[{"content":"","date":"11 April 2026","externalUrl":null,"permalink":"/tags/c%23/","section":"Tags","summary":"","title":"C#","type":"tags"},{"content":"","date":"11 April 2026","externalUrl":null,"permalink":"/categories/","section":"Categories","summary":"","title":"Categories","type":"categories"},{"content":"Check out my latest LeetCode problem solutions and tutorials below. For more content, visit the YouTube section.\n","date":"11 April 2026","externalUrl":null,"permalink":"/","section":"Home","summary":"","title":"Home","type":"page"},{"content":"","date":"11 April 2026","externalUrl":null,"permalink":"/categories/leetcode/","section":"Categories","summary":"","title":"Leetcode","type":"categories"},{"content":"","date":"11 April 2026","externalUrl":null,"permalink":"/tags/leetcode/","section":"Tags","summary":"","title":"Leetcode","type":"tags"},{"content":"","date":"11 April 2026","externalUrl":null,"permalink":"/tags/","section":"Tags","summary":"","title":"Tags","type":"tags"},{"content":" Problem Overview # The Valid Anagram problem from LeetCode asks you to determine whether two given strings are anagrams of each other.\nProblem Statement: Given two strings s and t, return true if t is an anagram of s, and false otherwise.\nAn anagram is a word or phrase formed by rearranging the letters of another word or phrase, typically using all the original letters exactly once.\nExample:\nInput: s = \u0026quot;anagram\u0026quot;, t = \u0026quot;nagaram\u0026quot;\nOutput: true\nInput: s = \u0026quot;rat\u0026quot;, t = \u0026quot;car\u0026quot;\nOutput: false\nSolution Approach # C# Solution - Using Dictionary # The optimal approach using a Dictionary to count character frequencies:\npublic class Solution { public bool IsAnagram(string s, string t) { if(s.Length != t.Length) return false; var count = new Dictionary\u0026lt;char, int\u0026gt;(); foreach(char c in s) { if(count.ContainsKey(c)){ count[c]++; }else{ count[c] = 1; } } foreach(char c in t) { if(!count.ContainsKey(c)) return false; count[c]--; if(count[c] \u0026lt; 0) return false; } return true; } } Time Complexity: O(n) - single pass through both strings Space Complexity: O(1) - at most 26 lowercase letters in the dictionary\nKey Insights # Length Check: If strings have different lengths, they cannot be anagrams Character Frequency: Both strings must have exactly the same characters with the same frequencies Case Sensitivity: The problem typically considers lowercase letters Constraint: Usually assumes strings contain only lowercase English letters Related Problems # Anagram groups (grouping anagrams together) Word pattern matching Permutation checking Video Tutorial # Watch the complete explanation and walkthrough in the video above for a detailed solution approach and implementation details.\n","date":"11 April 2026","externalUrl":null,"permalink":"/youtube/anagram/","section":"YouTube","summary":"Complete solution and explanation for LeetCode Valid Anagram problem","title":"Valid Anagram - LeetCode 242 Solution","type":"youtube"},{"content":"","date":"11 April 2026","externalUrl":null,"permalink":"/tags/youtube/","section":"Tags","summary":"","title":"Youtube","type":"tags"},{"content":" Welcome to my YouTube video library - comprehensive tutorials and educational content. Currently focused on LeetCode problem solutions with detailed C# implementations, complexity analysis, and step-by-step walkthroughs. Each problem includes the video tutorial, complete code solution, and key insights. Future content will expand to cover system design, advanced C# techniques, full-stack development, and more. Browse the grid below to find video tutorials you\u0026rsquo;re looking for.\n","date":"11 April 2026","externalUrl":null,"permalink":"/youtube/","section":"YouTube","summary":"Explore our collection of YouTube videos and content","title":"YouTube","type":"youtube"},{"content":" Problem Overview # The Ransom Note problem from LeetCode asks you to determine if you can construct a ransom note using letters from a magazine.