LeetCode 271: Length-Prefixed String Encoding

LeetCode Problem 1461: "Check if a binary string contains all binary codes of size k": Given a binary string s and an integer k, return true if every binary code of length k is a substring of s. Otherwise, return false. Approach: I simply used a sliding window and a hash map where hash map stores the seen substrings of size k. Two pointers keep track of each substring of size k. A count variable is initialized to 0 which is incremented each time when a unique unseen substring of size k is seen, if this count reaches the max possible number of unique binary codes of size k (i.e. 2^k) the function returns True else False. #LeetCode #Python #BitManipulation #Strings #SlidingWindow #DataStructures #Algorithms #CompetitiveProgramming #ProblemSolving #OptimalSolution

LeetCode POTD 💫: Description: Given an integer n, return the decimal value of the binary string formed by concatenating the binary representations of 1 to n in order, modulo 10**9 + 7. Here are my solutions: Using String Manipulation: https://lnkd.in/g4UpENeF Using Bit Manipulation: https://lnkd.in/gw9PZc7n #Python #DSA #Leetcode #DailyChallenge

✅ Day 6 of #DSAPrep > Problem: Reverse Integer > Platform: LeetCode > Concept: Number Manipulation + Overflow Handling Reversed an integer using modulo (%) and integer division (//). Handled: Negative numbers using a sign variable 32-bit integer overflow constraint Returned 0 if the reversed number goes outside the 32-bit signed integer range > Time Complexity: O(log n) > Space Complexity: O(1) #DSAPrep #Algorithms #Python #ProblemSolving #CodingJourney

Solved leetcode problem: Longest Substring Without Repeating Characters Approach: Use a set to maintain unique characters in the current window. Expand with r. If duplicate appears → shrink from l until valid. Track max window length. Consistent pattern practice > memorizing solutions. #LeetCode #SlidingWindow #Python #DSA #CodingInterview

Solved the Minimum Window Substring problem using the Sliding Window technique. Goal: Find the smallest substring in s that contains all characters of p (including duplicates). Key Idea • Maintain character frequency • Expand window to include required characters • Shrink window to get the minimum valid substring Time Complexity: O(n) #geekstreak60 #npci #algorithms #datastructures #python #slidingwindow #codingpractice #problemSolving #100DaysOfCode

LeetCode POTD 💫: Description: You are given a binary string s, and an integer k. In one operation, you must choose exactly k different indices and flip each '0' to '1' and each '1' to '0'. Return the minimum number of operations required to make all characters in the string equal to '1'. If it is not possible, return -1. Here's my solution: https://lnkd.in/gQquvaeD #Python #DSA #Leetcode #DailyChallenge #Hard

💻Move All Zeroes to End Focused on in-place array manipulation using the two-pointer technique. Optimized to O(n) time and O(1) space while maintaining relative order. #geekstreak60 #npci #Python #DSA #Algorithms #ProblemSolving #CodingJourney #Consistency #TechGrowth #DeveloperLife

🚀 Day 56 — LeetCode 451: Sort Characters by Frequency using HashMap + Sorting. 💡 Key Learnings: • Counting character frequency using dictionary • Custom sorting with key=lambda • Efficient string building using "".join() • Understanding time vs space complexity trade-offs ⏱️ Complexity: Time — O(n + k log k) Space — O(k) #Day56 #LeetCode #Python #DataStructures #DSA #CodingJourney #ProblemSolving

LeetCode Problem 1980: "Find Unique Binary String": Given an array of strings nums containing n unique binary strings each of length n, return a binary string of length n that does not appear in nums. If there are multiple answers, you may return any of them. The below implementation in Python correctly resolves this using bit manipulation functions and loops. #Python #LeetCode #Strings #BitManipulation #CompetitiveProgramming #Algorithms #DataStructures #ProblemSolving

Day 4/60 ✅ Solved Form the Largest Number. The key idea was customizing the sorting logic. Instead of normal numerical sorting, I compared numbers as strings by checking which combination — a+b or b+a — forms a larger value. This ensures the final concatenation produces the maximum possible number. Time Complexity: O(n log n) A good reminder that sometimes the problem isn’t about complex logic — just about choosing the right comparison strategy. #60DayChallenge #DSA #ProblemSolving #Python