100 Days LeetCode Challenge: Binary Concatenation Problem

✅ Day 63 of 100 Days LeetCode Challenge Problem: 🔹 #1758 – Minimum Changes To Make Alternating Binary String 🔗 https://lnkd.in/gSY9BDhw Learning Journey: 🔹 Today’s problem focused on converting a binary string into an alternating pattern with minimum changes. 🔹 There are only two valid alternating patterns: starting with '0' (0101...) or starting with '1' (1010...). 🔹 I counted mismatches for both patterns while iterating through the string. 🔹 The answer is the minimum number of flips required between the two possibilities. Concepts Used: 🔹 String Traversal 🔹 Pattern Matching 🔹 Greedy Counting 🔹 Parity Indexing Key Insight: 🔹 When only two structural patterns are possible, evaluate both and choose the minimum cost. 🔹 Index parity (i % 2) is a clean way to enforce alternating constraints. 🔹 Comparing against expected patterns avoids unnecessary string reconstruction. #LeetCode #DataStructures #Algorithms #CodingInterview #SoftwareEngineering #SoftwareDeveloper #ProblemSolving #Programming #ComputerScience #TechCareers #100DaysOfCode #DailyCoding #Consistency #LearningInPublic #Python #BackendDevelopment #InterviewPreparation #TechCommunity

✅ Day 61 of 100 Days LeetCode Challenge Problem: 🔹 #260 – Single Number III 🔗 https://lnkd.in/gKWtdDrb Learning Journey: 🔹 Today’s problem focused on finding the two elements that appear only once in an array where all others appear twice. 🔹 I used a frequency counting approach with Counter to track occurrences. 🔹 Then, I iterated through the dictionary and collected elements with frequency equal to 1. 🔹 This straightforward method ensures correctness with clear logic. Concepts Used: 🔹 Hash Map / Dictionary 🔹 Frequency Counting 🔹 Array Traversal 🔹 Filtering Logic Key Insight: 🔹 Hash-based counting simplifies duplicate detection problems. 🔹 When constraints allow extra space, clarity can be prioritized over bit-level optimization. 🔹 This problem also has an advanced XOR-based O(1) space solution, but counting keeps the implementation simple and readable. #LeetCode #DataStructures #Algorithms #CodingInterview #SoftwareEngineering #SoftwareDeveloper #ProblemSolving #Programming #ComputerScience #TechCareers #100DaysOfCode #DailyCoding #Consistency #LearningInPublic #Python #BackendDevelopment #InterviewPreparation #TechCommunity

✅ Day 59 of 100 Days LeetCode Challenge Problem: 🔹 #3857 – Minimum Cost to Split into Ones 🔗 https://lnkd.in/g-7ARYJV Learning Journey: 🔹 Today’s problem focused on minimizing the total cost of splitting an integer n into n ones. 🔹 Each split divides a number x into a and b such that a + b = x, with cost a × b. 🔹 The key observation is that splitting as evenly as possible at every step minimizes the total accumulated cost. 🔹 This creates a divide-and-conquer pattern where balanced partitions reduce overall multiplication cost. Concepts Used: 🔹 Greedy Strategy 🔹 Divide and Conquer 🔹 Mathematical Cost Analysis 🔹 Stack / Simulation Approach Key Insight: 🔹 Balanced splits minimize product cost at each stage. 🔹 The total minimum cost follows a structured pattern based on recursive halving. 🔹 Recognizing optimal partitioning significantly simplifies the solution logic. #LeetCode #WeeklyContest #DataStructures #Algorithms #CodingInterview #SoftwareEngineering #ProblemSolving #100DaysOfCode #CompetitiveProgramming #Python #TechCommunity

LeetCode Problem 1680: "Concatenation of consecutive binary numbers": 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 109 + 7. Approach: Simply initialize a string variable, run loop upto the given value of n, calculate binary representation of each number, convert it to string and concatenate it to the string variable. Convert this result into decimal using int() function and then return value after modulo 10^9 + 7. #Python #LeetCode #CompetitiveProgramming #DailyCoding #Programming #DataStructures #Algorithms #Math #BitManipulation #ProblemSolving