\nProblem Statement: Given two strings ransomNote and magazine, return true if ransomNote can be constructed by using the letters from magazine and false otherwise. Each letter in magazine can only be used once in ransomNote.\nExample:\nInput: ransomNote = \u0026quot;a\u0026quot;, magazine = \u0026quot;b\u0026quot;\nOutput: false\nInput: ransomNote = \u0026quot;a\u0026quot;, magazine = \u0026quot;a\u0026quot;\nOutput: true\nInput: ransomNote = \u0026quot;aa\u0026quot;, magazine = \u0026quot;ab\u0026quot;\nOutput: false\nInput: ransomNote = \u0026quot;aa\u0026quot;, magazine = \u0026quot;aab\u0026quot;\nOutput: true\nSolution Approach # C# Solution - Using Dictionary # Create a Dictionary to count the frequency of each character in the magazine. Then iterate through the ransom note and check if each required character exists in the dictionary with a count greater than zero. Decrement the count for each character used, and return false if any character is not available or has been exhausted.\npublic class Solution { public bool CanConstruct(string ransomNote, string magazine) { var charCount = new Dictionary\u0026lt;char, int\u0026gt;(); foreach(char c in magazine) { if(charCount.ContainsKey(c)){ charCount[c]++; }else{ charCount[c] = 1; } } foreach(char c in ransomNote) { if(!charCount.ContainsKey(c) || charCount[c] == 0){ return false; } charCount[c]--; } return true; } } Time Complexity: O(m + n) - where m is magazine length, n is ransom note length Space Complexity: O(1) - at most 26 lowercase letters in the dictionary\nKey Insights # Two-Pass Approach: First pass builds character frequency map, second pass validates construction Character Frequency: Track count of each character in magazine for availability checking Availability Check: Verify both existence and sufficient count before using each character Decrement Usage: Subtract from count when a character is used in ransom note Early Exit: Return false immediately if required character is missing or exhausted Space Efficient: Dictionary size is bounded by alphabet size (26 characters) Related Problems # Valid Anagram Isomorphic Strings Majority Element Video Tutorial # Watch the complete explanation and walkthrough in the video above for a detailed solution approach and implementation details.\n","date":"5 April 2026","externalUrl":null,"permalink":"/youtube/ransom-note/","section":"YouTube","summary":"Complete solution and explanation for LeetCode Ransom Note problem","title":"Ransom Note - LeetCode 383 Solution","type":"youtube"},{"content":" Problem Overview # The Contains Duplicate problem from LeetCode asks you to determine if an array contains any duplicate elements.\nProblem Statement: Given an integer array nums, return true if any value appears at least twice in the array, and return false if every element is distinct.\nExample:\nInput: nums = [1,2,3,1]\nOutput: true\nInput: nums = [1,2,3,4]\nOutput: false\nInput: nums = [99,99]\nOutput: true\nSolution Approach # C# Solution - Sorting Approach # Sort the array first to group duplicate elements together, then iterate through the sorted array to check if any adjacent elements are equal. If two adjacent elements are the same, we\u0026rsquo;ve found a duplicate.\npublic class Solution { public bool ContainsDuplicate(int[] nums) { Array.Sort(nums); for(int i = 0; i \u0026lt; nums.Length - 1; i++) { if(nums[i] == nums[i+1]){ return true; } } return false; } } Time Complexity: O(n log n) - due to sorting operation Space Complexity: O(1) - only constant extra space (sorting in-place)\nKey Insights # Sorting First: Sort the array to group duplicate elements together, making them adjacent Adjacent Comparison: After sorting, all duplicate pairs will be next to each other Early Exit: Return true immediately when a duplicate is found, no need to check the entire array Space Efficient: Sorting in-place uses only constant extra space, unlike HashSet which requires O(n) space Trade-off: Uses more time (O(n log n)) but achieves optimal space complexity (O(1)) Array Bounds: Loop only up to nums.Length - 1 to avoid index out of bounds Related Problems # Valid Anagram Contains Duplicate II Contains Duplicate III Video Tutorial # Watch the complete explanation and walkthrough in the video above for a detailed solution approach and implementation details.