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10 More Patterns to Solve 1000 More LeetCode Problems - Part II

by Thomas CherickalAugust 22nd, 2024

Too Long; Didn't Read

In our previous article, I introduced 10 powerful patterns that can help solve LeetCode problems efficiently. In this follow-up article, I present 10 more essential patterns that, when combined with the previous 10, form a comprehensive toolkit for conquering LeetCode. These patterns include Divide and Conquer, Bit Manipulation, Linked List, Interval, Trie, Reservoir Sampling, Monotonic Stack, Topological Sort, and Union Find.

featured image - 10 More Patterns to Solve 1000 More LeetCode Problems - Part II

Introduction

In our previous article, available on this link, I introduced 10 powerful patterns that can help solve a wide variety of LeetCode problems efficiently.

These patterns, including Two Pointers, Sliding Window, Binary Search, Depth-First Search, and Breadth-First Search, provide a structured approach to tackling common algorithmic challenges.

However, as you progress through LeetCode and encounter more complex problems, mastering just those 10 patterns will not be enough.

To truly excel at solving LeetCode problems and become an expert problem-solver, it's crucial to expand your knowledge and learn additional patterns.

In this follow-up article, we present 10 more essential patterns that, when combined with the previous 10, form a comprehensive toolkit for conquering LeetCode.

These patterns include Divide and Conquer, Bit Manipulation, Linked List, Interval, Trie, Heap, Reservoir Sampling, Monotonic Stack, Topological Sort, and Union Find.

By mastering all 20 patterns, you'll be equipped with a powerful arsenal of techniques to solve a vast majority of LeetCode problems.

Each pattern is accompanied by a detailed explanation, a sample problem, a solution, and 10 similar LeetCode problems to practice.

This comprehensive coverage ensures that you have a solid understanding of each pattern and can apply it effectively in various problem-solving scenarios.

Remember, learning these patterns is just the beginning.

The true mastery comes from consistent practice and applying these techniques to a wide range of problems.

As you work through more LeetCode problems, you'll gain a deeper understanding of when to apply each pattern and how to optimize your solutions.

Embrace the challenge, practice diligently, and let these 20 patterns be your guide on your journey to becoming a LeetCode master.

With dedication and persistence, you'll be well on your way to solving 1000+ LeetCode problems and beyond.

Furthermore, pay attention to the Time Complexity and Space Complexity Analysis.

That is a definite interview question on any one and every one of these patterns!

11. Union Find (Disjoint Set)

Explanation

Union Find, also known as Disjoint Set Union (DSU), is a data structure that keeps track of elements partitioned into a number of disjoint (non-overlapping) sets. It is particularly useful for solving problems that involve grouping elements, detecting cycles in graphs, and managing connectivity queries. The primary operations supported by Union Find are:

Find: Determine which set a particular element belongs to. This operation can be optimized using path compression, which flattens the structure of the tree whenever find is called, ensuring that all nodes directly point to the root.
Union: Merge two sets into a single set. This operation can be optimized using union by rank, which attaches the smaller tree to the root of the larger tree to keep the overall tree flat.

The efficiency of Union Find makes it suitable for problems in graph theory, such as detecting cycles in undirected graphs or finding connected components.

Sample Problem

**Problem:**200. Number of Islands
Given anm x n 2D binary grid grid where 0 represents water and 1 represents land, and returns the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically.

Solution

class UnionFind:
    def __init__(self, grid):
        m, n = len(grid), len(grid[0])
        self.parent = [-1] * (m * n)
        self.rank = [0] * (m * n)
        self.count = 0
        for i in range(m):
            for j in range(n):
                if grid[i][j] == '1':
                    self.parent[i * n + j] = i * n + j
                    self.count += 1

    def find(self, x):
        if self.parent[x] != x:
            self.parent[x] = self.find(self.parent[x])
        return self.parent[x]

    def union(self, x, y):
        x_root = self.find(x)
        y_root = self.find(y)
        if x_root == y_root:
            return
        if self.rank[x_root] < self.rank[y_root]:
            self.parent[x_root] = y_root
        elif self.rank[x_root] > self.rank[y_root]:
            self.parent[y_root] = x_root
        else:
            self.parent[y_root] = x_root
            self.rank[x_root] += 1

def numIslands(grid):
    if not grid:
        return 0

    uf = UnionFind(grid)
    directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]
    
    for i in range(len(grid)):
        for j in range(len(grid[0])):
            if grid[i][j] == '1':
                for d in directions:
                    ni, nj = i + d[0], j + d[1]
                    if 0 <= ni < len(grid) and 0 <= nj < len(grid[0]) and grid[ni][nj] == '1':
                        uf.union(i * len(grid[0]) + j, ni * len(grid[0]) + nj)

    return uf.count

Time and Space Complexity

Time Complexity: O(V + E), where V is the number of vertices (land cells), and E is the number of edges (connections between land cells). Each union and find operation is nearly constant time due to path compression and union by rank, making the amortized time complexity very efficient.
Space Complexity: O(V), as we need to maintain the parent and rank arrays. The space used is proportional to the number of vertices in the grid.