✅ Day 70 of 100 Days LeetCode Challenge Problem: 🔹 #1051 – Height Checker 🔗 https://lnkd.in/g7bmj62W Learning Journey: 🔹 Today’s problem required identifying how many students are not standing in the correct order according to their heights. 🔹 I first created a sorted version of the heights array to represent the expected order. 🔹 Then I compared each element of the original list with the corresponding element in the sorted list. 🔹 Every mismatch indicates a student standing in the wrong position, so I incremented a counter for each difference. Concepts Used: 🔹 Sorting 🔹 Array Comparison 🔹 Iteration Key Insight: 🔹 Instead of rearranging elements, simply comparing the original array with its sorted version reveals how many indices differ. 🔹 Each mismatch directly contributes to the final count. Complexity: 🔹 Time: O(n log n) 🔹 Space: O(n) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #SoftwareEngineering #Python #ProblemSolving #LearningInPublic #TechCareers

✅ Day 68 of 100 Days LeetCode Challenge Problem: 🔹 #1207 – Unique Number of Occurrences 🔗 https://lnkd.in/g2vjwwft Learning Journey: 🔹 Today’s problem required checking whether the frequency of each element in an array is unique. 🔹 I first used Counter to count how many times each number appears in the array. 🔹 Then I extracted the occurrence counts using .values(). 🔹 To verify uniqueness, I compared the length of the occurrences list with the length of a set created from those values. 🔹 If both lengths match, it means all occurrence counts are unique. Concepts Used: 🔹 Hash Map / Frequency Counting 🔹 Python Counter from collections 🔹 Sets for Uniqueness Check Key Insight: 🔹 Converting values to a set removes duplicates automatically. 🔹 If the size of the set equals the number of frequency values, then all occurrence counts are distinct. Complexity: 🔹 Time: O(n) 🔹 Space: O(n) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #SoftwareEngineering #Python #ProblemSolving #LearningInPublic #TechCareers

🚀 #Day41 of #100DaysOfCode 📌 Problem: 80. Remove Duplicates from Sorted Array II Today’s challenge was about handling duplicates in a sorted array while keeping the array in-place. 💡 Key Idea: The array is already sorted, so duplicates appear next to each other. We need to ensure that each number appears at most twice. Extra duplicates beyond two should be removed while maintaining the original order. 🧠 Approach: 1️⃣ Traverse the array from left to right. 2️⃣ Keep track of how many times the current number appears consecutively. 3️⃣ Allow insertion only if the occurrence count is ≤ 2. 4️⃣ Maintain an index pointer to place valid elements in the correct position. ⚡ Complexity: Time Complexity: O(n) Space Complexity: O(1) (in-place modification) #leetcode #Python #programming #coding #softwaredevelopment #100daysofcode