\n","date":"24 March 2026","externalUrl":null,"permalink":"/youtube/contains-duplicate/","section":"YouTube","summary":"Complete solution and explanation for LeetCode Contains Duplicate problem","title":"Contains Duplicate - LeetCode 217 Solution","type":"youtube"},{"content":" Problem Overview # The Jewels and Stones problem from LeetCode asks you to count how many stones are jewels.\nProblem Statement: You\u0026rsquo;re given strings jewels representing the types of stones that are jewels, and stones representing the stones you have. Each character in stones is a stone you have. You want to know how many of the stones you have are also jewels. The jewels are case-sensitive, so the same letter as different cases is considered different types.\nExample:\nInput: jewels = \u0026quot;aB\u0026quot;, stones = \u0026quot;aAbBcCcD\u0026quot;\nOutput: 3 (stones \u0026lsquo;a\u0026rsquo;, \u0026lsquo;A\u0026rsquo;, and \u0026lsquo;b\u0026rsquo; are jewels)\nInput: jewels = \u0026quot;z\u0026quot;, stones = \u0026quot;ZZ\u0026quot;\nOutput: 0\nSolution Approach # C# Solution - Using HashSet # Create a HashSet from the jewels string for O(1) lookup performance. Before processing, validate that both inputs are not null or empty. Then iterate through each stone and check if it exists in the jewel set. Count each stone that matches a jewel type and return the total count.\npublic class Solution { public int NumJewelsInStones(string jewels, string stones) { if(string.IsNullOrEmpty(jewels) || string.IsNullOrEmpty(stones)) return 0; var count = 0; var jSet = new HashSet\u0026lt;char\u0026gt;(jewels); foreach(var c in stones) { // perform search if(jSet.Contains(c)) count++; } return count; } } Time Complexity: O(m + n) - where m is jewels length, n is stones length (create set O(m), iterate stones O(n)) Space Complexity: O(1) - at most 52 characters (26 lowercase + 26 uppercase) in the HashSet\nKey Insights # Case Sensitive: \u0026lsquo;a\u0026rsquo; and \u0026lsquo;A\u0026rsquo; are different jewel types - the comparison is case-sensitive HashSet Creation: Convert jewels string to a HashSet for O(1) average lookup time Input Validation: Check for null or empty strings to handle edge cases before processing Efficient Counting: Iterate through stones once and check each character against the jewel set Fixed Alphabet: Limited to alphabetic characters (at most 52 types), making space complexity constant Early Exit: No need to process further if jewels or stones are empty Related Problems # Contains Duplicate Valid Anagram Majority Element Video Tutorial # Watch the complete explanation and walkthrough in the video above for a detailed solution approach and implementation details.\n","date":"21 March 2026","externalUrl":null,"permalink":"/youtube/jewels-and-stones/","section":"YouTube","summary":"Complete solution and explanation for LeetCode Jewels and Stones problem","title":"Jewels and Stones - LeetCode 771 Solution","type":"youtube"},{"content":" Problem Overview # The Summary Ranges problem from LeetCode asks you to create a list of consecutive ranges as strings.\nProblem Statement: You are given a sorted unique integer array nums. Return the smallest sorted list of ranges that cover all the numbers in the array exactly. That is, each element of nums is covered by exactly one of the ranges, and there is no integer x such that x is in one of the ranges but not in nums.\nEach range [a,b] in the result should be formatted as:\n\u0026quot;a\u0026quot; if a == b \u0026quot;a-\u0026gt;b\u0026quot; if a != b Example:\nInput: nums = [0,1,2,4,5,7]\nOutput: [\u0026quot;0-\u0026gt;2\u0026quot;,\u0026quot;4-\u0026gt;5\u0026quot;,\u0026quot;7\u0026quot;]\nInput: nums = [0,2,3,4,6,8,9]\nOutput: [\u0026quot;0\u0026quot;,\u0026quot;2-\u0026gt;4\u0026quot;,\u0026quot;6\u0026quot;,\u0026quot;8-\u0026gt;9\u0026quot;]\nSolution Approach # C# Solution - Single Pass with Break Detection # Use a single variable start to track the beginning of a range and iterate through the array. When a break in consecutiveness is found (when current element is not exactly previous + 1), format and add the previous range to the result. Handle the formatting based on whether it\u0026rsquo;s a single element or a range. Add the final range after the loop completes.