12. Divide and Conquer

Explanation

The divide and conquer technique is a powerful algorithmic paradigm that involves breaking a problem down into smaller, more manageable subproblems, solving each subproblem independently, and then combining the results to solve the original problem. This approach is particularly useful for problems that can be divided into independent parts, allowing for efficient processing and often leading to significant reductions in time complexity.In practice, the divide and conquer strategy typically follows three steps:

Divide: Split the problem into several smaller subproblems that are similar to the original problem.
Conquer: Solve each of the subproblems recursively. If the subproblem sizes are small enough, solve the subproblem as a base case.
Combine: Merge the results of the subproblems into the final solution for the original problem.

This method is widely used in sorting algorithms (like Merge Sort and Quick Sort), searching algorithms, and many optimization problems.

Sample Problem

**Problem:**148. Sort List
Given the head of a linked list, sort the list using merge sort.

Solution

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def sortList(head):
    if not head or not head.next:
        return head

    # Find the middle of the linked list
    slow, fast = head, head.next
    while fast and fast.next:
        slow = slow.next
        fast = fast.next.next
    mid = slow.next
    slow.next = None

    # Recursively sort the left and right halves
    left = sortList(head)
    right = sortList(mid)

    # Merge the sorted halves
    return merge(left, right)

def merge(left, right):
    dummy = ListNode()
    curr = dummy
    while left and right:
        if left.val < right.val:
            curr.next = left
            left = left.next
        else:
            curr.next = right
            right = right.next
        curr = curr.next
    if left:
        curr.next = left
    if right:
        curr.next = right
    return dummy.next

Time and Space Complexity

Time Complexity: O(n log n), where n is the number of nodes in the linked list. The merge sort algorithm divides the list into halves and merges them back together, resulting in a time complexity of O(n log n).
Space Complexity: O(log n) for the recursive call stack. The merge sort algorithm uses recursion to sort the left and right halves, and the maximum depth of the recursion stack is proportional to the height of the linked list, which is O(log n) for a balanced list.

13. Bit Manipulation

Explanation

Bit manipulation is a technique that involves directly manipulating bits or binary digits within a number. This approach is useful for a variety of problems, particularly those that require efficient storage, fast computations, or operations on binary representations. Common bit manipulation operations include AND, OR, XOR, NOT, left shift, and right shift.Bit manipulation can help solve problems related to:

Counting bits (e.g., counting the number of 1s in a binary representation)
Swapping values without using a temporary variable
Checking if a number is a power of two
Finding unique elements in a collection where all elements except one appear twice

This technique is often favored for its efficiency, as operations on bits are generally faster than arithmetic operations on integers.

Sample Problem

**Problem:**191. Number of 1 Bits
Write a function that takes an unsigned integer and returns the number of '1' bits it has (also known as the Hamming weight).

Solution

def hammingWeight(n):
    count = 0
    while n:
        count += n & 1  # Increment count if the least significant bit is 1
        n >>= 1  # Right shift n to process the next bit
    return count

Time and Space Complexity

Time Complexity: O(log n), where n is the number of bits in the integer. In the worst case, the number of iterations is equal to the number of bits in the integer.
Space Complexity: O(1), as we are using a constant amount of extra space.

14. Linked List

Explanation

A linked list is a linear data structure where each element (node) contains a value and a reference (or pointer) to the next node in the sequence. This structure allows for efficient insertion and deletion of nodes, as elements do not need to be contiguous in memory. Linked lists are particularly useful for dynamic memory allocation, where the size of the data structure can change during runtime.Common operations on linked lists include:

Traversing the list to access or modify elements
Inserting new nodes at various positions (head, tail, or specific index)
Deleting nodes from the list
Reversing the linked list
Detecting cycles within the list

The flexibility of linked lists makes them ideal for implementing stacks, queues, and other abstract data types.

Sample Problem

**Problem:**206. Reverse Linked List
Given the head of a singly linked list, reverse the list, and return the reversed list.

Solution

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def reverseList(head):
    prev = None
    curr = head
    while curr:
        next_node = curr.next  # Store the next node
        curr.next = prev  # Reverse the link
        prev = curr  # Move prev to current
        curr = next_node  # Move to the next node
    return prev  # New head of the reversed list

Time and Space Complexity

Time Complexity: O(n), where n is the number of nodes in the linked list. We traverse through the list once.
Space Complexity: O(1), as we are using a constant amount of extra space for the pointers.

