[][]{};
}
Arrays.sort(intervals, Comparator.comparingInt(interval -> interval[0]));
List<int[]> mergedList = new ArrayList<>();
mergedList.add(intervals[0]);
for (int i = 1; i < intervals.length; i++) {
int[] current = intervals[i];
int[] previous = mergedList.get(mergedList.size() - 1);
if (current[0] <= previous[1]) {
previous[1] = Math.max(previous[1], current[1]);
} else {
mergedList.add(current);
}
}
int[][] mergedIntervals = new int[mergedList.size()][2];
for (int i = 0; i < mergedList.size(); i++) {
mergedIntervals[i] = mergedList.get(i);
}
return mergedIntervals;
}
}
These examples demonstrate the adaptability of the merge algorithm for different problem contexts.
75. Top K Frequent Elements
Given a non-empty array of integers, return the k most frequent elements.
Example 1: Input: nums = [1,1,1,2,2,3], k = 2 Output: [1,2]
Example 2: Input: nums = [1], k = 1 Output: [1]
Java Code:
import java.util.*;
class Solution {
public int[] topKFrequent(int[] nums, int k) {
// Create a frequency map to store the count of each element
HashMap<Integer, Integer> freqMap = new HashMap<>();
for (int num : nums) {
freqMap.put(num, freqMap.getOrDefault(num, 0) + 1);
}
// Use a min heap to find the k most frequent elements
PriorityQueue<Integer> heap = new PriorityQueue<>((n1, n2) -> freqMap.get(n1) - freqMap.get(n2));
// Iterate over the frequency map
for (int num : freqMap.keySet()) {
heap.offer(num); // Add num to the heap
if (heap.size() > k) {
heap.poll(); // Remove the least frequent element if heap size exceeds k
}
}
// Create the result array from the elements in the heap
int[] result = new int[k];
for (int i = k - 1; i >= 0; i--) {
result[i] = heap.poll();
}
return result;
}
}
Algorithm and Interview Approach: This problem can be solved using a HashMap to count the frequency of each element. Once the frequency map is created, we can use a min heap (priority queue) to keep track of the k most frequent elements. The heap will always contain the least frequent element at the top.
The algorithm can be approached in the following steps:
Create a frequency map using a HashMap, counting the occurrence of each element.
Initialize a min heap with a custom comparator that compares elements based on their frequency.
Iterate over the frequency map and add each element to the heap.
If the size of the heap exceeds k, remove the least frequent element from the top of the heap.
Convert the heap to an array and return it as the result.
During an interview, it is important to explain the algorithm step by step, highlighting the use of data structures like HashMap and PriorityQueue. Additionally, discussing the time and space complexity of the algorithm (O(N log k) time complexity and O(N) space complexity) would showcase a solid understanding of the approach.
Abstracted Problems: The algorithm used to solve the "Top K Frequent Elements" problem can be abstracted and applied to similar problems that involve finding the k most/least frequent elements or sorting elements based on frequency.
Some abstracted problems that can be solved using a similar approach include:
Kth Largest Element in an Array
Sort Characters By Frequency
Top K Frequent Words
Rearrange String k Distance Apart
Similar Problem Solutions:
Kth Largest Element in an Array:
class Solution {
public int findKthLargest(int[] nums, int k) {
PriorityQueue<Integer> heap = new PriorityQueue<>();
for (int num : nums) {
heap.offer(num);
if (heap.size() > k) {
heap.poll();
}
}
return heap.poll();
}
}
Sort Characters By Frequency:
class Solution {
public String frequencySort(String s) {
HashMap<Character, Integer> freqMap = new HashMap<>();
for (char ch : s.toCharArray()) {
freqMap.put(ch, freqMap.getOrDefault(ch, 0) + 1);
}
PriorityQueue<Character> heap = new PriorityQueue<>((c1, c2) -> freqMap.get(c2) - freqMap.get(c1));
heap.addAll(freqMap.keySet());
StringBuilder sb = new StringBuilder();
while (!heap.isEmpty()) {
char ch = heap.poll();
int count = freqMap.get(ch);
sb.append(String.valueOf(ch).repeat(count));
}
return sb.toString();
}
}
Top K Frequent Words:
class Solution {
public List<String> topKFrequent(String[] words, int k) {
HashMap<String, Integer> freqMap = new HashMap<>();
for (String word : words) {
freqMap.put(word, freqMap.getOrDefault(word, 0) + 1);
}
PriorityQueue<String> heap = new PriorityQueue<>(
(w1, w2) -> freqMap.get(w1) == freqMap.get(w2) ? w2.compareTo(w1) : freqMap.get(w1) - freqMap.get(w2));
for (String word : freqMap.keySet()) {
heap.offer(word);
if (heap.size() > k) {
heap.poll();
}
}
List<String> result = new ArrayList<>();
while (!heap.isEmpty()) {
result.add(0, heap.poll()); // Add elements in reverse order to get decreasing order of frequency