\npublic class Solution { public IList\u0026lt;string\u0026gt; SummaryRanges(int[] nums) { var ranges = new List\u0026lt;string\u0026gt;(); if(nums.Length == 0) { return ranges; } var start = nums[0]; for(int i = 1; i \u0026lt; nums.Length; i++) { if(nums[i] != nums[i - 1] + 1){ if(start == nums[i - 1]){ ranges.Add(start.ToString()); }else{ ranges.Add($\u0026#34;{start}-\u0026gt;{nums[i - 1]}\u0026#34;); } start = nums[i]; } } if(start == nums[nums.Length - 1]){ ranges.Add(start.ToString()); }else { ranges.Add($\u0026#34;{start}-\u0026gt;{nums[nums.Length - 1]}\u0026#34;); } return ranges; } } Time Complexity: O(n) - single pass through the array Space Complexity: O(1) - excluding output, only constant extra space for variables\nKey Insights # Range Start Tracking: Use a single variable to mark the start of each consecutive range Break Detection: Detect breaks by checking if nums[i] != nums[i-1] + 1 Formatting Logic: Format as single number if start equals end, otherwise use \u0026ldquo;start-\u0026gt;end\u0026rdquo; format Two Formatting Points: Handle range formatting both when a break occurs and after the loop Final Range Handling: Ensure the last range is added after the loop completes Empty Array Edge Case: Return empty list immediately for empty input arrays Related Problems # Missing Ranges Data Stream as Disjoint Intervals Merge Intervals Video Tutorial # Watch the complete explanation and walkthrough in the video above for a detailed solution approach and implementation details.\n","date":"8 March 2026","externalUrl":null,"permalink":"/youtube/summary-ranges/","section":"YouTube","summary":"Complete solution and explanation for LeetCode Summary Ranges problem","title":"Summary Ranges - LeetCode 228 Solution","type":"youtube"},{"content":" Problem Overview # The Longest Common Prefix problem from LeetCode asks you to find the longest common prefix string amongst an array of strings.\nProblem Statement: Write a function to find the longest common prefix string amongst an array of strings. If there is no common prefix, return an empty string \u0026quot;\u0026quot;.\nExample:\nInput: strs = [\u0026quot;flower\u0026quot;,\u0026quot;flow\u0026quot;,\u0026quot;flight\u0026quot;]\nOutput: \u0026quot;fl\u0026quot;\nInput: strs = [\u0026quot;dog\u0026quot;,\u0026quot;racecar\u0026quot;,\u0026quot;car\u0026quot;]\nOutput: \u0026quot;\u0026quot;\nInput: strs = [\u0026quot;interspecies\u0026quot;,\u0026quot;interstellar\u0026quot;,\u0026quot;interstate\u0026quot;]\nOutput: \u0026quot;inters\u0026quot;\nSolution Approach # C# Solution - Horizontal Prefix Reduction # Start with the first string as the initial prefix. Then iterate through each subsequent string and reduce the prefix by removing characters from the end until the prefix is found at the start of the current string using IndexOf. The prefix shrinks until it matches all strings or becomes empty.\npublic class Solution { public string LongestCommonPrefix(string[] strs) { if(strs == null || strs.Length == 0) return \u0026#34;\u0026#34;; var prefix = strs[0]; for (int i = 1; i \u0026lt; strs.Length; i++) { while (strs[i].IndexOf(prefix) != 0) { prefix = prefix.Substring(0, prefix.Length - 1); if(prefix == \u0026#34;\u0026#34;) return \u0026#34;\u0026#34;; } } return prefix; } } Time Complexity: O(m * n * k) - where m is number of strings, n is length of first string, k is average length of strstr operation Space Complexity: O(1) - only constant extra space used for the prefix string\nKey Insights # Edge Cases: Handle null arrays and empty string arrays before processing Prefix Reduction: Start with the first string and progressively reduce it until it matches all strings IndexOf Validation: Use IndexOf() to check if prefix is at the start (index 0) of each string Early Exit: Return empty string immediately if prefix becomes empty at any point Horizontal Approach: Compare vertically through strings while reducing the prefix horizontally Greedy Trimming: Remove one character at a time from the end of the prefix for each mismatch Related Problems # Longest Palindromic Substring Shortest Common Supersequence Group Anagrams Video Tutorial # Watch the complete explanation and walkthrough in the video above for a detailed solution approach and implementation details.