15. Interval

Explanation

Interval problems involve working with a collection of intervals, which are typically represented as pairs of start and end points. These problems often require merging overlapping intervals, finding gaps between intervals, or determining whether a new interval can fit within existing ones.Key concepts in working with intervals include:

Merging Intervals: Combining overlapping intervals into a single interval.
Finding Gaps: Identifying spaces between intervals where no elements exist.
Inserting New Intervals: Determining where a new interval can be placed in a sorted list of existing intervals.

Efficiently handling intervals often involves sorting the intervals based on their starting points, allowing for linear scans to process them.

Sample Problem

**Problem:**56. Merge Intervals
Given an array of intervals whereintervals[i] = [starti, endi], merge all overlapping intervals, and return an array of the merged intervals.

Solution

def merge(intervals):
    intervals.sort(key=lambda x: x[0])  # Sort intervals based on start time
    merged = []
    for interval in intervals:
        if not merged or merged[-1][1] < interval[0]:
            merged.append(interval)  # No overlap, add interval
        else:
            merged[-1][1] = max(merged[-1][1], interval[1])  # Merge intervals
    return merged

Time and Space Complexity

Time Complexity: O(n log n), where n is the number of intervals. Sorting the intervals takes O(n log n) time, and the merging process takes O(n) time in the worst case.
Space Complexity: O(n), as we are creating a new list to store the merged intervals.

16. Trie (Prefix Tree)

Explanation

A trie, or prefix tree, is a specialized tree data structure used to store a dynamic set of strings, where each node represents a common prefix shared by some strings. Tries are particularly useful for problems involving prefix matching, autocomplete features, and dictionary implementations.Key operations in a trie include:

Insertion: Adding a new string to the trie by creating nodes for each character.
Search: Checking if a string exists in the trie by traversing the nodes according to the characters in the string.
Prefix Search: Finding all strings that start with a given prefix, which is efficient due to the structure of the trie.

Tries offer efficient retrieval times, making them suitable for applications like spell-checking, IP routing, and auto-suggestions.

Sample Problem

**Problem:**208. Implement Trie (Prefix Tree)
Implement a trie data structure that supports the following operations:insert, search, and startsWith.

Solution

class TrieNode:
    def __init__(self):
        self.children = {}
        self.is_end_of_word = False

class Trie:
    def __init__(self):
        self.root = TrieNode()

    def insert(self, word: str) -> None:
        node = self.root
        for char in word:
            if char not in node.children:
                node.children[char] = TrieNode()
            node = node.children[char]
        node.is_end_of_word = True

    def search(self, word: str) -> bool:
        node = self.root
        for char in word:
            if char not in node.children:
                return False
            node = node.children[char]
        return node.is_end_of_word

    def startsWith(self, prefix: str) -> bool:
        node = self.root
        for char in prefix:
            if char not in node.children:
                return False
            node = node.children[char]
        return True

Time and Space Complexity

Time Complexity:
- insert: O(k), where k is the length of the word. We traverse through each character in the word.
- search: O(k), where k is the length of the word. We traverse through each character in the word.
- startsWith: O(k), where k is the length of the prefix. We traverse through each character in the prefix.
Space Complexity: O(k * n), where k is the average length of the words and n is the number of words. In the worst case, each character of each word will have a separate node in the trie.

17. Heap (Priority Queue)

Explanation

A heap is a specialized tree-based data structure that satisfies the heap property, where the value of each node is either greater than or equal to (max-heap) or less than or equal to (min-heap) the values of its children. Heaps are commonly used to implement priority queues, which allow for efficient retrieval of the highest (or lowest) priority element.Key operations in heaps include:

Insertion: Adding a new element while maintaining the heap property.
Deletion: Removing the root element (highest or lowest) and re-structuring the heap.
Peek: Accessing the root element without removing it.

Heaps are particularly useful for problems that require frequent access to the largest or smallest elements, such as finding the top K elements or implementing Dijkstra's algorithm for shortest paths.

Sample Problem

**Problem:**703. Kth Largest Element in a Stream
Design a class to find the kth largest element in a stream. Note that it is the kth largest element in the sorted order, not the kth distinct element.