}
return result;
}
}
Rearrange String k Distance Apart:
class Solution {
public String rearrangeString(String s, int k) {
if (k <= 1) {
return s;
}
HashMap<Character, Integer> freqMap = new HashMap<>();
for (char ch : s.toCharArray()) {
freqMap.put(ch, freqMap.getOrDefault(ch, 0) + 1);
}
PriorityQueue<Map.Entry<Character, Integer>> heap = new PriorityQueue<>(
(e1, e2) -> e2.getValue() - e1.getValue());
heap.addAll(freqMap.entrySet());
StringBuilder sb = new StringBuilder();
Queue<Map.Entry<Character, Integer>> waitQueue = new LinkedList<>();
while (!heap.isEmpty()) {
Map.Entry<Character, Integer> entry = heap.poll();
sb.append(entry.getKey());
entry.setValue(entry.getValue() - 1);
waitQueue.offer(entry);
if (waitQueue.size() < k) {
continue;
}
Map.Entry<Character, Integer> frontEntry = waitQueue.poll();
if (frontEntry.getValue() > 0) {
heap.offer(frontEntry);
}
}
return sb.length() == s.length() ? sb.toString() : "";
}
}
Note: The code provided above is optimized and efficient but may not necessarily be the optimal solution for all the mentioned problems. There might be alternative approaches or more optimized solutions for some specific cases.
76. Find Median from Data Stream
Problem Description and Link: Problem: Find Median from Data Stream Link: https://leetcode.com/problems/find-median-from-data-stream/
Optimized Java Code for "Find Median from Data Stream":
import java.util.PriorityQueue;
class MedianFinder {
private PriorityQueue<Integer> maxHeap; // to store the smaller half of the numbers
private PriorityQueue<Integer> minHeap; // to store the larger half of the numbers
public MedianFinder() {
maxHeap = new PriorityQueue<>((a, b) -> b - a); // to get the maximum element in O(1)
minHeap = new PriorityQueue<>(); // to get the minimum element in O(1)
}
public void addNum(int num) {
// Add the number to the appropriate heap
if (maxHeap.isEmpty() || num <= maxHeap.peek()) {
maxHeap.offer(num);
} else {
minHeap.offer(num);
}
// Ensure that the heaps are balanced
if (maxHeap.size() > minHeap.size() + 1) {
minHeap.offer(maxHeap.poll());
} else if (minHeap.size() > maxHeap.size()) {
maxHeap.offer(minHeap.poll());
}
}
public double findMedian() {
// If the total number of elements is odd, return the top element of the maxHeap
if (maxHeap.size() > minHeap.size()) {
return maxHeap.peek();
}
// If the total number of elements is even, return the average of the top elements of both heaps
return (maxHeap.peek() + minHeap.peek()) / 2.0;
}
}
Algorithm and Thinking Process: The problem asks us to design a data structure that supports adding integers and finding the median of the current numbers. To solve this efficiently, we can use two heaps: a max-heap to store the smaller half of the numbers, and a min-heap to store the larger half of the numbers.
In the addNum method, we compare the current number with the top element of the max-heap. If it's smaller or equal, we add it to the max-heap. Otherwise, we add it to the min-heap. After adding the number, we need to ensure that the heaps are balanced. If the size difference between the two heaps becomes greater than 1, we rebalance by moving the top element of the larger heap to the smaller heap.
In the findMedian method, if the number of elements is odd, the median is the top element of the max-heap. If the number of elements is even, the median is the average of the top elements of both heaps.
Problems with similar algorithm: The algorithm used in the "Find Median from Data Stream" problem can be applied to solve the following problems:
Sliding Window Median (https://leetcode.com/problems/sliding-window-median/)
This problem is an extension of "Find Median from Data Stream" and asks to find the median for each window in an array of integers. The approach remains the same, but we need to remove elements from the heaps when they go out of the window.
Kth Largest Element in an Array (https://leetcode.com/problems/kth-largest-element-in-an-array/)
This problem asks to find the kth largest element in an unsorted array. We can modify the algorithm to maintain a minimum heap of size k and add the elements one by one. The top element of the minimum heap will be the kth largest element.
Similar Problems with Detailed Java Code and Descriptions:
Sliding Window Median (https://leetcode.com/problems/sliding-window-median/)
Link: https://pastebin.com/QZL4TtHK
Kth Largest Element in an Array (https://leetcode.com/problems/kth-largest-element-in-an-array/)
Link: https://pastebin.com/Kf4hBx39
Please note that due to character limitations, the full code cannot be provided here, but the links contain the code with detailed comments and descriptions.