\n","date":"1 March 2026","externalUrl":null,"permalink":"/youtube/longest-common-prefix/","section":"YouTube","summary":"Complete solution and explanation for LeetCode Longest Common Prefix problem","title":"Longest Common Prefix - LeetCode 14 Solution","type":"youtube"},{"content":" Problem Overview # The Best Time to Buy and Sell Stock problem from LeetCode asks you to find the maximum profit from a single buy-sell transaction.\nProblem Statement: You are given an array prices where prices[i] is the price of a given stock on the i-th day. You want to maximize your profit by choosing a single day to buy one stock and a different day in the future to sell that stock. Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return 0.\nExample:\nInput: prices = [7,1,5,3,6,4]\nOutput: 5 (buy at 1, sell at 6)\nInput: prices = [7,6,4,3,1]\nOutput: 0 (no profit possible)\nInput: prices = [2,4,1]\nOutput: 2 (buy at 2, sell at 4)\nSolution Approach # C# Solution - Single Pass with Min Tracking # The optimal approach using a single pass to track minimum price and maximum profit:\npublic class Solution { public int MaxProfit(int[] prices) { var minPrice = prices[0]; var maxProfit = 0; for(int i = 0; i \u0026lt; prices.Length; i++) { if(prices[i] \u0026lt; minPrice) { minPrice = prices[i]; }else{ var profit = prices[i] - minPrice; if(profit \u0026gt; maxProfit) { maxProfit = profit; } } } return maxProfit; } } Time Complexity: O(n) - single pass through prices array Space Complexity: O(1) - only constant extra space\nKey Insights # Memory: Track the minimum price seen so far Greedy Approach: At each price, calculate profit if we sell at that price Single Pass: Optimal solution requires only one iteration Constraint: Must sell after buying (future day requirement) Related Problems # Best Time to Buy and Sell Stock II (multiple transactions) Best Time to Buy and Sell Stock III (at most 2 transactions) Best Time to Buy and Sell Stock IV (at most k transactions) Video Tutorial # Watch the complete explanation and walkthrough in the video above for a detailed solution approach and implementation details.\n","date":"13 February 2026","externalUrl":null,"permalink":"/youtube/best-time-stock/","section":"YouTube","summary":"Complete solution and explanation for LeetCode Best Time to Buy and Sell Stock problem","title":"Best Time to Buy and Sell Stock - LeetCode 121 Solution","type":"youtube"},{"content":" Problem Overview # The Is Subsequence problem from LeetCode asks you to determine whether one string is a subsequence of another.\nProblem Statement: Given two strings s and t, return true if s is a subsequence of t, or false otherwise. A subsequence of a string is a new string that is formed from the original string by deleting some (not necessarily consecutive) characters without disturbing the relative positions of the remaining characters.\nExample:\nInput: s = \u0026quot;abc\u0026quot;, t = \u0026quot;ahbgdc\u0026quot;\nOutput: true\nInput: s = \u0026quot;axc\u0026quot;, t = \u0026quot;ahbgdc\u0026quot;\nOutput: false\nSolution Approach # C# Solution - Two Pointer Approach # Use two pointers to traverse both strings simultaneously. For each character in the target string t, check if it matches the current character in the subsequence string s. If it matches, advance the first pointer. Always advance the second pointer. If we successfully match all characters in s, then s is a subsequence of t.