Solution

import heapq

class KthLargest:
    def __init__(self, k: int, nums: List[int]):
        self.k = k
        self.heap = nums
        heapq.heapify(self.heap)
        while len(self.heap) > k:
            heapq.heappop(self.heap)

    def add(self, val: int) -> int:
        if len(self.heap) < self.k:
            heapq.heappush(self.heap, val)
        elif val > self.heap[0]:
            heapq.heappop(self.heap)
            heapq.heappush(self.heap, val)
        return self.heap[0]

Time and Space Complexity

Time Complexity:
- __init__: O(n log k), where n is the number of elements in nums. Building the heap takes O(n) time, and removing (k-1) elements takes O(k log k) time.
- add: O(log k), as we push or pop an element from the heap of size k.
Space Complexity: O(k), as we store at most k elements in the heap.

18. Reservoir Sampling

Explanation

Reservoir sampling is a family of randomized algorithms designed for randomly choosing a sample of k items from a list S containing n items, where n is either a very large or unknown number. This technique is particularly useful when dealing with streams of data or large datasets where it is impractical to store all elements. The algorithm works by maintaining a "reservoir" of size k. As each element is processed, it has a chance of replacing an existing element in the reservoir, ensuring that each element has an equal probability of being included in the final sample. This method is efficient and allows for sampling without needing to know the total number of items in advance.

Sample Problem

**Problem:**382. Linked List Random Node
Given a singly linked list, return a random node's value from the linked list. Each node must have the same probability of being chosen.

Solution

import random

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

class Solution:
    def __init__(self, head: Optional[ListNode]):
        self.head = head

    def getRandom(self) -> int:
        reservoir = None
        current = self.head
        i = 0
        while current:
            i += 1
            if random.randint(1, i) == 1:
                reservoir = current
            current = current.next
        return reservoir.val

Time and Space Complexity

Time Complexity: O(n), where n is the number of nodes in the linked list. We traverse through the entire list once.
Space Complexity: O(1), as we are using a constant amount of extra space for the reservoir variable.

19. Monotonic Stack

Explanation

A monotonic stack is a data structure that maintains elements in either strictly increasing or strictly decreasing order. This structure is particularly useful for solving problems that require finding the nearest greater or smaller element for each element in an array, as well as for calculating areas in histograms or determining the maximum span of elements. The key advantage of using a monotonic stack is that it allows for efficient retrieval of the next greater or smaller elements in linear time, which is particularly beneficial in problems involving arrays or sequences.

Sample Problem

**Problem:**84. Largest Rectangle in Histogram
Given an array of non-negative integersheights representing the histogram's bar height where the width of each bar is 1, return the area of the largest rectangle in the histogram.

Solution

def largestRectangleArea(heights):
    stack = []
    max_area = 0
    for i in range(len(heights) + 1):
        while stack and (i == len(heights) or heights[i] < heights[stack[-1]]):
            height = heights[stack.pop()]
            width = i if not stack else i - stack[-1] - 1
            max_area = max(max_area, height * width)
        stack.append(i)
    return max_area

Time and Space Complexity

Time Complexity: O(n), where n is the number of elements in the input array. Each element is pushed and popped from the stack at most once.
Space Complexity: O(n), as we are using a stack to store the indices. In the worst case, the stack can store all the indices.

20. Topological Sort

Explanation

Topological sorting is an algorithm used to order the vertices of a directed acyclic graph (DAG) in such a way that for every directed edge from vertex A to vertex B, vertex A comes before vertex B in the ordering. This technique is particularly useful for scheduling tasks, resolving dependencies, and managing prerequisites. Topological sort can be implemented using either Depth-First Search (DFS) or Kahn's algorithm (which uses in-degree counting). The key characteristic of topological sorting is that it only applies to directed acyclic graphs, as cycles would prevent a valid ordering.

Sample Problem

**Problem:**207. Course Schedule
There are a total ofnumCourses courses you have to take, labeled from 0 to numCourses - 1. You are given an array prerequisites where prerequisites[i] = [ai, bi] indicates that you must take course bi first if you want to take course ai.Return true if you can finish all courses. Otherwise, return false.

Solution

from collections import defaultdict

def canFinish(numCourses, prerequisites):
    graph = defaultdict(list)
    indegree = [0] * numCourses
    for course, prereq in prerequisites:
        graph[prereq].append(course)
        indegree[course] += 1

    queue = [course for course in range(numCourses) if indegree[course] == 0]
    topological_order = []

    while queue:
        course = queue.pop(0)
        topological_order.append(course)
        for neighbor in graph[course]:
            indegree[neighbor] -= 1
            if indegree[neighbor] == 0:
                queue.append(neighbor)

    return len(topological_order) == numCourses

Time and Space Complexity

Time Complexity: O(V + E), where V is the number of vertices (courses) and E is the number of edges (prerequisites). We traverse each vertex and each edge once.
Space Complexity: O(V), as we use an adjacency list to represent the graph and an array to store the in-degrees.