Day 76: Sorting Colors with Selection Sort

✅ Day 77 of 100 Days LeetCode Challenge Problem: 🔹 #77 – Combinations 🔗 https://lnkd.in/gNFFZQYw Learning Journey: 🔹 Today’s problem required generating all possible combinations of k numbers from the range [1, n]. 🔹 I used a backtracking approach to build combinations step by step. 🔹 At each step, I added a number to the current path and recursively explored further choices. 🔹 Once the size of the current combination reached k, I added it to the result. 🔹 I also applied a pruning condition to stop unnecessary recursion when there aren’t enough elements left to complete a combination. Concepts Used: 🔹 Backtracking 🔹 Recursion 🔹 Combinatorics 🔹 Pruning Optimization Key Insight: 🔹 Backtracking systematically explores all possible combinations while maintaining constraints. 🔹 Pruning helps reduce unnecessary recursive calls, improving efficiency significantly. Complexity: 🔹 Time: O(C(n, k)) 🔹 Space: O(k) (excluding output) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #SoftwareEngineering #Python #ProblemSolving #LearningInPublic #TechCareers

✅ Day 80 of 100 Days LeetCode Challenge Problem: 🔹 #3876 – Construct Uniform Parity Array II 🔗 https://lnkd.in/gENuCXJc Learning Journey: 🔹 Today’s problem focused on determining whether we can construct an array where all elements are either all even or all odd using allowed operations. 🔹 The key observation was around parity (odd/even nature) of numbers. 🔹 Since we can subtract elements, the parity depends on whether differences can produce consistent odd or even values. 🔹 I simplified the logic by checking: • If the minimum element is odd → we can make all elements odd • Otherwise, verify if all elements are already even Concepts Used: 🔹 Parity (Odd/Even Analysis) 🔹 Mathematical Observation 🔹 Greedy Logic Key Insight: 🔹 Subtracting numbers preserves or changes parity in predictable ways. 🔹 The smallest element plays a crucial role in determining whether all elements can be transformed into the same parity. Complexity: 🔹 Time: O(n) 🔹 Space: O(1) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #SoftwareEngineering #Python #ProblemSolving #LearningInPublic #TechCareers

✅ Day 78 of 100 Days LeetCode Challenge Problem: 🔹 #3110 – Score of a String 🔗 https://lnkd.in/g9jgxcdK Learning Journey: 🔹 Today’s problem involved calculating the score of a string based on the absolute differences between ASCII values of adjacent characters. 🔹 I iterated through the string starting from index 1 and computed the difference between the current and previous character using ord(). 🔹 For each pair, I added the absolute difference to a running total. 🔹 This approach efficiently accumulates the score in a single pass. Concepts Used: 🔹 String Traversal 🔹 ASCII Value Conversion (ord()) 🔹 Absolute Difference Calculation Key Insight: 🔹 The problem reduces to comparing adjacent elements, making it a straightforward linear scan. 🔹 Using ord() allows direct access to ASCII values, simplifying the computation. Complexity: 🔹 Time: O(n) 🔹 Space: O(1) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #SoftwareEngineering #Python #ProblemSolving #LearningInPublic #TechCareers

✅ Day 81 of 100 Days LeetCode Challenge Problem: 🔹 #1897 – Redistribute Characters to Make All Strings Equal 🔗 https://lnkd.in/gZ7wZ7JE Learning Journey: 🔹 Today’s problem required checking if we can redistribute characters so that all strings become identical. 🔹 I combined all strings and used a frequency counter to count occurrences of each character. 🔹 Since we can freely move characters between strings, the only requirement is that each character count must be divisible by the number of strings. 🔹 If any character fails this condition, it’s impossible to evenly distribute it. Concepts Used: 🔹 Hash Map / Frequency Counting (Counter) 🔹 String Concatenation 🔹 Divisibility Check Key Insight: 🔹 The exact arrangement doesn’t matter—only the total frequency distribution matters. 🔹 If every character can be equally divided among all strings, a valid configuration always exists. Complexity: 🔹 Time: O(n * k) (n = number of strings, k = average length) 🔹 Space: O(1) (bounded by alphabet size) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #SoftwareEngineering #Python #ProblemSolving #LearningInPublic #TechCareers

✅ Day 95 of 100 Days LeetCode Challenge Problem: 🔹 #869 – Reordered Power of 2 🔗 https://lnkd.in/gkNXaSFM Learning Journey: 🔹 Today’s problem focused on checking whether the digits of a number can be rearranged to form a power of 2. 🔹 I created a helper function to extract and store the digits of a number. 🔹 Then I sorted the digits of the input number for comparison. 🔹 Next, I generated powers of 2 iteratively and compared their sorted digit lists with the input. 🔹 If any match was found, I returned True. Otherwise, continued until the digit length exceeded the input. Concepts Used: 🔹 Digit Extraction 🔹 Sorting 🔹 Simulation of Powers of 2 🔹 Brute Force Optimization Key Insight: 🔹 Instead of generating permutations (which is expensive), sorting digits allows quick comparison. 🔹 Any valid rearrangement must have the same digit frequency as some power of 2. Complexity: 🔹 Time: O(log n * d log d) 🔹 Space: O(d) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #Python #ProblemSolving #LearningInPublic #TechCareers