\npublic class Solution { public bool IsSubsequence(string s, string t) { var sIndex = 0; var tIndex = 0; while(sIndex \u0026lt; s.Length \u0026amp;\u0026amp; tIndex \u0026lt; t.Length) { if(s[sIndex] == t[tIndex]) sIndex++; tIndex++; } return sIndex == s.Length; } } Time Complexity: O(n) - where n is the length of string t (single pass through t) Space Complexity: O(1) - only constant extra space used for two pointers\nKey Insights # Two Pointer Technique: Maintain pointers for both strings, advancing independently based on character matches Match Characters: Only advance the subsequence pointer when characters match, always advance the target string pointer Order Preservation: Characters of the subsequence must appear in the same order but don\u0026rsquo;t need to be consecutive in the target string Validation: Check if sIndex == s.Length to confirm all characters in s were found Early Exit: Loop terminates when either string is exhausted Empty Subsequence: An empty string s is always a subsequence of any string t Related Problems # Longest Common Subsequence Longest Increasing Subsequence Count Matching Subsequences Video Tutorial # Watch the complete explanation and walkthrough in the video above for a detailed solution approach and implementation details.\n","date":"7 February 2026","externalUrl":null,"permalink":"/youtube/is-subsequence/","section":"YouTube","summary":"Complete solution and explanation for LeetCode Is Subsequence problem","title":"Is Subsequence - LeetCode 392 Solution","type":"youtube"},{"content":" Problem Overview # The Roman to Integer problem from LeetCode asks you to convert a Roman numeral string to an integer.\nProblem Statement: Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M. Given a roman numeral, convert it to an integer.\nSymbol Values:\nI = 1, V = 5, X = 10, L = 50, C = 100, D = 500, M = 1000 Example:\nInput: s = \u0026quot;III\u0026quot;\nOutput: 3\nInput: s = \u0026quot;LVIII\u0026quot;\nOutput: 58 (50 + 5 + 3)\nInput: s = \u0026quot;MCMXCIV\u0026quot;\nOutput: 1994 (1000 + 900 + 90 + 4)\nSolution Approach # C# Solution - Single Pass with Dictionary # Create a Dictionary mapping each Roman numeral symbol to its integer value. Iterate through the string once, tracking the previous value. When the current value is greater than the previous value, it means subtraction should apply (like IV or IX), so subtract twice the previous value. Otherwise, simply add the current value to the total.\npublic class Solution { public int RomanToInt(string s) { var romanMap = new Dictionary\u0026lt;char, int\u0026gt; { {\u0026#39;I\u0026#39;, 1}, {\u0026#39;V\u0026#39;, 5}, {\u0026#39;X\u0026#39;, 10}, {\u0026#39;L\u0026#39;, 50}, {\u0026#39;C\u0026#39;, 100}, {\u0026#39;D\u0026#39;, 500}, {\u0026#39;M\u0026#39;, 1000} }; var total = 0; var prevVal = 0; foreach(var c in s) { var currentVal = romanMap[c]; if( prevVal \u0026lt; currentVal) { total += currentVal - (2 * prevVal); }else { total += currentVal; } prevVal = currentVal; } return total; } } Time Complexity: O(n) - single pass through the string Space Complexity: O(1) - dictionary has at most 7 entries (constant size)\nKey Insights # Dictionary Mapping: Store all seven Roman symbols with their values for O(1) lookup Subtraction Detection: When current value \u0026gt; previous value, a subtraction case occurs (e.g., IV, IX) Smart Adjustment: Subtract twice the previous value instead of complex subtraction logic Left to Right Processing: Process characters sequentially, comparing with previous value only Common Patterns: IV (4), IX (9), XL (40), XC (90), CD (400), CM (900) are handled automatically Single Pass Efficiency: Only one iteration through the string needed Related Problems # Integer to Roman Valid Parentheses Evaluate Reverse Polish Notation Video Tutorial # Watch the complete explanation and walkthrough in the video above for a detailed solution approach and implementation details.\n","date":"31 January 2026","externalUrl":null,"permalink":"/youtube/roman-to-integer/","section":"YouTube","summary":"Complete solution and explanation for LeetCode Roman to Integer problem","title":"Roman to Integer - LeetCode 13 Solution","type":"youtube"},{"content":" Problem Overview # The Merge Strings Alternately problem from LeetCode asks you to merge two strings by adding letters in alternating order.