Given two strings s and t, return true if t is an anagram of s (same characters in same frequencies). Otherwise, return false. Example: 'anagram' and 'nagaram' are anagrams.

Word Pattern" problem: Given a pattern string & a string s, return true if each char maps to exactly one word & vice versa. Use two hash maps to ensure consistency. Time complexity: O(n), space complexity: O(1).

Two strings s and t are isomorphic if each character in s can be replaced by a unique character in t, preserving order. Example: 'e' maps to 'a', 'g' maps to 'd'. JavaScript solution uses two hash maps to ensure consistent mapping.

Can you construct ransomNote using letters from magazine? Count chars in magazine, then validate against ransomNote. Time complexity: O(n+m), space complexity: O(k). Example solutions in JavaScript.

Solve LeetCode 289: Game of Life by updating board in place with state encoding (2: live->dead, 3: dead->live) and two passes through the matrix.

Solved LeetCode 73: Set Matrix Zeroes in O(1) space complexity by marking first row & column as markers, then updating matrix accordingly.

Rotate a matrix by 90 degrees clockwise in place. Transpose the matrix, then reverse each row. Example: [[1,2,3],[4,5,6],[7,8,9]] -> [[7,4,1],[8,5,2],[9,6,3]]. Time complexity: O(n^2), space complexity: O(1).

Solve LeetCode 54: Spiral Matrix by defining boundaries (top, bottom, left, right) that shrink as you traverse the matrix in spiral order. Time complexity is O(m⋅n), space complexity is O(1).

We'll validate a 9x9 Sudoku board based on rules: each row, column & 3x3 sub-box must contain unique digits 1-9.

Minimum Window Substring: Find smallest substring of s containing all chars from t. Use sliding window & hash maps for efficient solution.

Master the Sliding Window Algorithm: Solve problems efficiently with this simple yet powerful technique, reducing time complexity and memory usage 🚀💡

Solve substring with concatenation of all words using sliding window & hash map. Precompute word frequencies, slide window, shrink or expand based on matches. Time complexity: O(n⋅k), space complexity: O(m+k).

Longest Substring Without Repeating Characters" solved using sliding window & Set, O(n) time, O(k) space. Optimize with hash map for better perf.

Solve Minimum Size Subarray Sum problem with O(n) sliding window or O(nlogn) binary search techniques, returning minimal length of subarray whose sum is >= target.

The 3Sum problem is a classic interview challenge that combines sorting, two-pointer techniques, and efficient duplicate handling to find all unique triplets in an integer array that sum up to zero.

Lessons from 2024: Resilience is key, collaboration matters, time management crucial & community just as important as code. Mastering cloud-native dev, launching SaaS side project & shaping future with AI, serverless arch & Web3.

Maximize water area between two lines using two-pointer technique. Move pointers inward based on shorter height for optimal container capacity in O(n) time and O(1) space.

Two-pointer technique solves LeetCode 392: Is Subsequence in O(n+m) time, traversing strings s and t to check if s is a subsequence of t.

Valid Palindrome problem: remove non-alphanumeric chars, convert to lowercase. Use two pointers to check if string is palindrome. Time complexity O(n), space complexity O(1).

Solved LeetCode 68: Text Justification with greedy algorithms, distributing spaces evenly across words and adding extra to the left when needed.

Find first occurrence of substring in string efficiently using built-in indexOf() or manual sliding window approach with O(n-m+1)*m time complexity.

Reverse Words in a String": Optimize solution without split or trim. Traverse string, skip spaces, extract words & prepend to result. Handle single spaces between words. Time complexity: O(n), space complexity: O(n).

Solve LeetCode 135: Candy with a greedy algorithm. Assign 1 candy to each child, then traverse left-to-right and right-to-left to ensure higher-rated children get more candies.

Gas Station problem: Can travel circuit if total gas >= total cost. Greedy approach finds starting station with net gain >= 0.

Maximize stock profit by buying & selling as many times as needed. Greedy approach: add profit for every upward slope, ignoring declines. Time complexity: O(n), space complexity: O(1).

Remove duplicates from sorted array in-place: use two-pointer technique, initialize k=1, iterate through array & copy unique elements to nums[k], return k for count of unique elements.