✅ Day 97 of 100 Days LeetCode Challenge Problem: 🔹 #1281 – Subtract the Product and Sum of Digits of an Integer 🔗 https://lnkd.in/gxTAZc6U Learning Journey: 🔹 Today’s problem involved extracting digits of a number and performing two operations simultaneously. 🔹 I initialized two variables: one for product (pr) and one for sum (sm). 🔹 Using a while loop, I extracted each digit using n % 10. 🔹 Updated the product by multiplying the digit and updated the sum by adding it. 🔹 Reduced the number using integer division (n //= 10) after each step. 🔹 Finally returned the difference between product and sum. Concepts Used: 🔹 Digit Extraction 🔹 While Loop 🔹 Arithmetic Operations 🔹 Number Manipulation Key Insight: 🔹 Both product and sum can be computed in a single traversal of digits. 🔹 Efficient use of modulus and division avoids converting the number to a string. Complexity: 🔹 Time: O(d) 🔹 Space: O(1) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #Python #ProblemSolving #LearningInPublic #TechCareers

✅ Day 83 of 100 Days LeetCode Challenge Problem: 🔹 #832 – Flipping an Image 🔗 https://lnkd.in/ghfEcHHT Learning Journey: 🔹 Today’s problem involved two operations on a binary matrix: horizontal flip and bit inversion. 🔹 First, I reversed each row to achieve the horizontal flip. 🔹 Then, I inverted every bit (0 → 1 and 1 → 0) using a helper function. 🔹 This approach ensured the transformation was done in-place without extra space. Concepts Used: 🔹 Array Manipulation 🔹 Matrix Traversal 🔹 Two-Pointer / In-place Operations 🔹 Bit Manipulation Key Insight: 🔹 Instead of treating flipping and inversion as separate heavy operations, they can be efficiently combined or done sequentially in-place. 🔹 Using simple transformations keeps the solution clean and optimal. Complexity: 🔹 Time: O(n × m) 🔹 Space: O(1) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #SoftwareEngineering #Python #ProblemSolving #LearningInPublic #TechCareers

✅ Day 88 of 100 Days LeetCode Challenge Problem: 🔹 #1252 – Cells with Odd Values in a Matrix 🔗 https://lnkd.in/g5eE9A-e Learning Journey: 🔹 Today’s problem focused on simulating row and column increment operations on a matrix. 🔹 I initialized an m × n matrix with all zeros. 🔹 For each index [ri, ci], I incremented: • All elements in row ri • All elements in column ci 🔹 After applying all operations, I traversed the matrix to count how many cells contain odd values. Concepts Used: 🔹 Matrix Simulation 🔹 Nested Loops 🔹 Conditional Counting 🔹 Basic Array Manipulation Key Insight: 🔹 Direct simulation works efficiently due to small constraints. 🔹 Each operation affects an entire row and column, so tracking increments carefully is key. Complexity: 🔹 Time: O(m × n + k × (m + n)) 🔹 Space: O(m × n) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #Python #ProblemSolving #LearningInPublic #TechCareers

🚀 LeetCode Win: Missing Number (Beats 100%) Solved this problem using a simple mathematical insight instead of overcomplicating the logic ⚡ 🧠 Core Idea: If a list contains numbers from 0 → n, one number is missing. 👉 So instead of searching for it… I compared: Expected sum (0 → n) Actual sum (given array) ⚙️ Approach: Generate range → 0 to n Compute expected sum Subtract actual array sum ✅ Missing Number = Expected Sum - Actual Sum 🔥 Why this approach stands out: ⏱️ O(n) Time Complexity 💾 O(1) Space Complexity ❌ No extra loops ❌ No sorting ✔️ Clean & efficient 📊 Result: ⚡ Runtime: 0 ms (Beats 100%) 📉 Memory: Optimized 💭 Key Learning: The best solutions aren’t always complex. Sometimes, stepping back and thinking mathematically changes everything. 🚀 Consistency check: Showing up, solving daily, improving 1% 💪 #LeetCode #Algorithms #DataStructures #ProblemSolving #Python #CodingJourney #100DaysOfCode #DeveloperLife #WomenWhoCode #TechGrowth #CodeDaily #ProgrammingLife #SoftwareEngineer #ConsistencyWins

✅ Day 87 of 100 Days LeetCode Challenge Problem: 🔹 #2840 – Check if Strings Can be Made Equal With Operations II 🔗 https://lnkd.in/gY73RBb5 Learning Journey: 🔹 Today’s problem extended the previous one, allowing swaps between indices where the difference is even. 🔹 I observed that this again partitions the string into two independent groups: • Even indices (0, 2, 4, …) • Odd indices (1, 3, 5, …) 🔹 I extracted characters from even and odd positions separately for both strings. 🔹 Then, I sorted these groups and compared them between s1 and s2. 🔹 If both even-index groups and odd-index groups match, the strings can be made equal. Concepts Used: 🔹 String Manipulation 🔹 Index Grouping (Parity-based) 🔹 Sorting 🔹 Greedy Observation Key Insight: 🔹 Since swaps are allowed only between indices with even distance, characters can only move within their parity group. 🔹 Therefore, the problem reduces to checking if both parity groups have identical character distributions. Complexity: 🔹 Time: O(n log n) 🔹 Space: O(n) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #SoftwareEngineering #Python #ProblemSolving #LearningInPublic #TechCareers