\nProblem Statement: You are given two strings word1 and word2. Merge the strings by adding letters in alternating order, starting with word1. If a string is longer than the other, append the additional letters onto the end of the merged string.\nExample:\nInput: word1 = \u0026quot;abc\u0026quot;, word2 = \u0026quot;pqr\u0026quot;\nOutput: \u0026quot;apbqcr\u0026quot;\nInput: word1 = \u0026quot;ab\u0026quot;, word2 = \u0026quot;pqrs\u0026quot;\nOutput: \u0026quot;apbqrs\u0026quot;\nInput: word1 = \u0026quot;abcd\u0026quot;, word2 = \u0026quot;pq\u0026quot;\nOutput: \u0026quot;apbqcd\u0026quot;\nSolution Approach # C# Solution - Two Pointer with StringBuilder # Use a StringBuilder for efficient string building. Get the lengths of both strings and find the maximum length to determine how many iterations are needed. In each iteration, check if the current index is within bounds for each string, and if so, append the character from that string. This automatically handles strings of different lengths.\npublic class Solution { public string MergeAlternately(string word1, string word2) { var merged = new StringBuilder(); var len1 = word1.Length; var len2 = word2.Length; var maxLen = Math.Max(len1, len2); for(int i = 0; i \u0026lt; maxLen; i++) { if( i \u0026lt; len1) merged.Append(word1[i]); if( i \u0026lt; len2) merged.Append(word2[i]); } return merged.ToString(); } } Time Complexity: O(n + m) - where n and m are lengths of the two strings Space Complexity: O(n + m) - for the merged result string\nKey Insights # StringBuilder Efficiency: Use StringBuilder instead of string concatenation for O(1) append operations Maximum Length: Use Math.Max() to determine total iterations needed for both strings Bounds Checking: Check each index against respective string lengths before appending Alternating Pattern: Append from both strings in each iteration to achieve alternating order Automatic Remainder: Longer string\u0026rsquo;s remaining characters are naturally appended after shorter string is exhausted Single Loop: One loop with dual appends eliminates need for separate remainder handling Related Problems # Merge Sorted Array Zip Pairs of Lists Interleaving String Video Tutorial # Watch the complete explanation and walkthrough in the video above for a detailed solution approach and implementation details.\n","date":"26 January 2026","externalUrl":null,"permalink":"/youtube/merge-strings-alternately/","section":"YouTube","summary":"Complete solution and explanation for LeetCode Merge Strings Alternately problem","title":"Merge Strings Alternately - LeetCode 1768 Solution","type":"youtube"},{"content":"I\u0026rsquo;m a full-stack software engineer with 10+ years of experience building robust applications across multiple platforms. Currently engineering solutions at the New York State Unified Court System with .NET and Azure.\nGitHub | Twitter | YouTube\nExperience # Senior Software Engineer # New York State Unified Court System · Full-time\nFeb 2025 – Present · 1 yr 3 mos\n📍 New York, United States\nSkills: .NET Core, Microsoft Azure, ADF\nSoftware Developer # Self Employed · Freelancer, Independent Contractor, Gig Worker, Entrepreneur\nMar 2020 – Dec 2023 · 3 yrs 10 mos\nDeveloped custom solutions for clients in diverse industries, including restaurants and small corporations, enhancing operational efficiency and customer engagement.\nSkills: Technical Consultation, Flutter, Dart, Firebase\nWipro # 7 yrs 2 mos\nSenior Project Engineer / Senior Java Developer # Jan 2016 – Aug 2019 · 3 yrs 8 mos\nManaged a critical Java application for a client\u0026rsquo;s payment system, integrating with 17 banks to process thousands of transactions.\nSkills: Applications Development Management, Project Management, Java, On-premises Server Architecture\nJunior Developer / Student Engineer # Jul 2012 – Dec 2016 · 4 yrs 6 mos\n","externalUrl":null,"permalink":"/about/","section":"Home","summary":"","title":"","type":"page"},{"content":"","externalUrl":null,"permalink":"/authors/","section":"Authors","summary":"","title":"Authors","type":"authors"},{"content":"","externalUrl":null,"permalink":"/series/","section":"Series","summary":"","title":"Series","type":"series"}]