LeetCode 1680: Binary Concatenation Modulo 10^9 + 7

LeetCode Problem 1009: "Complement of base 10": The complement of an integer is the integer you get when you flip all the 0's to 1's and all the 1's to 0's in its binary representation. For example, The integer 5 is "101" in binary and its complement is "010" which is the integer 2. Given an integer n, return its complement. Approach: First get the binary string representation of given number using bin() and str() functions, second iterate through the string and append '0' to a new string variable if you get '1' and '1' if you get '0', last return the integer value of the new string variable using the int() function with base 2. #Python #Strings #LeetCode #CompetitiveProgramming #DataStructures #DSA #Algorithms #Programming #ProblemSolving

🚀 #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

Output: [9, 1, 1, 25, 81] 🧠 filter() + map() chained together! Step 1 — filter() keeps only odd numbers: [3, 1, 1, 5, 9] Step 2 — map() squares each one: [9, 1, 1, 25, 81] ✅ A cleaner, Pythonic way using list comprehension: result = [x**2 for x in nums if x % 2 != 0] Same result, more readable! 🎉 #python #pythonprogramming #coding #programming

✅ 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 58 of 100 Days LeetCode Challenge Problem: 🔹 #1680 – Concatenation of Consecutive Binary Numbers 🔗 https://lnkd.in/gt22hnfF Learning Journey: 🔹 Today’s problem involved concatenating the binary representations of numbers from 1 to n. 🔹 I generated binary strings using bin(i)[2:] and appended them sequentially. 🔹 After forming the final binary string, I converted it back to an integer. 🔹 Since the result can be very large, I applied modulo 10**9+7. Concepts Used: 🔹 Binary Representation 🔹 String Manipulation 🔹 Modular Arithmetic 🔹 Simulation Key Insight: 🔹 Direct string concatenation works but is not optimal for large n. 🔹 Efficient solutions avoid large intermediate strings by using bit shifts and modular arithmetic at each step. 🔹 Understanding bit-length growth is crucial for optimization in binary construction problems. #LeetCode #DataStructures #Algorithms #CodingInterview #SoftwareEngineering #SoftwareDeveloper #ProblemSolving #Programming #ComputerScience #TechCareers #100DaysOfCode #DailyCoding #Consistency #LearningInPublic #Python #BackendDevelopment #InterviewPreparation #TechCommunity

🚀 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

🚀 Day 60 of #DSAJourney — Pascal’s Triangle II Today’s problem: Pascal’s Triangle II 📌 Given an integer rowIndex, return the rowIndex-th row of Pascal’s Triangle. 💡 Key Insight: Instead of generating the whole triangle, we can build the row in-place using the relation: 👉 Each element = sum of two elements from previous row 👉 Update from right to left to avoid overwriting values 🧠 Approach: Start with [1] Iteratively build each row Update elements backward to maintain correctness ⚡ Complexity: Time: O(n²) Space: O(n) (optimized, no extra triangle) 📈 What I Learned: In-place updates can optimize space significantly Backward traversal is key in many DP problems Simple math patterns → powerful solutions #Day60 #LeetCode #DSA #CodingJourney #Python #Programming

🔹 Problem Statement Given a string s, reverse only the vowels in the string and return the updated string. Vowels include a, e, i, o, u and they may appear in both lowercase and uppercase. 🔹 Example Input: IceCreAm Output: AceCreIm Explanation: The vowels in the string are ['I', 'e', 'e', 'A']. After reversing their order, the string becomes AceCreIm while all consonants remain in their original positions. 📌 Key Concepts Practiced String traversal List operations Stack behavior using pop() Problem decomposition Consistently practicing coding problems helps strengthen logic building and algorithmic thinking. #Python #Programming #DataStructures #ProblemSolving #CodingPractice

LeetCode Problem 1582: "Special Positions in a Binary Matrix": Given an m x n binary matrix mat, return the number of special positions in mat. A position (i, j) is called special if mat[i][j] == 1 and all other elements in row i and column j are 0 (rows and columns are 0-indexed). Approach: Initialize two arrays, one stores the number of 1s in each row of the matrix and other stores the number of 1s in each column, by traversing the matrix for filling each of the arrays. Now re-traverse the matrix and if mat[i][j]==1, check for the number of 1s in corresponding row and column, if number of 1s in that row and column is equal to one, we get our required position, calculate the total number of such positions. #Python #LeetCode #Matrices #ProblemSolving #CompetitiveProgramming #DataStructures #Algorithms #Arrays #Coding

Day 3 / 100 🚀 Solved “Reverse Integer” — a problem that looks simple but actually tests how carefully you handle edge cases. At first, reversing digits feels straightforward. But the real challenge is handling 32-bit overflow without using extra space. 💡 Key learning: Before updating the result, always check if multiplying by 10 will exceed the allowed range. Core idea: rev * 10 + digit must stay within [-2³¹, 2³¹ - 1] Highlights: • Time Complexity: O(log n) • Space Complexity: O(1) • Correctly handles negative numbers and overflow This problem reinforced a critical habit: Don’t just make the logic work — validate boundary conditions. #100DaysOfCode #LeetCode #DSA #Python #ProblemSolving #CodingInterview