🚀 𝐃𝐚𝐲 20/60 – 60-𝐃𝐚𝐲 𝐏𝐲𝐭𝐡𝐨𝐧 𝐂𝐡𝐚𝐥𝐥𝐞𝐧𝐠𝐞 🦾 Today's topic is "𝐬𝐭𝐚𝐧𝐝𝐚𝐫𝐝 𝐦𝐨𝐝𝐮𝐥𝐞𝐬" Using standard modules in Python 𝐦𝐚𝐭𝐡, 𝐫𝐚𝐧𝐝𝐨𝐦, and 𝐩𝐥𝐚𝐭𝐟𝐨𝐫𝐦  lets you perform common, portable tasks without reinventing the wheel. The math module provides precise mathematical functions like 𝒔𝒒𝒓𝒕 and 𝒔𝒊𝒏, the random module offers simple randomness utilities such as randint and choice, and the platform module helps you inspect the interpreter and OS details to write portable code. 𝐄𝐱𝐚𝐦𝐩𝐥𝐞: 𝘪𝘮𝘱𝘰𝘳𝘵 𝘮𝘢𝘵𝘩, 𝘳𝘢𝘯𝘥𝘰𝘮, 𝘱𝘭𝘢𝘵𝘧𝘰𝘳𝘮 # math 𝘱𝘳𝘪𝘯𝘵("𝘗𝘪:", 𝘮𝘢𝘵𝘩.𝘱𝘪) 𝘱𝘳𝘪𝘯𝘵("𝘚𝘲𝘳𝘵(16):", 𝘮𝘢𝘵𝘩.𝘴𝘲𝘳𝘵(16)) # random 𝘱𝘳𝘪𝘯𝘵("𝘙𝘢𝘯𝘥𝘰𝘮 𝘪𝘯𝘵𝘦𝘨𝘦𝘳 𝘣𝘦𝘵𝘸𝘦𝘦𝘯 1 𝘢𝘯𝘥 10:", 𝘳𝘢𝘯𝘥𝘰𝘮.𝘳𝘢𝘯𝘥𝘪𝘯𝘵(1, 10)) # platform 𝘱𝘳𝘪𝘯𝘵("𝘗𝘺𝘵𝘩𝘰𝘯 𝘷𝘦𝘳𝘴𝘪𝘰𝘯:", 𝘱𝘭𝘢𝘵𝘧𝘰𝘳𝘮.𝘱𝘺𝘵𝘩𝘰𝘯_𝘷𝘦𝘳𝘴𝘪𝘰𝘯()) 𝘱𝘳𝘪𝘯𝘵("𝘖𝘚:", 𝘱𝘭𝘢𝘵𝘧𝘰𝘳𝘮.𝘴𝘺𝘴𝘵𝘦𝘮()) Understanding these operators made me realize how programs make decisions and perform actions based on logic. They may look like simple symbols, but they are essential for writing meaningful code. Step by step, building stronger logic. #learning #python #consistency #challenge #60days #coding #programming #modules

✅ Day 72 of 100 Days LeetCode Challenge Problem: 🔹 #3868 – Minimum Cost to Equalize Arrays Using Swaps 🔗 https://lnkd.in/gwbcmecy Learning Journey: 🔹 Today’s problem involved making two arrays identical with the minimum number of cross-array swaps. 🔹 Swapping within the same array is free, but swapping elements between arrays costs 1 operation. 🔹 I used Counter to count the frequency of elements in both arrays. 🔹 Then I combined the counters to check the total occurrences of each element. 🔹 If any element has an odd total frequency, it’s impossible to distribute it equally between both arrays. 🔹 Otherwise, I calculated the difference in counts between the two arrays to determine how many elements must be swapped. Concepts Used: 🔹 Frequency Counting (Counter) 🔹 Hash Maps 🔹 Greedy Counting Logic 🔹 Swap Balancing Key Insight: 🔹 For the arrays to become identical, every element must appear an even number of times across both arrays. 🔹 The imbalance of each element indicates how many swaps are required, and dividing appropriately accounts for pairwise swaps. Complexity: 🔹 Time: O(n) 🔹 Space: O(n) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #SoftwareEngineering #Python #ProblemSolving #LearningInPublic #TechCareers

🚀 Day 45 of #100DaysOfCode 📌 Problem: Rotate Array (LeetCode 189) 🔹 Problem Statement: Given an integer array nums, rotate the array to the right by k steps, where k is non-negative. 🔹 Approach: Instead of rotating the array one step at a time, we use the Reverse Technique. Steps: 1️⃣ First calculate k % n to handle large rotations. 2️⃣ Reverse the first part of the array (0 → n-k-1). 3️⃣ Reverse the second part (n-k → n-1). 4️⃣ Finally reverse the entire array. This gives the rotated array efficiently. 🔹 Time Complexity: ⏱️ O(n) 🔹 Space Complexity: 📦 O(1) (In-place rotation) 💡 Key Learning: Using array reversal is a very efficient way to rotate arrays without extra space. #DSA #LeetCode #Python #CodingChallenge #Programming